{"1": {"fulltext": "Q/. 531\\n.W4\\nCopy 2", "height": "3782", "width": "2461", "jp2-path": "completetrigonom00well_0001.jp2"}, "2": {"fulltext": "Class Q/4 531\\nBookiWi^\\nCDPffilGHT DEPOSm", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0002.jp2"}, "3": {"fulltext": "", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0003.jp2"}, "4": {"fulltext": "", "height": "3676", "width": "2254", "jp2-path": "completetrigonom00well_0004.jp2"}, "5": {"fulltext": "COMPLETE TRIGONOMETRY\\n^-2 65\\ni\\nBY\\nWEBSTER WELLS, S.B.\\nPROFESSOR OF M A.THEMATICS IN THE MASSACHUSETTS\\nINSTITUTE OF TECHNOLOGY\\n3 0:^C\\nBOSTON, U.S.A.\\nD. C. HEATH CO., PUBLISHERS\\n1900", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0005.jp2"}, "6": {"fulltext": "40355\\nIJt r\u00c2\u00ab(ry of Congress\\nTwo CofiES Received\\nAUG 30 1900\\nAUG 30 19U0\\nFIRST copy.\\n2114 Co^l Odiverodi t*\\nORDER DlVtSlOH\\nSFP n 1900\\nQ A r 3 1\\nCopyright, 1900,\\nBy WEBSTER WELLS.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0006.jp2"}, "7": {"fulltext": "r- \\\\v PREFACE.\\nThe present volume is a revision of the author s Essen-\\ntials of Trigonometry.\\nIn preparing the new edition, many improvements have\\nbeen effected; the attention of teachers is specially invited\\nto the following features of the work\\n1. The proofs of the functions of 120\u00c2\u00b0, 135\u00c2\u00b0, 150\u00c2\u00b0, etc.;\\n27.\\n2. The proofs of the functions of (-A), and (90\u00c2\u00b0\\nin terms of those of \u00c2\u00a7\u00c2\u00a728, 29.\\n3. The method of solution employed in the examples of\\n33 and 34.\\n4. The general demonstration of the formulae\\ntan X and sin^a; cos^o? 1,\\nCOSOJ\\nin 36 and 38; the four cases being considered together.\\n5. The general demonstrations of the formulae\\ncot a; sec^o^ 1 tannic, and csc^x 1 cot^a;\\nsince\\nin 37 and 39.\\n6. The proofs of the formulae for sin {x-{-y) and cos (x-{-i/),\\nwhen X and y are acute the two cases when x-\\\\- y is acute\\nor obtuse being considered together 41.\\n7. The proofs of the formulae for tan i x and cot i aj 48.\\n8. The illustrative examples in 49.", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0007.jp2"}, "8": {"fulltext": "\u00e2\u0080\u00a2IV\\nPreface.\\n9. The solution of right triangles by Natural Functions\\nsee Ex. 1, page 66.\\nThe new work contains a great many more examples than\\nthe old they have been selected with great care, and most\\nof them are new. It is not expected that every class will\\nsolve all the examples they are sufficiently numerous to\\nfurnish a variety in successive years.\\nAttention is specially invited to the sets in 96, 114, 157,\\nand 160.\\nIn 112 will be found a set of miscellaneous examples in\\nthe solution of plane oblique triangles, and in 155 a set in\\nspherical oblique.\\nThe results have been worked out by aid of the author s\\nNew Four Place Logarithmic Tables, which contain also\\nTables of Natural Functions.\\nWEBSTER WELLS.\\nMassachusetts Institute of\\nTechnology, 1900.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0008.jp2"}, "9": {"fulltext": "CONTENTS.\\nPLANE TRIGONOMETRY.\\nPAGE\\nI. Trigonometric Functions of Acute Angles 1\\n11. Trigonometric Functions of Angles in General 7\\nIII. General Formula 26\\nIV. Miscellaneous Theorems 38\\nV. Logarithms 50\\nProperties of Logarithms 52\\nApplications 58\\nVI. Solution of Right Triangles .65\\nFormulae for the Area of a Right Triangle 72\\nVII. General Properties of Triangles 75\\nFormulae for the Area of an Oblique Triangle 80\\nVIII. Solution of Oblique Triangles 82\\nSPHERICAL TRIGONOMETRY.\\nIX. Geometrical Principles 97\\nX. Right Spherical Triangles 101\\nSolution of Right Spherical Triangles 108", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0009.jp2"}, "10": {"fulltext": "vi Contents.\\nPAGE\\nXL Oblique Spherical Triangles 118\\nGeneral Properties of Spherical Triangles 118\\nNapier s Analogies 124\\nSolution of Oblique Spherical Triangles 127\\nXII. Applications 139\\nFormula\\nPlane Trigonometry 144\\nSpherical Trigonometry 147\\nAnswers\\nUse of the Tables", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0010.jp2"}, "11": {"fulltext": "PLAICE TRIGONOMETRY.\\n35*JC\\nI. TRIGONOMETRIC FUNCTIONS OP ACUTE\\nANGLES.\\n1. Trigonometry treats of the properties and measure-\\nment of angles and triangles.\\nIn Plane Trigonometry, we consider plane figures only.\\n2. Definitions of the Trigonometric Functions of Acute\\nAngles.\\nB\\nLet BAC be any acute angle.\\nFrom any point in either side, as B, draw line BC per-\\npendicular to AC, forming right triangle ABC.\\nWe then have the following definitions, applicable to\\neither of the acute angles A ov B:\\nIn any right triangle,\\nThe sine of either acute angle is the ratio of the opposite\\nside to the hypotenuse.\\nThe cosine is the ratio of the adjacent side to the hypotenuse.\\nThe tangent is the ratio of the opposite side to the adjacent\\nThe cotangent is the ratio of the adjacent side to the oppo-\\nsite side.", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0011.jp2"}, "12": {"fulltext": "Plane Trigonometry.\\nThe secant is the ratio of the hypotenuse to the adjacent\\nThe cosecant is the ratio of the hypotenuse to the opposite\\nWe also have the following definitions\\nThe versed sine of an angle is 1 minus the cosine of the\\nangle.\\nThe coversed sine is 1 minus the sine.\\nThe eight ratios defined above are called the Trigono-\\nmetric Functions of the angle.\\nRepresenting sides EC, CA, and AB, by a, b, and c,\\nrespectively, and employing the usual abbreviations, we\\nhave:\\nA\\nsm A\\nc\\ntan^\\nb\\nvers A l\\nb\\nc\\nA\\ni[ COS A\\nc\\nC0t-4\\na\\nA C\\nCSC\\na\\ncovers A l\\na\\nc\\nsinB l.\\ntan.B\\na\\nsec -B\\na\\nvers B l\\na\\nc\\nG0SB\\nc\\ncot 5\\nb\\ncovers B l\\nb\\nc\\n3. It is important to observe that the values of the trigo-\\nnometric functions depend solely on the magnitude of the\\nangle, and are entirely independent of the lengths of the\\nsides of the right triangle which contains it.\\nFor let B and B be any two points in side AD of angle\\nDAE, and draw lines BC and B C perpendicular to AE.", "height": "3681", "width": "2444", "jp2-path": "completetrigonom00well_0012.jp2"}, "13": {"fulltext": "Trigonometric Functions.\\nThen, by the definitions of 2,\\nsinA and sm^^\\nBut right triangles ABC and AB C are similar, since\\nthey have angle A common.\\nWhence, by Geometry,\\nBC^B C\\nAB AB\\nThus the two values found for sin A are equal.\\nThe same may be proved true of each of the remaining\\nfunctions.\\n4. We have from 2,\\nsin A cos B. sec A esc B.\\ntan A cot B. vers A covers B.\\nBut B is the complement of\\nHence, the sine, tangent, secant, and versed sine of any\\nacute angle are, respectively, the cosine, cotangent, cosecant,\\nand coversed sine of the complement of the angle.\\n5. From 2, sin cos B, and cos J. sin B.\\nC G\\nWhence,\\na c sin A c cos B, and h c sin B c cos A.\\nThat is, in any right triangle, either side about the right\\n{ingle is equal to the sine of the opposite angle, or the cosine\\nof the adjacent angle, multiplied by the hypotenuse,\\n6. To find the Values of the Other Seven Functions of an\\nAcute Angle, when the Value of Any One is Given,\\n1. Given csc^ 3; find the values of the remaining\\nfunctions of A.", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0013.jp2"}, "14": {"fulltext": "Plane Trigonometry.\\nB\\nWe may write the equation esc A=\\nSince the cosecant is the hypotenuse divided by the opposite side,\\nwe may regard A as one of the acute angles of right triangle ABC-,\\nin which hypotenuse AB 3, and opposite side BC =1.\\nBy Geometry, AC AB^ -BC \\\\/9 1 V8 2 V2.\\nThen by the definitions of 2,\\n1\\nsin J.\\ncos^\\n3\\n2\\\\/2\\ntan^\\n1\\n2\\\\/2\\ncot r= 2 \\\\/2.\\n1 2\\nsec A :2\\n2V2\\nvers A= 1\\n2V2\\n3\\n2. Given vers A\\ncovers A\\n2\\n1\\n3 3\\nfind the value of cot A.\\nB\\nSince vers 1 cos we have cos A\\n5 5\\nThen, in right triangle ABC, we take adjacent side AC S, and\\nhypotenuse AB 5.\\nWhence, BC VaB^ ^C^ V25 9 VI6 4.\\n3\\nThen, by definition, cot A", "height": "3704", "width": "2370", "jp2-path": "completetrigonom00well_0014.jp2"}, "15": {"fulltext": "Trigonometric Functions.\\nEXAMPLES.\\nIn each of the following, find the values of the remaining\\nfunctions\\n3.\\nsin^\\n6.\\nCSC A 7.\\n9.\\nsec A=: X.\\n4.\\nvers A.\\n7.\\ncosA- f.\\n14\\n10.\\ntan A\\nX\\n5.\\ncot^\\n8.\\n2\\ncovers A\\n11.\\nsm A\\n12. Given cot j find sin A.\\n13. Given esc A\\n41\\n40\\nfind cos A.\\n14. Given sec J. 5 find cot A.\\n21\\n15. Given cos A find esc A.\\n29\\n16. Given tan A find sec A.\\n17. Given sinJ. find vers A\\n7. Functions of 45\u00c2\u00b0.\\nLet ABC be an isosceles right triangle, C being the right\\nangle, and sides AC and BC being each equal to 1.\\nThen, ZA=4.5\u00c2\u00b0, and AB=-^AC -i-BC =-\\\\/l+l V2.", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0015.jp2"}, "16": {"fulltext": "6 Plane Trigonometry.\\nWhence, by definition,\\nsec45\u00c2\u00b0=V2.\\nsin45\u00c2\u00b0 A: +V2.\\nV2\\ncos45\u00c2\u00b0- iV2.\\nV2\\ncsc45\u00c2\u00b0=V2.\\nvers 45\u00c2\u00b0 1-1 V2\\ncovers45\u00c2\u00b0 l-iV2\\n2-V2^\\n2\\n2-V2\\nLet ^57) be an equilateral triangle having each side equal\\nto 2, and draw line AC perpendicular to BD.\\nBy Geometry, BC=\\\\BD 1, Z BAC =i Z BAD 30\u00c2\u00b0.\\nAlso, AC ^AB -BC V4.-1=^^.\\nThen from right triangle ABC, by definition,\\nsin 30\u00c2\u00b0 cos 60\u00c2\u00b0.\\ncos 30\u00c2\u00b0=^= sin 60\u00c2\u00b0.\\ntan 30^ 1 V3 cot 60\u00c2\u00b0. cot 30\u00c2\u00b0 V3 tan 60=\\nV3\\nsec 30^^\\nV3\\n2 V3 CSC 60\u00c2\u00b0. CSC 30\u00c2\u00b0 2 sec 60\u00c2\u00b0.\\nV3\\nvers 30\u00c2\u00b0 1 covers 60\u00c2\u00b0.\\n2\\ncovers 30\u00c2\u00b0 1 vers 60\u00c2\u00b0.\\n2 2", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0016.jp2"}, "17": {"fulltext": "Trigonometric Functions.\\nII. TRIGONOMETRIC FUNCTIONS OF\\nANGLES IN GENERAL.\\n9. In Geometry, we are, as a rule, concerned with angles\\nless than two right angles but in Trigonometry it is con-\\nvenient to consider them as unrestricted in magnitude.\\nLet be the centre, and XX and YY a pair of perpen-\\ndicular diameters, of circle AY^ OY being above, and Y\\nbelow, XX when OX is horizontal and extends to the\\nright, and OX to the left, of 0.\\nLet radius OA (Fig. 1) start from the position OX, and\\nrevolve about point as a pivot towards the position OY.\\nWhen OA coincides with OY, it has generated an angle\\nof 90\u00c2\u00b0; when it coincides with OX of 180\u00c2\u00b0; with OY\\nof 270\u00c2\u00b0; with OX, its first position, of 360\u00c2\u00b0; with OF again,\\nof 450\u00c2\u00b0 and so on.\\nHence, a meaning may be attached to a positive angle of\\nany number of degrees.\\n10. We may also conceive of a negative angle of any\\nnumber of degrees.\\nThus, if a positive angle indicates revolution from the\\nposition OX towards OY a negative angle may be taken as\\nindicating revolution from the position OX in the opposite\\ndirection, towards OY", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0017.jp2"}, "18": {"fulltext": "8 Plane Trigonometry.\\nThus, if radius OA (Fig. 2) starts from the position OX,\\nand revolves about point as a pivot towards the position\\nY when it coincides with T\\\\ it has generated an angle\\nof 90\u00c2\u00b0; when it coincides with 0X\\\\ of 180\u00c2\u00b0; with OT,\\nof -270\u00c2\u00b0; and so on.\\nNote. It is immaterial which direction we consider the positive\\ndirection of rotation but having at the outset adopted a certain direc-\\ntion as positive, our subsequent operations must be in accordance.\\n11. In generating a positive or negative angle of any\\nnumber of degrees, the line from which the rotation is sup-\\nposed to commence is called the initial line of the angle, and\\nthe final position of the rotating radius the tei^minal line.\\n12. To designate an angle, we always write first the letter\\nat the extremity of the initial line.\\nThus, in designating the angle formed by the lines OX\\nand OA, if we regard OX as the initial line, we should call\\nit XOA and if we regard OA as the initial line, we should\\ncall it AOX.\\n13. There are always two angles less than 360\u00c2\u00b0 in abso-\\nlute value, one positive and the other negative, having the\\nsame initial and terminal line.\\nThus there are formed by OX and OA (Fig. 2) the posi-\\ntive angle XOA between 270\u00c2\u00b0 and 360\u00c2\u00b0, and the negative\\nangle XOA between 0\u00c2\u00b0 and 90\u00c2\u00b0.\\nWe shall distinguish between these angles by referring to\\nthem as Hhe positive angle XOA,^^ and the negative\\nangle XOJ. respectively.\\n14. Rectangular Co-ordinates.\\nLet XX^ and YY^ be two straight lines intersecting at\\nright angles at O the letters being arranged as in the fig-\\nures of 9.\\nLet Pi be any point in the plane of XX and YY\\\\ and\\ndraw line P^M perpendicular to XX", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0018.jp2"}, "19": {"fulltext": "Trigonometric Functions. 9\\nThen OM and P^M are called the rectangular co-ordinates\\nof Pi OM is called the abscissa, and PiM the ordinate.\\na\\n?i (b,.cO\\nP3\\nP4 (Py-Ci)\\nThe lines of reference, XX and FF are called the axis\\nof X and the axis of Y, respectively, and is called the\\norigin.\\nIt is customary to express the fact that the abscissa of a\\npoint is h, and its ordinate a, by saying that for the point\\nin question x h and y a; or, more concisely, we may\\nrefer to the point as the point (b, a), where the first term\\nin the parenthesis is understood to be the abscissa, and the\\nsecond term the ordinate.\\n15. If, in the figure of 14, M and JSf be points on OX\\nand OX respectively, such that 0M= ON=b, and lines\\nP1P4 and P2P3 be drawn through M and N, respectively,\\nperpendicular to XX making P^M P^JSF P.^JSF P^M a,\\neach of the points Pi, P^, P3, and P4 will have its abscissa\\nequal to b, and its ordinate equal to a.\\nTo avoid this ambiguity, abscissas measured to the right\\nof are considered positive, and to the left, negative; and\\nordinates measured above XX are considered positive, and\\nbelow, negative.\\nThen the co-ordinates of the points will be as follows\\nPi, (b, a) P\u00e2\u0080\u009e b, a); P\u00e2\u0080\u009e b, a); P^ (b, a).", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0019.jp2"}, "20": {"fulltext": "lo Plane Trigonometry.\\nNote 1. It is understood, in the above convention with regard to\\nsigns, that the figure is so placed that OX is horizontal, and extends\\nto the right from 0.\\nNote 2. In all the figures of the present chapter, the small letters\\nare understood as denoting the lengths of the lines to luhich they are\\nattached, without regard to their algebraic signs; hence they always\\nrepresent positive quantities.\\n16. If a point lies upon XX its ordinate is zero; and\\nif it lies upon YY its abscissa is zero.\\n17. General Definitions of the Functions.\\nWe will now give general definitions of the trigonometric\\nfunctions, applicable to any angle whatever.\\nConstruct axes in such a way that the initial line of the\\nangle shall be the positive direction of the axis of X, and\\nthe vertex the origin.\\nFrom any point in the terminal line drop a perpendicular\\nto the axis of X, and find the co-ordinates of this point.\\nThen, the sine of the arigle is the ratio of the ordinate of the\\npoint to its distance from the origin.\\nThe cosine is the ratio of the abscissa to the distance.\\nThe tangent is the ratio of the ordinate to the abscissa.\\nThe cotangent is the ratio of the abscissa to the ordinate.\\nThe secant is the ratio of the distance to the abscissa.\\nThe cosecant is the ratio of the distance to the ordinate.\\nNote. The above definitions include those of 2.\\nThe definitions of the versed sine and coversed sine, given in 2,\\nare sufficiently general to apply to any angle whatever.\\n18. We will now apply the definitions of 17 in the fol-\\nlowing figures.\\nIn each case, we construct axes in such a way that the\\ninitial line of the angle shall be the positive direction of the\\naxis of X, and the vertex the origin.", "height": "3636", "width": "2307", "jp2-path": "completetrigonom00well_0020.jp2"}, "21": {"fulltext": "Trigonometric Functions.\\nI. Let XOP2 be any angle between 90\u00c2\u00b0 and 180\u00c2\u00b0.\\nY\\nI I\\nLet P2 be any point on the terminal line, and draw P2M\\nperpendicular to XX let P2M a, OM b, and OP2 c.\\nThen the co-ordinates of P2 are b, a).\\nWhence, by definition.\\nsin XOPo\\ntan XOP,\\nsec XOPo\\n-b\\ncos XOP,\\ncot XOP.\\nCSC XOP.2\\nc\\n-b\\nh^_b_\\nc\\n11. Let XOP3 be any angle between 180\u00c2\u00b0 and 270\u00c2\u00b0.\\nP3(-Z)-a)\\nLet Pg be any point on the terminal line, and draw P^M\\nperpendicular to XX letP3lf=a, OM=b, and OP^ e.\\nThen the co-ordinates of P3 are (\u00e2\u0080\u0094b, a).", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0021.jp2"}, "22": {"fulltext": "12 Plane Trigonometry\\nWhence, by definition,\\ntan XOP^\\na _a\\nsec XOPo\\ncos XOP3\\ncot XOP3\\nCSC XOPo z=\\n-b^ b\\nc c\\n-b^b\\na a\\n_c__ _ c\\na a\\nIII. Let XOP4 be any angle between 270\u00c2\u00b0 and 360\u00c2\u00b0.\\nY\\nPA brO\\nLet P4 be any point in the terminal line, and draw P^M\\nperpendicular to XX let P^M a, OM b, and OP4 c.\\nThen the co-ordinates of P4 are (6, a).\\nWhence, by definition,\\nsinX0P4\\nc\\ntanX0P4-^^^\\nb\\nsec XOP4\\ncos XOP4\\ncotXOP4\\na\\na a\\n19. It is evident that the terminal lines of any two\\nangles which differ by a multiple of 360\u00c2\u00b0 are coincident,\\nand hence the trigonometric functions of two such angles\\nare identical.", "height": "3681", "width": "2470", "jp2-path": "completetrigonom00well_0022.jp2"}, "23": {"fulltext": "Trigonometric Functions.\\n13\\nThus, the functions of 50\u00c2\u00b0, 410\u00c2\u00b0, 770\u00c2\u00b0, 310\u00c2\u00b0, etc., are\\nidentical.\\n20. If the initial line of an angle coincides with OX, and\\nits terminal line lies between OX and Y, the angle is said\\nto be in the Jlrst quadrant; if the terminal line lies between\\nT and OX the angle is said to be in the second quadrant\\nif between OX and OY in the third quadrant; if between\\nOF and OX, in the fourth quadrant.\\nThus, any positive angle between 0\u00c2\u00b0 and 90\u00c2\u00b0, or 360\u00c2\u00b0 and\\n450\u00c2\u00b0, or any negative angle between 270\u00c2\u00b0 and 360\u00c2\u00b0, is\\nin the first quadrant any positive angle between 90\u00c2\u00b0 and\\n180\u00c2\u00b0, or 450\u00c2\u00b0 and 540\u00c2\u00b0, or any negative angle between 180\u00c2\u00b0\\nand 270\u00c2\u00b0, is in the second quadrant.\\n21. It follows from the definitions of 17 that, for any\\nangle in the first quadrant, all the functions are positive.\\nIt is also evident by inspection of the results of 18 that\\nIn the second quadrant, the sine and cosecant are positive,\\nand the cosine, tangent, cotangent, and secant are negative.\\nIn the third quadrant, the tangent and cotangent are positive,\\nand the sine, cosine, secant, and cosecant are negative.\\nIn the fourth quadrant, the cosine and secant are positive,\\nand the sine, tangent, cotangent, and cosecant are negative.\\nIt is customary to express the above in tabular form, as\\nfollows\\nFunctions.\\nFirst\\nQuad.\\nSecond\\nQuad.\\nThird\\nQuad.\\nFourth\\nQuad.\\nSine and cosecant\\nCosine and secant\\nTangent and cotangent\\n-f-", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0023.jp2"}, "24": {"fulltext": "14 Plane Trigonometry.\\n22. Functions of 0\u00c2\u00b0 and 360\u00c2\u00b0.\\na P(a.O\\\\\\nThe terminal line of 0\u00c2\u00b0 coincides with the initial line OX.\\nLet P be a point on OX such that OP a.\\nThen by 16, the co-ordinates of P are (a, 0).\\nWhence by definition,\\nsinO\\n0.\\ntan 0\\n0.\\nsec 0=\\n1.\\ncos 0\u00c2\u00b0 1.\\ncotO\u00c2\u00b0 ^-oo.\\ncscO\u00c2\u00b0=:- oo.\\nBy 19, the functions of 360\u00c2\u00b0 are the same as those of 0\\n23. Functions of 90\u00c2\u00b0.\\nT\\nP(0,a)\\na\\nf^90o\\nFor the angle 90\u00c2\u00b0, OF is the terminal line.\\nLet P be a point on OY, such that OP a.\\nThen the co-ordinates of P are (0, a).\\nWhence by definition.\\nsin 90\u00c2\u00b0 1.\\na\\ncos 90^\\n0.\\ntan 90\u00c2\u00b0 00.\\ncot 90\u00c2\u00b0 0.\\na\\nsec 90\u00c2\u00b0 00.\\nCSC 90\u00c2\u00b0 1.\\na", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0024.jp2"}, "25": {"fulltext": "Trigonometric Functions.\\n24. Functions of 180\u00c2\u00b0.\\n15\\n180=\\nP(-a,0)\\nFor the angle 180\u00c2\u00b0, OX is the terminal line.\\nLet P be a point on OX such that OP a.\\nThen the co-ordinates of P are (\u00e2\u0080\u0094a, 0).\\nWhence by definition,\\nsin 180\u00c2\u00b0 =0.\\na\\ncos 180\u00c2\u00b0 1.\\na\\ntan 180\u00c2\u00b0 0.\\na\\ncot 180\u00c2\u00b0 00.\\nsec 180\u00c2\u00b0 1.\\na\\nCSC 180\u00c2\u00b0 =00.\\n25. Functions of 270\u00c2\u00b0.\\n270\u00c2\u00b0\\nX-\\nT (0,-a)\\nFor the angle 270\u00c2\u00b0, OT is the terminal line.\\nLet P be a point on OT such that OP a.\\nThen the co-ordinates of P are (0, a).", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0025.jp2"}, "26": {"fulltext": "i6\\nPlane Trigonometry.\\nWhence by definition,\\nsin 270\u00c2\u00b0\\na\\ntan270\u00c2\u00b0=-=p=:oo.\\ncos 270\u00c2\u00b0 =0.\\na\\ncot 270\u00c2\u00b0\\na\\n0.\\nsec 270^\\n00.\\nesc 270\u00c2\u00b0=-^\\na\\nNote. No absolute meaning can be attached to such a result as\\ncot 0\u00c2\u00b0 00 it merely signifies that as an angle approaches 0\u00c2\u00b0, its co-\\ntangent increases without limit.\\nA similar interpretation must be given to the equations esc 0\u00c2\u00b0 cxj,\\ntan 90\u00c2\u00b0 GO, etc.\\n26. Given the value of one function of an angles to find the\\nvalues of the remaining functions. (Compare 6.)\\n3\\n1. Given sin A required the values of the remain-\\ning functions of A.\\nThe example may be solved by a method similar to that of 6\\nsince the sine is the ratio of the ordinate to the distance, v^e may\\nregard the point of reference as having its ordinate equal to 3, and\\nits distance equal to 5.\\nThere are tvm points, P and P which are 3 units below the axis\\nof X, and distant 5 units from O.\\nP(-4,-3)\\nP (4,-3)\\nThere are then two angles^ XOP and XOP in the third and fourth\\nquadrants, respectively, either of which may be the angle A.\\nNow, OM OM ^OP^ PM^ V25 -9 4.\\nThen co-ordinates of Pare (\u00e2\u0080\u00944, 3), and of P (4, 3).", "height": "3704", "width": "2370", "jp2-path": "completetrigonom00well_0026.jp2"}, "27": {"fulltext": "Trigonometric Functions.\\nWhence by definition\\n17\\nAngle.\\nCos.\\nTan\\nCo..\\nSec.\\nCsc.\\nXOP\\n4\\n5\\n3\\n4\\n4\\n3\\n5\\n4\\n5\\n3\\nXOP\\n4\\n5\\n_3\\n4\\n4\\n3\\n5\\n4\\n5\\n3\\nThus the two solutions to the problem are\\ncosA=T tan A cot A sec =F csc\\n5 4 3 4 3\\nwhere the upper signs refer to XOP, and the lower signs to XOP\\n2. Given cot A 3-, required the values of tlie remain-\\ning functions of A.\\nThe equation may be written in the forms\\n3 Q\\ncot A or cot A\\nWe may then regard the point of reference as having its abscissa\\nequal to 3 and its ordinate equal to 1, or as having its abscissa equal\\nto 3 and its ordinate equal to 1.\\nThere are two angles, XOP and XOP in the first and third quad-\\nrants, respectively, either of which satisfies the given condition.\\nP^(-3,-l)\\nThen, OP OP 03f P3f \\\\/9 1 VlO.", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0027.jp2"}, "28": {"fulltext": "1 8 Plane Trigonometry.\\nWhence by definition\\nAngle.\\nSin.\\nCos.\\nTan.\\nSec.\\nCsc.\\nXOP\\n1\\n3\\n1\\nVlO\\nVlO\\nVlO\\nVlO\\n3\\n3\\nXOP\\n1\\n3\\n1\\nVTo\\n-VlO\\nVlO\\nVlO\\n3\\n3\\nThus the two solutions are\\nsin cos tan A\\nVlO VlO 3\\nsec A\\n10\\n3\\nCSC VlO.\\nNote. It must be clearly borne in mind, in examples like the\\nabove, that the distance is always positive.\\nEXAMPLES.\\nIn each of the following, find the values of the remaining\\nfunctions\\n3. secJ.\\n4\\n7. csc\\n25\\n4. cot^--\u00e2\u0080\u0094 8. tSinA\\n5 40\\n5. sin A\\n15\\n9. sec^\\n21\\n6. GOsA 10. sin^\\n29 5\\n27. Functions of 120\u00c2\u00b0, 135\u00c2\u00b0, 150\u00c2\u00b0, etc.\\n11. tanJ^=-7.\\n12. csc 3.\\n13. cos^\\nb\\n14. cot A x.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0028.jp2"}, "29": {"fulltext": "Trigonometric Functions.\\n19\\nLet 0PM be a right triangle having OP, OM, and PM\\nequal to 2, 1, and VS, respectively, and Z POM =60\u00c2\u00b0.\\n(Compare 8.)\\nThen Z XOP 120\u00c2\u00b0, and co-ordinates of P are (-1, V3).\\nWhence by definition,\\nsin 120\u00c2\u00b0\\ncos 120\u00c2\u00b0\\ntanl20\u00c2\u00b0 -V3.\\ncot 120\u00c2\u00b0= ^V3.\\nV3 3\\nsec 120\u00c2\u00b0 -2.\\nCSC 120\u00c2\u00b0= =^V3.\\nV3 3\\nIn like manner may be proved the remaining values given\\nin the following table, which are left as exercises for the\\nstudent\\nAngle.\\nSin.\\nCos.\\nTan.\\nCoiJ.\\nSec.\\n(7w.\\n120\u00c2\u00b0\\niV3\\n-4\\n-V3\\n-iV3\\n-2\\nfV3\\n135\u00c2\u00b0\\niV2\\n-iV2\\n-1\\n1\\n-V2\\nV2\\n150\u00c2\u00b0\\n1\\n-iV3\\n-iV3\\n-V3\\n-fV3\\n2\\n210\u00c2\u00b0\\n-i\\n-iV3\\niV3\\nV3\\n-fV3\\n-2\\n225\u00c2\u00b0\\n-1V2\\n-4V2\\n1\\n1\\n-V2\\n-V2\\n240\u00c2\u00b0\\n-iV3\\n-4\\nV3\\n|V3\\n-2\\n-fV3\\n300\u00c2\u00b0\\n-iV3\\ni\\n-V3\\n-iV3\\n2\\n-|V3\\n315\u00c2\u00b0\\n-iV2\\n1V2\\n-1\\n-1\\nV2\\n-V2\\n330\u00c2\u00b0\\n-4\\niV3\\n-1V3\\n-V3\\n|V3\\n-2\\n28. Functions of (\u00e2\u0080\u0094-4) in terms of those of A.\\nTo prove the formulm\\nsin J.) sin A,\\ntan A^= tan A,\\nsec(\u00e2\u0080\u0094 sec^,\\nfor any value of A.\\ncos(\u00e2\u0080\u0094 cos^,\\ncot(\u00e2\u0080\u0094 cot^,\\nCSC J.) CSC A,\\n(1)", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0029.jp2"}, "30": {"fulltext": "20\\nPlane Trigonometry.\\nThere may be four cases A in the first quadrant (Fig. 1),\\nA in the second quadrant (Fig. 2). A in the third quadrant\\n(Fig. 3), or A in the fourth quadrant (Fig. 4).\\nFig. 1.\\nTig. 3.\\nIn each figure, let the positive angle XOP represent the\\nangle A, and the negative angle XOP the angle A.\\nDraw PM perpendicular to XX and produce it to meet\\nOP at P\\nIn right triangles 0PM and OP M, side OM is common,\\n2.rLd.Z.P0M=AP 0M.\\nThen the triangles are equal, and PM=P M and\\n0P 0P\\nTherefore, in each figure,\\nabscissa P abscissa P,\\nordinate P ordinate P,\\nand\\ndistance P distance P.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0030.jp2"}, "31": {"fulltext": "Then, sin(\u00e2\u0080\u0094\\ncos(\u00e2\u0080\u0094\\ntan(\u00e2\u0080\u0094\\ncot(- J)\\nTrigonometric Functions,\\nord. P ord. P\\n21\\ndist.P\\nabs. F\\ndist.P\\nord. P\\nabs. P\\nabs.P\\nord. P\\ndist. P\\nabs.P\\ndist.P\\nord.P\\nabs.P\\nabs. P\\nord.P\\nsin A,\\ncos A.\\ntan A.\\ncot A\\n.X dist.P dist.P\\nsec(\u00e2\u0080\u0094 sec^.\\ncsc(\u00e2\u0080\u0094\\nabs. P\\ndist. P\\nord. P\\nabs.P\\ndist.P\\nord. P\\nesc A\\n29. Functions of (90\u00c2\u00b0 -f in terms of those of A,\\nTo prove the formulce\\nsin (90\u00c2\u00b0 cos^\\ntan(90\u00c2\u00b0 ^)=-cot^,\\nsec(90\u00c2\u00b0 -csc^,\\nfor any value of A.\\nPC\\ncos(90\u00c2\u00b0 -sin^\\ncot(90\u00c2\u00b0 ^)=-tan^,\\nesc (90\u00c2\u00b0 sec^, J\\n(2)", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0031.jp2"}, "32": {"fulltext": "22\\nPlane Trigonometry.\\nI\\nFig. 3.\\nThere may be four cases A in the first quadrant (Fig. 1),\\nA in the second quadrant (Fig. 2), A in the third quadrant\\n(Fig. 3), or A in the fourth quadrant (Fig. 4).\\nIn each figure, let the positive angle XOP represent the\\nangle A, and the positive angle XOP the angle 90\u00c2\u00b0 -f A.\\nTake OP OP, and draw PM and P M perpendicular\\nto XX\\nSince OP is perpendicular to OP, and OJf to P M\\\\\\nZPOM=ZOPM\\nThen right triangles 0PM and OP M have the hypote-\\nnuse and an acute angle of one equal, respectively, to the\\nhypotenuse and an acute angle of the other, and are equal.\\nWhence, PM= OM and 0M= P M\\nTherefore, in each, figure,\\nordinate P abscissa P,\\nabscissa P ordinate P,\\nand distance P distance P.\\nord. P abs. P\\nThen, sin (90\u00c2\u00b0 -\\\\-A)\\ncos (90\u00c2\u00b0 -\\\\-A)\\ndist. P\\nabs. P\\ndist. P\\ndist. P\\nord. P\\ndist. P\\ncos A.\\nsin A.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0032.jp2"}, "33": {"fulltext": "Trigonometric Functions. 23\\ntan (90\u00c2\u00b0 2\u00c2\u00a3d^ _ 5]^ cot A\\nord. P\\ncot (90\u00c2\u00b0 A)= tan A.\\nV y D, abs. P\\nabs.\\nP\\nabs.\\nP\\nord.\\nP\\ndist. P\\nabs.\\nP\\ndist. P\\nsec(90\u00c2\u00b0 T^ -^^^r^ -cscA\\nV .u. Df ord. P\\ncsc(90\u00c2\u00b0 ^T\\\\ distP^ gg^^^\\nord. P abs. P\\n30. The results of 29 may be stated as follows\\nThe sine, cosine, tangent, cotangent, secant, and cosecant of\\nany angle are equal, respectively, to the cosine, minus the sine,\\nminus the cotangent, minus the tangent, minus the cosecant,\\nand the secant, of an angle 90\u00c2\u00b0 less.\\n31. Functions of (90\u00c2\u00b0 A) in terms of those of A.\\nBy \u00c2\u00a730, sin (90\u00c2\u00b0-^)= cos cos 28).\\ncos (90\u00c2\u00b0 sin sin A.\\ntan (90\u00c2\u00b0 cot cot A.\\ncot(90\u00c2\u00b0-^)=-tan(-^) tanA\\nsec (90\u00c2\u00b0 CSC CSC A.\\nCSC (90\u00c2\u00b0 -A)= seG{-A) sec A.\\nThese formulse were proved for acute angles in 4.\\n32. Functions of (180\u00c2\u00b0 A) in terms of those of A.\\nBy 30, sin (180\u00c2\u00b0 -A)= cos (90\u00c2\u00b0-^) sin A 31).\\ncos (180\u00c2\u00b0 -A) sin (90\u00c2\u00b0-^) cos A.\\ntan (180\u00c2\u00b0 -A) cot (90\u00c2\u00b0-^) tan A.\\ncot (180\u00c2\u00b0 -A)=- tan (90\u00c2\u00b0-^) cot A.\\nsec (180\u00c2\u00b0 -A)=- CSC (90\u00c2\u00b0-^) sec A.\\nCSC (180\u00c2\u00b0 -A)= sec (90\u00c2\u00b0 -A)= esc A.", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0033.jp2"}, "34": {"fulltext": "24\\nPlane Trigonometry.\\n33. By successive applications of the theorem of 30,\\nany function of a multiple of 90\u00c2\u00b0, plus or minus A, may be\\nexpressed as a function of\\n1. Express sin (270\u00c2\u00b0 J.) as a function of A.\\nBy 30, sin (270 cos (180\u00c2\u00b0+^) sin (90\u00c2\u00b0+^) cos A.\\nIf the multiple of 90\u00c2\u00b0 is greater than 270\u00c2\u00b0, we may sub-\\ntract 360\u00c2\u00b0, or any multiple of 360\u00c2\u00b0, from the angle, in\\naccordance with 19.\\n2. Express sec (990\u00c2\u00b0 J.) as a function of A.\\nSubtracting twice 360\u00c2\u00b0, or 720\u00c2\u00b0, from the angle, we have\\nsec (990\u00c2\u00b0-^) sec (270\u00c2\u00b0-^).\\nAnd by 30, sec (270\u00c2\u00b0 -A) esc (180\u00c2\u00b0 esc 32)\\nIf the multiple of 90\u00c2\u00b0 is negative, we may add 360\u00c2\u00b0, or\\nany multiple of 360\u00c2\u00b0, to the angle.\\n3. Express tan 180\u00c2\u00b0 as a function of A.\\nAdding 360\u00c2\u00b0 to the angle, we have\\ntan 180\u00c2\u00b0 A tan (180\u00c2\u00b0 A).\\nAnd by 30, tan (180\u00c2\u00b0 -hA) cot (90\u00c2\u00b0 A tan A.\\nEXAMPLES.\\nExpress each of the following as a function of A\\n4. sin (180\u00c2\u00b0\\n5. cos (270\u00c2\u00b0-^).\\n6. cot (450\u00c2\u00b0\\n7. CSC (360\u00c2\u00b0-^).\\n8. tan (540\u00c2\u00b0\\n9. sec (630\u00c2\u00b0 4-^).\\n10. tan 270\u00c2\u00b0\\n11. CSC (-90\u00c2\u00b0-^).\\n12. cot 180\u00c2\u00b0\\n13. sin 630\u00c2\u00b0\\n14. tan 450\u00c2\u00b0\\n15. cos 900\u00c2\u00b0\\n16. sin (810\u00c2\u00b0-^).\\n17. CSC (1080\u00c2\u00b0\\n18. sec (1260\u00c2\u00b0", "height": "3702", "width": "2370", "jp2-path": "completetrigonom00well_0034.jp2"}, "35": {"fulltext": "Trigonometric Functions. 25\\n34. By means of the theorem of 30, any function of\\nany angle, positive or negative, may be expressed as a func-\\ntion of a certain acute angle.\\n1. Express sin 317\u00c2\u00b0 as a function of an acute angle.\\nBy 30, sin 317\u00c2\u00b0 cos 227\u00c2\u00b0 sin 137\u00c2\u00b0 cos 47\u00c2\u00b0.\\nSince the complement of 47\u00c2\u00b0 is 43\u00c2\u00b0, another form of the result is\\n-sin 43\u00c2\u00b0 (\u00c2\u00a74).\\nNote. As in the examples of 33, 360\u00c2\u00b0, or any multiple of 360\u00c2\u00b0,\\nmay be added to, or subtracted from, the angle.\\nEXAMPLES.\\nExpress each of the following as a function of an acute\\nangle\\n2. cos 322\u00c2\u00b0. 4. sec 559\u00c2\u00b0. 6. cot 378\u00c2\u00b0).\\n3. tan 208\u00c2\u00b0. 5. esc 803\u00c2\u00b0 45 7. sin 139\u00c2\u00b0 5\\nIt is evident from the above that any function of any\\nangle can be expressed as a function of a certain acute\\nangle less than 45\u00c2\u00b0.\\nExpress each of the following as a function of an acute\\nangle less than 45\u00c2\u00b0\\n8. cot 155\u00c2\u00b0. 10. sec 457\u00c2\u00b0. 12. tan (-681\u00c2\u00b0).\\n9. sin 1138\u00c2\u00b0 36 11. cos 496\u00c2\u00b0 20 13. esc 257\u00c2\u00b0).\\n14. Eind the numerical value of esc 210\u00c2\u00b0).\\nAdding 360\u00c2\u00b0 to the angle, we have\\nCSC (-210\u00c2\u00b0)= CSC 150\u00c2\u00b0.\\nAnd by 30, csc 150\u00c2\u00b0 sec 60\u00c2\u00b0 2 8).\\nEind the numerical values of the following\\n15. cot 405\u00c2\u00b0. 17. CSC 600\u00c2\u00b0. 19. cos (-420\u00c2\u00b0).\\n16. sin 480\u00c2\u00b0. 18. tan 690\u00c2\u00b0. 20. sec (-225\u00c2\u00b0).", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0035.jp2"}, "36": {"fulltext": "26\\nPlane Trigonometry.\\nIII. GENERAL FORMULiB.\\n35. It follows directly from the definitions of 17 that,\\nif X is any angle,\\nsm X\\ncos X\\nCSC X\\n1\\ntanic\\ncot X\\ncot it\\n1\\nsec X\\nsec X tan x\\n36. To iwove the formula\\ncscx\\nCOS X\\n1\\nsm a;\\ntan a\\nsm a?\\nCOSiC\\n(3)\\n(4)\\nJlf\\nPio. 2.\\nThere may be four cases x in the first quadrant (Fig. 1),\\nX in the second quadrant (Fig. 2), x in the third quadrant\\n(Fig. 3), or x in the fourth quadrant (Fig. 4).\\nIn each case, let the positive angle XOP represent the\\nangle x, and draw PM perpendicular to XX", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0036.jp2"}, "37": {"fulltext": "General Formulae. 27\\nThen in each figure, by the definitions of 17,\\nord. P\\ntan X\\nord. P\\nabs. P\\ndist. P\\nabs. P\\ndist. P\\nsin X\\ncos a?\\n37.\\nTop\\\\\\nrove the\\nformula\\nGotx\\ncos a?\\nsin .1?\\nBy\u00c2\u00a7\\n35,\\ncot X\\n_ 1 _\\ntan a;\\nsm a; sin a.-\\n(5)\\ncos a.\\n38. To prove the formula\\nsin^ X cos^ x 1. (6)\\nNote. Sin2 aj signifies (sin x)^ that is, the square of the sine of x.\\nTliere may be four cases x in the first quadrant, x in the\\nsecond quadrant, x in the third quadrant, or x in the fourth\\nquadrant.\\nIn each figure of 36, we have by Geometry,\\npm om 0P\\\\\\nDividing by OP 1.\\nop op\\nBut in each figure,\\nPM s2 a OM .2\\n(sm xY, and (cos a^)^\\nOP op\\nPM PM PM\\nfor, whether sin x equals or its square is\\nOP OP Qp^\\nWhence, sin^a^ cos^a^^ 1.\\n39. Formula (6) may be written in the forms\\nsin^ x l cos^ x, and cos^ x l sin^ x.", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0037.jp2"}, "38": {"fulltext": "28\\nPlane Trigonometry.\\n40. To prove the formulce\\nsec^ ic 1 tan^ x,\\nand csc^ x 1 -j- cot^ x.\\nBy (6), 1 cos^ X sin^ x.\\n(7)\\n(8)\\n(A)\\nDividing by cos^ x,\\n1\\nWhence by (3) and (4), sec^ x 1 tan^ x.\\nAgain, dividing (A) by sin^ x, we have\\n1\\ncos^a?\\nsnr a? sm a?\\nWhence by (3) and (5), csc^ x l-[- cot^ x.\\n41. To express sin (a? y) and cos (a? y) in terms of the\\nsines and cosines of x ayid y.\\nI. When\\nX\\nand y are acute\\nE\\nB\\nA\u00c2\u00bb\\nA D\\nFic\\n1\\nThere may be two cases x-\\\\-y acute (Fig. 1), and x y\\nobtuse (Fig. 2).\\nIn each figure, let Z DOS aj and Z BOC y.\\nThen, ZDOC=x-^y.\\nFrom any point (7 in 00 draw lines CA and OjB perpen-\\ndicular to OD and 0-B, respectively also, draw lines BD\\nand BE perpendicular to OD and AC, respectively.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0038.jp2"}, "39": {"fulltext": "General Formulae. 29\\nSince EC is perpendicular to OD, and BC to OB, angles\\nBCE and DOB are equal tliat is, Z BCE x.\\nIn either figure, by 17,\\n^C BD CE BD CE\\n^BD OB CE BC\\nOB OC BC OC\\nThen, sin {x y)= sin a? cos 2/ cos x sin (9)\\nAgain, by 17, in Fig. 1, cos (aj y) -g^,\\nOA\\nand in Fig. 2, cos (a?\\nThen in either figure,\\nOD BE OD BE\\ncos (x-\\\\-y)\\nOC OC OC\\n^Op qB_BE BC\\nOB OC BC OC\\nThen, cos (x -}-y)= cos x cos y sin x sin y. (10)\\n42. Formulae (9) and (10) are very important, and it is\\nnecessary to prove them for all values of x and y.\\nThey have already been proved when x and y are any\\ntwo acute angles or, what is the same thing, when they\\nare any two angles in the first quadrant.\\nNow let a and b be any two angles in the first quadrant.\\nBy 29, sin [90\u00c2\u00b0 (a 6)] cos (a 6),\\nand cos [90\u00c2\u00b0 (a 6)] sin (a -f b).\\nWhence, by (9) and (10),\\nsin [90\u00c2\u00b0 (a-\\\\-b) cos a cos b sin a sin b, (A)\\nand cos [90\u00c2\u00b0 (a sin a cos b cos a sin b. (B)", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0039.jp2"}, "40": {"fulltext": "30 Plane Trigonometry.\\nBy 29, cos a sin (90\u00c2\u00b0+ a), and sin a cos (90\u00c2\u00b0+ a).\\nThen, (A) and (B) may be written in the forms\\nsin [(90\u00c2\u00b0 a) 6] sin (90\u00c2\u00b0 a) cos b cos (90\u00c2\u00b0 a) sin b,\\ncos [(90\u00c2\u00b0 a) 6] cos (90\u00c2\u00b0 a) cos b sin (90\u00c2\u00b0 a) sin b\\nwhich are in accordance with (9) and (10).\\nBut 90\u00c2\u00b0 a is an angle in the second quadrant.\\nTherefore, (9) and (10) hold when one of the angles is\\nin the second quadrant, and the other in the first.\\nIn like manner, by supposing a to be any angle in the\\nfirst quadrant, and b any angle in the second, (9) and (10)\\nmay be proved to hold when both angles are in the second\\nquadrant.\\nAgain, by supposing a and b to be any two angles in the\\nsecond quadrant, (9) and (10) may be proved to hold when\\none angle is in the second quadrant and the other in the\\nthird; and so on.\\nHence, (9) and (10) hold for any values of x and y what-\\never, positive or negative.\\n43. Putting, in (9) and (10), y in place of y,\\nsin (x y)= sin a?. cos y)-\\\\- cos a? sin y)\\nsin X cos y cos x(\u00e2\u0080\u0094 sin y), by 2S,\\nsin X cos y cos x sin y. (11)\\ncos (x y)= cos X cos y) sin x sin y)\\ncos X cos y sin x(\u00e2\u0080\u0094 sin y)\\ncos xGosy sin x sin y. (12)\\n44. By (4),\\ntan(a. 2/)=^H^^\\ncos {x y)\\nsin X cos 2/ cos X sin .v^\\nCOS X COS y sin x sin y", "height": "3670", "width": "2370", "jp2-path": "completetrigonom00well_0040.jp2"}, "41": {"fulltext": "General Formulae. 31\\nDividing each, term of tlie fraction by cos .v cos y,\\nsin X cos y cos 2 sin y\\ncos X COS If cos X COS\\ntan (2 v) T-^\\nCOS 2* COS sm .r sm 2/\\ncos X COS ;y cos x cos\\n_ tan X tan\\n1 tan X tan\\nIn like manner, we may prove\\n(13)\\n4- tan .r tan?/ x_^.\\ntan (a; (14)\\nl tan.i tan?/\\nAgain, by (5), cot {,c\\nsm (x- y)\\n_ cos COS sin a- sin\\nsin X cos cos x sin\\nDividing each term of the fraction by sin x sin y,\\ncos X cos sin a. sin y\\nsin X sin sin x sin v\\ncot (iC\\nsm.rcos^ cos.i sm^\\nsin X sin y sin x sin\\n_ cot X cot 1\\ncot y cot :c\\nIn like manner, vre may prove\\nX C0t.TC0t?/ l\\ncot ix y) (16)\\ncot y cot x\\n45. From (9), (10), (11), and (12), we have\\nsin h) sin a cos h cos a sin h. (A)\\nsin (a \u00e2\u0080\u00941)) sin o cos 6 cos a sin 5. (B)\\ncos (a cos a cos 6 sin a sin 6. (C)\\ncos {a\u00e2\u0080\u0094 h) cos a cos 6 sin a sin 6. (D)\\n(15)", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0041.jp2"}, "42": {"fulltext": "32 Plane Trigonometry.\\nAdding and subtracting (A) and (B), and then (C) and (D),\\nsin (a sin (a b) 2 sin a cos b.\\nsin (a sin (a 6) 2 cos asin b.\\ncos (a 6) 4- cos (a 6) 2 cos a cos 6.\\ncos (a 5) cos (a 5) 2 sin a sin b.\\nLet a 6 a?, and a b y.\\nAdding, 2a==x-\\\\-y, or a, -J- (x\\nSubtracting, 2b x y, or 6 (ic\\nSubstituting these values, we have\\nsin a? sin 2/ 2 sm^(x-\\\\-y)GOS^{x y). (17)\\nsin X sin 2 cos -J (a? 2/) sin i(x~ y). (18)\\ncos a? cos 2cos-J-(a^ 2/) cos-2-(ic 2/). (19)\\ncos a? cos 2/ 2 sin J (cc 2/) sin (a? 2/). (20)\\n46. By (17) and (18), we have\\nsin X sin y _ 2 sin -J- (a? 2/) cos ^(x y)\\nsin ic sin 2/ 2 cos i (a^ 2/) sin {x y)\\ntan|-(a; y) cot ^(a; 2/)\\n47. Functions of 2 x.\\nPutting 2/ in (9)j we have\\nsin 2 i\u00c2\u00bb sin x cos cc cos x sin ic\\n2 sin ic cos a?. (22)\\nPutting 2/ a? in (10), we obtain\\ncos 2x cos a? cos x sin a? sin a;\\ncos^ X sin^ X. (23)", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0042.jp2"}, "43": {"fulltext": "General Formulae.\\nWe also have by 39,\\ncos 2 ic (1 sin^ x) sin^ ic 1 2 sin^ x. (24)\\ncos 2x cos^ X\u00e2\u0080\u0094 (1 cos^ 0?) 2 cos^ x\u00e2\u0080\u00941. (25)\\nPutting 2/ a? in (13) and (15),\\no tan X tan x 2 tan x ,_ _^\\ntan 2 a? (26)\\n1 tan X tan ic 1 tan- x\\no cot 07 cot X 1 COt^ X 1 /\u00c2\u00ab\u00e2\u0080\u009ex\\ncot 2 a? (27)\\ncot X cot X 2 cot X\\n48. Functions of a?.\\nFrom (24) and (25) we have, by transposition,\\n2 sin^ x l cos 2 a?, and 2 cos^ a? 1 cos 2 a?.\\nPutting J X in place of a;, and therefore x in place of 2 a?,\\nwe have\\n2 sin^ aj 1 cos x, (28)\\n2 cos^ X 1 cos X. (29)\\nAgain, putting -J-x in place of x in (22),\\n2 sin a; cos ix sin x. (A)\\nDividing (28) by (A),\\n2 sin^ ^x _ 1 cos x\\n2 sin -J- a? cos -J- a? sin x\\nWhence, by (4), tan -i x (30)\\nDividing (29) by (A),\\n2 cos^ ^x _ 1 cos a;\\n2 sin i x cos 4- a? sin x\\ncos a;\\nWhence, by (5), cot -J- x iJlidd^. (31)\\nsm X", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0043.jp2"}, "44": {"fulltext": "34 Plane Trigonometry.\\nEXERCISES.\\n49. 1. Prove the relation sec^ x csc^ x sec^ x csc^\\nBy (3), sec^x csc^x f by (6^\\ncos^ X sin2 X cos2 x sm^ x\\ncos^ X sin2 X cos2 x sin^ x\\n1 1\\ncos X sin2 X\\nsin 5 \u00c2\u00a3c\\ncos 5x cos X\\nsec2 X csc^ X.\\n2. Prove the relation tan 2 x.\\nBy (18) and (19),\\nsin 5 X sin x _ 2 cos ^(5x x) sin|(5x x) _ sin 2 x _\\ncos5x cosx 2 cos|(5x x) cos-|(5x x) cos2x\\nOX) ^1, 14.- tan (.T v) tan x\\n3. Prove the relation tan y.\\n1 tan (x y) tan x\\nBy (14). tan(x i/)-tanx tan [(x x] tan y.\\n1 +tan(x 2/)tanx lk ifj j y\\n4. Prove the relation sin 3x Z sin x 4 sin^ a:;.\\nBy (18), sin 3x sin x 2 cosi(3x x) sin 1(3 x x).\\nThen, sin 3 x sin x 2 cos 2 x sin x\\nsin X 2 (1 2 sin2 x) sin x, by (24)\\nsin X 2 sin x 4 sin^ x 3 sin x 4 sin^ x.\\nThe artifice used above is advantageous in finding the\\nsine or cosine of any odd multiple of x.\\nProve the following relations\\nK sin {x y) __ tan x tan y\\nsin (x y) tan x tan y\\ncos (x I/) _ cot X cot 1\\ncos (x y) cot a? cot 2/ 1\\ni~ cos a? cos y i j. i\\n7. cot i (it- y) cot i (a^ 2/)-\\ncos X cos 2/", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0044.jp2"}, "45": {"fulltext": "General Formulae. 35\\n8. sm3x sinx^^^^^^\\nCOS 3x cos X\\n9. sec^ A tan^ 5 sec^ B tan^ A.\\ncos a? cos y\\nt^n(A B) t^n(A~B)\\n1 tan (A 5) tan (A 5)\\n12. cos X cos (a? 2/) sin x sin (x y) cos 2/.\\n^13. sin 5 aj cos 2x cos 5 a? sin 2 x sin 3 ic.\\nM sin 3 a? sin 5 .t\\n14. cot 4 a;.\\ncos 3x cos 5 ic\\n1 c cot X tan a? o\\n15. sec2aj.\\ncot X tan X\\n16. cos 3x 4: cos^ X 3 cos ic.\\n17. tan^ia^^\\n1 cos a^\\n18. sin (x-\\\\-y z) sin .t cos cos 2 cos x sin cos 2\\nH- cos a? cos sin z sin a.- sin y sin 2.\\n19. cos (x y z) cos a^ cos y cos 2; sin x sin cos ;2\\nsin X cos sin z cos sin y sin 2.\\nrtfk tanflj+tanw+tans; tancc tan?/ tanz\\n20. tdi,ii(x-{-y^z)\\n1 tan 37 tan tan 2/ tan 2; tan 2; tan a;\\n21. Prove the relation of Ex. 4 by putting x 2x, and\\ny x, in (9).\\nProve the relations\\nnn o 3 tan X tan^ x\\n22. tan3a^==\u00e2\u0080\u0094\\n1-3 tan2 X\\n23. sin (x y) sin (x y)\u00e2\u0080\u0094 sin^ a? sin^ i/-\\n24. cos (a; i/) cos (x y) cos^ a) siu^ y.", "height": "3617", "width": "2367", "jp2-path": "completetrigonom00well_0045.jp2"}, "46": {"fulltext": "^6 Plane Trigonometry,\\n2 tana?\\n25. sin 2 x\\n26. cos2x\\n1 tan^ X\\n1 tan^ X\\n1 tan^ X\\n27. 4 cos X cos (60\u00c2\u00b0 x) cos (60\u00c2\u00b0 \u00e2\u0080\u0094x) cos 3 x.\\n28. tan (45\u00c2\u00b0 x) tan (45\u00c2\u00b0 ic) 2 tan 2 x.\\n29. cos^ (x-i-y)\u00e2\u0080\u0094 sin^ a? cos (2 x-\\\\-y) cos 2/.\\n30. cot^- cot 2^ CSC 2 A\\n31. sin4x sin_3x^^^^^^^\\ncos 3 X COS 4 X\\noo sin 50\u00c2\u00b0 sin 10\u00c2\u00b0 /p\\ncos 50\u00c2\u00b0 -cos 10\u00c2\u00b0\\n33. cos 80\u00c2\u00b0 cos 40\u00c2\u00b0 cos 20\u00c2\u00b0.\\n34. sin 4:X 4:smx cos x\u00e2\u0080\u0094S sin^ x cos x.\\n35. cos 4 a; 1 8 cos^ x-\\\\-S cos^ a?.\\n36. cos 5x 5 cos a? 20 cos^ x 16 cos^ x.\\n37. cos (2 X 2 sin aj sin (x-{-y) cos\\n38. By putting x 45\u00c2\u00b0 and y 30\u00c2\u00b0 in (11) and (12), prove\\nsin 15\u00c2\u00b0= cos 15\u00c2\u00b0==^ A\\n4 4\\n39. By putting x 30\u00c2\u00b0 in (30) and (31), prove\\ntan 15\u00c2\u00b0 2 -V3, cotl5\u00c2\u00b0 2+V3.\\n40. By putting x 15\u00c2\u00b0 in (3), and using the results of\\nEx. 38, prove\\nsecl5\u00c2\u00b0=V6-V2, cscl5\u00c2\u00b0 V6 V2.\\n41. By putting x 45\u00c2\u00b0 in (28) and (29), prove\\nsin 221\u00c2\u00b0 JV2-V2, cos 22i-\u00c2\u00b0 iV2 a/2.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0046.jp2"}, "47": {"fulltext": "General Formulae. 37\\n42. By putting x 45\u00c2\u00b0 in (30) and (31), prove\\ntan 22^\u00c2\u00b0 V2 1, cot 22^\u00c2\u00b0 V2 1.\\n43. By putting x 22j-\u00c2\u00b0 in (7) and (8), and using tlie\\nresults of Ex. 42, prove\\nsec22i\u00c2\u00b0 V4-2V2, esc 22^\u00c2\u00b0 V 4 -f- 2 V2.\\nProve the relations\\n44. cos^ X sin* x cos 2 x.\\ne CSC X cot X ,0 1\\n45. cot^ i X.\\nCSC X cot X\\n46. sin (90\u00c2\u00b0 A) sin (210\u00c2\u00b0 sin (210\u00c2\u00b0 0.\\n47. (sin X sin yY (cos a? cos y^ 4 cos^\\n48. tan 2 a? cot a^ 1 sec 2 a?.\\n49. tan x tan (60\u00c2\u00b0 x) tan (120\u00c2\u00b0 x) tan 3 x.\\nKQ 1 sin 2 X _ co s X sin .t\\ncos 2 a? cos a^ sin x\\n4 tan a^ 4 tan^ x\\n51. tan 4 a?\\n1 6 tan^ X tan* aj\\n52. ^tan(45\u00c2\u00b0-ix).\\ncosx\\n2 tan (45\u00c2\u00b0 -a^)\\n53. ..,0 cos 2 a?.\\n1 tan^ (45 x)\\n54. Prove the first result of Ex. 43 by putting x 221\\nin (3), and using the result of Ex. 41.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0047.jp2"}, "48": {"fulltext": "38 Plane Trigonometry.\\nIV. MISCELLANEOUS THEOREMS.\\n50. Circular Measure of an Angle.\\nAn angle is measured by finding its ratio to another angle,\\nadopted arbitrarily as the unit of measure.\\nThe usual unit of measure for angles is the degree, which\\nis an angle equal to the ninetieth part of a right angle.\\nAnother method of measuring angles, and one of great\\nimportance, is known as the Circular Method in which the\\nunit of measure is the angle at the centre of a circle subtended\\nby an arc ivhose length is equal to the radius.\\nNote. The unit of circular measure is called a radian.\\nThus, let AOB be any central angle, and AOC the unit\\nof circular measure that is, the angle at the centre sub-\\ntended by an arc whose length is equal to OA.\\nAOB\\nThen, circular measure AOB\\nBut by Geometry,\\n.AOC\\nZAOB arc .45 arc ^B\\nWhence, circular measure A OB\\nAOC diVcAC OA\\narc AB\\nOA\\nThat is, the circular measure of an angle is the ratio of its\\nsubtending arc to the radius of the circle.\\n51. By 50, the circular measure of a right angle is the\\nratio of one-fourth the circumference to the radius.\\nBut if R denotes the radius, the circumference is 2 irR,", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0048.jp2"}, "49": {"fulltext": "Miscellaneous Theorems. 39\\nWhence, circular measure 90\u00c2\u00b0 _-!I\\nR 2\\nIt follows from the above that the circular measure of\\n180\u00c2\u00b0 is TT of 60\u00c2\u00b0, of 45\u00c2\u00b0, etc.\\nThat is, an angle expressed in degrees may he reduced to\\ncircular measure by finding its ratio to 180\u00c2\u00b0, and midtiplying\\nthe result by ir.\\n23\\nThus, since 115\u00c2\u00b0 is of 180\u00c2\u00b0, the circular measure of\\n90 36\\n115\u00c2\u00b0 is\\n36\\n52. Conversely, an angle expressed in circular measure\\nmay be reduced to degrees by tnultiijlying by 180\u00c2\u00b0 and dividing\\nby IT or, more briefly, by substituting 180\u00c2\u00b0 for tt.\\nThus, I^ X of 180\u00c2\u00b0 84\u00c2\u00b0.\\n15 15\\n53. In the circular method, such expressions may occur\\nas the angle the angle 1, etc.\\nThese refer to the unit of circular measure; thus, the\\nangle signifies an angle whose subtending arc is two-thirds\\nof the radius.\\nThe angle 1 signifies the angle whose subtending arc is\\nequal to the radius, or the unit of circulai measure.\\nThe angle 1, reduced to degrees by the first rule of 52,\\ngives\\no-.!f^I =57.2958\u00c2\u00b0, approximately.\\nTT 3.14159\\nThen the rule of 52 may be modified as follows\\nAn angle expressed in circular measure may be reduced to\\ndegrees by multiplying by 57.2958\u00c2\u00b0.\\nThus, the angle f\\nI X 57.2958\u00c2\u00b0 38.1972\u00c2\u00b0 38\u00c2\u00b0 11 49.92", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0049.jp2"}, "50": {"fulltext": "40 Plane Trigonometry.\\nEXAMPLES.\\n54. Express each, of the following in circular measure\\n1. 120\u00c2\u00b0. 3. 67\u00c2\u00b0 30 5. 86\u00c2\u00b0 24 7. 163\u00c2\u00b0 7 30\\n2. 315\u00c2\u00b0. 4. 146\u00c2\u00b0 15 6. 53\u00c2\u00b0 20 8. 88\u00c2\u00b0 53 20\\nExpress each of the following in degree measure\\n9. 11. li^. 13. 5. 15.\\n6 81 2 3\\n10. y^. 12. 14. i. 16.\\n24 64 4 5\\n55. Inverse Trigonometric Functions.\\nThe expression sin~^cc, called the inverse sine of x, or the\\nanti-sine of x, signifies the angle ivhose sine is x.\\nThus, the statement that the sine of the angle x is equal\\nto y may be expressed in either of the ways\\nsin x y, or a? sin~-^ y.\\nIn like manner, cos~^ x signifies the angle whose cosine is\\nX tan~^ X, the angle whose tangent is x etc.\\nNote. The student must be careful not to confuse the above nota-\\ntion with the exponent 1 the 1 power of sin x is expressed\\n(sin x) i, and not sin-i x.\\nIt is evident that the sine of the angle whose sine is x is\\nX that is, sin (sin~^ x) x.\\nIn like manner, cos (cos x) x-, tan (tan~^ aj) x etc.\\n56. By aid of the principles of 55, we may derive from\\nany formula involving direct functions a relation between\\ninverse functions.\\n1. From the formula t2iii(x y)= tana; tan?/\\n1 tan X tan v\\n.tan-i a tan-^ b tan-^ -^\u00c2\u00b1A\u00e2\u0080\u009e\\n1 ab", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0050.jp2"}, "51": {"fulltext": "Miscellaneous Theorems.\\nLet tan x a, and tan y b.\\nThen by 55, x tan-i a, and y tan-i b.\\nSubstituting these values in the given formula,\\na b\\n41\\nWhence,\\ntan(tan-i a tan~^ 6)\\ntan-i a tan-i b tan\\n1 -ab\\n1 -ab\\n2. Prove the relation cot~^a\u00e2\u0080\u0094 sec cos\\nLet cot-i a x, and sec-i b =y.\\nThen, cot x a, and sec y b.\\nNow,, cos(x y)= cos X cos j/ sin x sin y. (A)\\nTo find the sines and cosines of x and y, we use the method of 6.\\nIn the right triangle containing angle x, the adjacent side is a, and\\nthe opposite side 1 then, the hypotenuse is Va^ 1.\\nIn the right triangle containing angle y, the hypotenuse is b, and\\nthe adjacent side 1 then, the opposite side is Vft^ _ i.\\nSubstituting the values of cosx, cosy, sin x, and siny in (A), we\\nhave\\na 1\\ncos(x y)\\nV^^ +l Va2 1\\nVa2 1\\nWhence, x y or cot i a sec-i cos-i\\nV\u00c2\u00a5\\nEXAMPLES.\\n3. From cot 2x\\ncot^a. 1\\nbVa^Tl\\nn nr\u00c2\u00bbf ~1\\nprove 2cot~^a=cot\\n2 cot i\u00c2\u00bb\\n4. From cos 2 a? 1 2 sin^ a;, prove\\n2sin-ia cos-\\\\l-2a2).\\n2a", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0051.jp2"}, "52": {"fulltext": "42 Plane Trigonometry.\\n5. From sin 2x 2 sin x cos x^ prove\\n2 cos^^ a sin~^ (2 aV 1 a^).\\n6. From cos (a? cos x cos y sin a? sin prove\\ncos~^ a cos~^ 6 cos~^(a? Vl a^Vl 6^).\\nProve the following relations\\n^7. cot-i a -h cot-i h cot-i\\n8. 2cos-ia cos-\\\\2a2-l).\\n9. sin~^ a sin~^ sin~^ (a Vl h^\u00e2\u0080\u0094 5 Vl a^).\\n10. 3 sin-i a sin-^ (3 a 4 a^). (Ex. 4, p. 34.)\\n11. tan-^ a cot-^ 6 tan-^ ^^5_\u00c2\u00b11.\\n6 a\\n12. cot-^a -h)- cot-i (a 6) cot-^ a^-^^ l\\n26\\n\u00e2\u0096\u00a0^13. tan~^a sin~^\\nVa^ 1\\n14. csc~^ a cos~^\\n15. sin- a cos- tan- a^ Vl aVl\\n5Vr^^^-aVl-\\n1 1 Va- 1 V6^ 1\\n#16. sec\u00e2\u0080\u0094 a csc-^ 6 cos\u00e2\u0080\u0094\\nah\\n1 K- 1 _i 1 1 a Va^ 1\\n17. tan~^a cos\u00e2\u0080\u0094 -^sm\\na-^d^ 1\\n18. 2 sin- a tan-\\n1 2a^\\n19. tan-\u00e2\u0080\u0094 ^-tan-^-^^i^ tan--i-\\na 1 a 2 a", "height": "3704", "width": "2370", "jp2-path": "completetrigonom00well_0052.jp2"}, "53": {"fulltext": "20. 2 sec a cot\\nMiscellaneous Theorems.\\n43\\n2 Va^ 1\\n57. The following table expresses the value of each of\\nthe six principal functions in terms of the other five\\nsin\\ncos\\ntan\\ncot\\nsec\\nCSC\\ntan\\n1\\nJ_\\nCSC\\nVsec2\\n1\\nVl cos^\\nsec\\nVl -f tan^\\n1\\nVl cof-\\ncot 1\\nVCSC2 1\\nVl sin2\\nsin\\nCSC\\n1\\nVl tan\\n1\\ntan\\nVl cof^\\n1\\ncot\\nsec\\nVl cos-\\nVsec2\\n1\\n1\\nVl sin^\\ncos\\ncos\\nVcSC2 1\\nVl sin^\\nVCSC2 1\\nCSC\\nsin\\n1\\nVl cos^\\ncos\\n1\\nVsec2\\nsec\\n1\\nVl C0t2\\nVl tan2\\nVl sin2\\n1\\nsin\\ncot\\nVCSC2 1\\na/1 _l t,P,n2\\nVl C0t2\\nVl C0S2\\ntan\\nVsec2\\nThe reciprocal forms were proved in 35.\\nThe others may be derived by aid of 35, 36, 37, 39,\\nand 40, and are left as exercises for the pupil.\\nAs an illustration, we will prove the formula\\ncos^\\nBy 39,\\nVcsc^ A\\ncsc^\\nCOS A V 1 sin^ A\\n4\\nVcscM-1\\ncsc^^\\ncsc^\\nThey may also be conveniently proved by the method of\\n6 thus, let it be required to prove the formula for each\\nof the other functions in terms of the secant.\\nWe have\\nsec^\\nsec J.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0053.jp2"}, "54": {"fulltext": "44\\nPlane Trigonometry.\\nB\\nSince the secant is the ratio of the hypotenuse to the\\nadjacent side, we take AB=s,qq,A, and AC 1.\\nWhence, 50 V^s -^O VsecM -1.\\nThen by definition,\\nVsec^ A 1\\nsm A\\nsec^\\ncos A -J\\nsec J.\\ncot^\\nCSC J.\\nVsecM-1\\nsec A^\\nVsecM-1\\ntan A V sec^ A 1,\\n58. Line Values of the Functions.\\nLet XOB be any angle.\\nWith as a centre, and a radius equal to 1, describe cir-\\ncumference AB, cutting OX at A, OB at B, and F at C.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0054.jp2"}, "55": {"fulltext": "Miscellaneous Theorems.\\n45\\nF\\nX\\ny\\nC\\n\u00e2\u0096\u00a0r\\n-\u00c2\u00bb-r\\nC\\ny\\n.Fvr\\ny-^\\nr\\ni\\nP\\n1\\nV\\nIK\\n^i\\nA\\nI\\nK\\nJ\\nA\\nI\\nV\\n1\\nB\\nLy\\ny. V^\\nr\\nF\\nFia\\n.3.\\nFig\\n.4.\\nDraw line BD perpendicular to XX also, lines AE and\\nCF perpendicular to OX and Y, respectively, meeting OB,\\nor OB produced, at E and F, respectively.\\nThen by 17, the functions of Z XOB are\\nSin.\\nCos.\\nTan.\\nCot.\\nSec.\\nCsc.\\nFig. 1.\\nBD\\nOB\\nOD\\nOB\\nBn\\non\\nOD\\nBD\\nOB\\nOD\\nOB\\nBD\\nFig. 2.\\nBD\\nOB\\nOD\\nOB\\nBD\\nOD\\nOD\\nBD\\nOB\\nOD\\nOB\\nBD\\nFig. 3.\\nBD\\nOB\\nOD\\nOB\\nBD\\nOD\\nOD\\nBD\\nOB\\nOD\\nOB\\nBD\\nFig. 4.\\nBD\\nOB\\nOD\\nOB\\nBD\\nOD\\nOD\\nBD\\nOB\\nOD\\nOB\\nBD\\nNow right triangles OBD, OEA, and OCF are similar,\\nsince their sides are parallel each to each.\\nThen, since OA OC 1, we have\\n?D^AE^^\\nOD OA\\n2^ ^=CF\\nBD 00\\nOB^OE\\nOD OA\\nOB^OF\\nBD OC\\nOE,\\nOF.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0055.jp2"}, "56": {"fulltext": "4.6 Plane Trigonometry.\\nWhence, since OB 1, the functions of Z XOB are\\nSin.\\nCos.\\nTan.\\nCot.\\nSec.\\nCsc.\\nFig. 1.\\nBD\\nOD\\nAE\\nCF\\nOE\\nOF\\nFig. 2.\\nBD\\n-OD\\n-AE\\n-CF\\n-OE\\nOF\\nFig. 3.\\n-BD\\n-OD\\nAE\\nCF\\n-OE\\n-OF\\nFig. 4.\\n-BD\\nOD\\n-AE\\n-CF\\nOE\\n-OF\\nThat is, iftJie radius of the circle is 1,\\nThe sine is the perpendicular drawn to XX from the\\nintersection of tire circumference with the terminal line.\\nThe cosine is the line drawn from the centre to the foot of\\nthe sine.\\nThe tangent is that portion of the geometrical tangent to\\nthe circle at the intersection of its circumference with OX\\nincluded between OX and the terminal line, produced if\\nnecessary.\\nThe cotangent is that portion of the geometrical tangent\\nto the circle at the intersection of its circumference with\\nOY included between OY and the terminal line, produced\\nif necessary.\\nThe secant is that portion of the terminal line, or terminal\\nline produced, included between the centre and the tangent.\\nThe cosecant is that portion of the terminal line, or termi-\\nnal line produced, included between the centre and the\\ncotangent.\\nAnd with regard to algebraic signs.\\nSines and tangents measured above XX are positive, and\\nbelow, negative cosines and cotangents measured to the right\\nof YY are positive, and to the left, negative; secants and\\ncosecants measured on the terminal line itself are positive,\\nand on the terminal \\\\mQ produced through 0, negative.\\nThe above are called the line values of the trigonometric\\nfunctions.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0056.jp2"}, "57": {"fulltext": "Miscellaneous Theorems.\\n47\\nThey simply represent the values of the functions when\\nthe radius is 1 that is, the mimerical value of the sine of an\\nangle is the same as the number which expresses the length\\nof the perpendicular drawn to XX from the intersection of\\nthe circumference and terminal line.\\n59. To trace the changes in the sine, cosine, tangent, cotan-\\ngent, secant, and cosecant of an angle as the angle increases\\nfrom 0\u00c2\u00b0 to 360\\n^4 -F3 c| Foj\\nLet AB^ be a circle whose radius is 1.\\nLet the terminal line start from the position OA, and re-\\nvolve about point as a pivot towards the position 0(7.\\nThen since the sine of the angle commences with the\\nvalue 0, and assumes in succession the values B^D^, B2D2,\\nOC, B^Dg, BJ)^, etc. 58), it is evident that, as the angle\\nincreases from 0\u00c2\u00b0 to 90\u00c2\u00b0, the sine increases from to 1\\nfrom 90\u00c2\u00b0 to 180\u00c2\u00b0, it decreases from 1 to from 180\u00c2\u00b0 to 270\u00c2\u00b0,\\nit decreases (algebraically) from to 1 and from 270\u00c2\u00b0 to\\n360\u00c2\u00b0, it increases from 1 to 0.\\nSince the cosine commences with the value OA, and\\nassumes in succession the values OD^, OD^, 0, OD^, OD^,\\netc., from 0\u00c2\u00b0 to 90\u00c2\u00b0, it decreases from 1 to from 90\u00c2\u00b0 to\\n180\u00c2\u00b0, it decreases from to 1 from 180\u00c2\u00b0 to 270\u00c2\u00b0, it in-\\ncreases from 1 to and from 270\u00c2\u00b0 to 360\u00c2\u00b0, it increases\\nfrom to 1.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0057.jp2"}, "58": {"fulltext": "48 Plane Trigonometry.\\nSince the tangent commences with the value 0, and as-\\nsumes in succession the values AEi, AE^, 00 (see Note to\\n25), -AE^, -AE^, etc., from 0\u00c2\u00b0 to 90\u00c2\u00b0, it increases from\\nto 00 from 90\u00c2\u00b0 to 180\u00c2\u00b0, it increases from 00 to from\\n180\u00c2\u00b0 to 270\u00c2\u00b0, it increases from to 00 and from 270\u00c2\u00b0 to\\n360\u00c2\u00b0, it increases from 00 to 0.\\nSince the cotangent commences at 00, and assumes in suc-\\ncession the values CF^, CF^, 0, GF^, CF^, etc., from 0\u00c2\u00b0\\nto 90\u00c2\u00b0, it decreases from 00 to from 90\u00c2\u00b0 to 180\u00c2\u00b0, it de-\\ncreases from to 00 from 180\u00c2\u00b0 to 270\u00c2\u00b0, it decreases from\\n00 to and from 270\u00c2\u00b0 to 360\u00c2\u00b0, it decreases from to 00.\\nSince the secant commences with the value OA, and as-\\nsumes in succession the values OEi, OE2, 00, \u00e2\u0080\u0094OE^, \u00e2\u0080\u0094OE^,\\netc., from 0\u00c2\u00b0 to 90\u00c2\u00b0, it increases from 1 to 00 from 90\u00c2\u00b0 to\\n180\u00c2\u00b0, it increases from 00 to 1 from 180\u00c2\u00b0 to 270\u00c2\u00b0, it\\ndecreases from 1 to 00 and from 270\u00c2\u00b0 to 360\u00c2\u00b0, it de-\\ncreases from 00 to 1.\\nSince the cosecant commences at 00, and assumes in suc-\\ncession the values OF^, OF^, OC, OF^, OF^, etc., from 0\u00c2\u00b0 to\\n90\u00c2\u00b0, it decreases from 00 to 1 from 90\u00c2\u00b0 to 180\u00c2\u00b0, it increases\\nfrom 1 to 00 from 180\u00c2\u00b0 to 270\u00c2\u00b0, it increases from 00 to\\n1 and from 270\u00c2\u00b0 to 360\u00c2\u00b0, it decreases from 1 to 00.\\n__ vr 1 p sm i^ tans?\\n60. Limiting Values of and\\nr^ ^7/. sin X tan x\\nTo find the limiting values of the fractions ana\\nwhen X is indefinitely decreased.\\nNote. We suppose x to be expressed in circular measure 50).\\nP", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0058.jp2"}, "59": {"fulltext": "Miscellaneous Theorems. 49\\nLet OPXP^ be a sector of a circle Z POP being 180\u00c2\u00b0.\\nDraw lines PT and PT tangent to the arc at P and P\\\\\\nrespectively; also, lines OT and PP^ intersecting at M.\\nBy Geometry, PT P T.\\nThen, OT bisects PP at right angles, and also bisects\\narc PP^ at X.\\nLet Z XOP Z XOP X.\\nBy Geometry, arc PP^ chord PP and PTP.\\nWhence, arc PX PJf, and PT\\nSiTcPX^PM ^PT\\nTherefore, ___ _, and\\nOr by 50, circ. meas. x sin x, and tan x.\\nEepresenting the circular measure of ic by cc simply, and\\ndividing through by sin a?, we have\\n1, and or -i- 36).\\nsm X sm X cos x\\n-rxn smx\\nWhence, 1, and cosic.\\nX\\nBut when x is indefinitely decreased, cos x approaches the\\nlimit 1 22).\\nHence, approaches the limit 1 when x is indefinitely\\nX\\ndecreased.\\ntana? sin a? since 1\\nAgain, X\\nX a? cos a? x cosic\\nsm X 1\\nBut and approach the limit 1 when x is indefi-\\nX cos a?\\nnitely decreased.\\nHence, approaches the limit 1 when x is indefinitely\\nX\\ndecreased.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0059.jp2"}, "60": {"fulltext": "50 Plane Trigonometry.\\nV. LOGARITHMS.\\n61. Every positive number may be expressed, exactly or\\napproximately, as a power of 10.\\nThus, 100 102; 13 10^-1139 e|-(,_\\nWhen thus expressed, the corresponding exponent is called\\nits Logarithm to the Base 10.\\nThus, 2 is the logarithm of 100 to the base 10 5 a relation\\nwhich is written logio 100 2, or simply log 100 2.\\n62. Logarithms of numbers to the base 10 are called\\nCommon Logarithms, and, collectively, form the Common\\nSystem.\\nThey are the only ones used for numerical computations.\\nAny positive number, except unity, may be taken as the\\nbase of a system of logarithms thus, if w m, where a\\nand m are positive numbers, then x log\u00e2\u0080\u009e m.\\nNote. A negative number is not considered as having a logarithm.\\n63. We have\\nby Algebra,\\n10^ 1,\\nio- i .i,\\n10^ 10,\\nio- ii-.= -oi.\\n10^ 100,\\n10-^ 4= -001, etc.\\n10^\\nWhence by the definition of 61,\\nlog 1 0, log .1 1 9 10,\\nlog 10 1, log .01 2 8 10,\\nlog 100 2, log .001 3 7 10, etc.\\nNote. The second form for log. 1, log .01, etc., is preferable in\\npractice. If no base is expressed, the base 10 is understood.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0060.jp2"}, "61": {"fulltext": "Logarithms. 51\\n64. It is evident from 63 that the logarithm of a num-\\nber greater than 1 is positive, and the logarithm of a num-\\nber between and 1 negative.\\n65. If a number is not an exact power of 10, its common\\nlogarithm can only be expressed approximately.\\nThe integral part of the logarithm is called the character-\\nistic, and the decimal part the mantissa.\\nFor example, log 13 1.1139.\\nHere, the characteristic is 1, and the mantissa .1139.\\nFor reasons which will appear hereafter, only the man-\\ntissa of the logarithm is given in a table of logarithms of\\nnumbers the characteristic must be found by aid of the\\nrules of Q)^ and 67.\\n66. It is evident from 63 that the logarithm of a num-\\nber between\\n1 and 10 is a decimal\\n10 and 100 is 1 a decimal\\n100 and 1000 is 2 a decimal etc.\\nTherefore, the characteristic of the logarithm of a number\\nwith one place to the left of the decimal point, is with\\ntwo places to the left of the decimal point, is 1 with three\\nplaces to the left of the decimal point, is 2 etc.\\nHence, the characteristic of the logarithm of a number\\ngreater than 1 is 1 less than the number of places to the left of\\nthe decimal point.\\nFor example, the characteristic of log 906328.5 is 5.\\n67. In like manner, the logarithm of a number between\\n1 and .1 is 9 a decimal 10\\n.1 and .01 is 8 H- a decimal 10\\n.01 and .001 is 7 a decimal 10 5 etc.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0061.jp2"}, "62": {"fulltext": "52 Plane Trigonometry.\\nTherefore, the characteristic of the logarithm of a decimal\\nwith no ciphers between the decimal point and first signifi-\\ncant figure, is 9, with 10 after the mantissa of a decimal\\nwith one cipher between the point and first significant figure\\nis 8, with 10 after the mantissa of a decimal with two\\nciphers between the point and first significant figure is 7,\\nwith 10 after the mantissa etc.\\nHence, to find the characteristic of the logarithm of a num-\\nber between a7id 1, subtract the number of ciphers between\\nthe decimal point and first significant figure from 9, writing\\n10 after the mantissa.\\nFor example, the characteristic of log .007023 is 7, with\\n10 written after the mantissa.\\nNote 1. It is customary in practice to omit the 10 after the man-\\ntissa of a negative logarithm but it should he allowed for in the result.\\nBeginners should always write it.\\nNote 2. Some writers combine the two portions of the character-\\nistic, and write the result as a negative characteristic before the\\nmantissa.\\nThus, instead of 7.6036 10, the student will frequently find 3.6036,\\na minus sign being written over the characteristic to denote that it\\nalone is negative, the mantissa being always positive.\\nPROPERTIES OF LOGARITHMS.\\n68. In any system, the logarithm of 1 is 0.\\nFor by Algebra, a* 1 whence by ^2, log\u00e2\u0080\u009e 1 0.\\n69. /n any system, the logarithm of the base is 1.\\nFor a} a whence, log\u00e2\u0080\u009e a 1.\\n70. In any system whose base is greater than 1, the logor\\nrithm ofOis oo.\\nFor if a is greater than 1, a 0.\\nWhence by 62, log\u00e2\u0080\u009e oo.\\nNote. No literal meaning can be attached to such a result as\\nlOga0=-0O.", "height": "3705", "width": "2370", "jp2-path": "completetrigonom00well_0062.jp2"}, "63": {"fulltext": "Logarithms. 53\\nIt must be interpreted as follows\\nIf, in any system whose base is greater than unity, a number ap-\\nproaches the limit 0, its logarithm is negative, and increases without\\nlimit in absolute value.\\n71. In any system, the logarithm of a product is equal to\\nthe sum of the logarithms of its factors.\\nAssume the equations\\nI whence by 62, ]\u00c2\u00b0S\\na^ n (y log^ n.\\nMultiplying the assumed equations,\\na X a^ mn, or a^+^ mn.\\nWhence, log\u00c2\u00ab mn a? 2/ log^ m log^ n.\\nIn like manner, the theorem may be proved for the prod-\\nuct of three or more factors.\\n72. By aid of 71, the logarithm of a composite number\\nmay be found when the logarithms of its factors are known.\\n1. Given log2 .3010 and log3 .4771 find log 72.\\nlog 72 log(2 X2x2x3x3)\\nlog2+ log2 log2 logs logs 71)\\n3 X log 2 2 X log 3 .90S0 .9542 1.8572.\\nEXAMPLES.\\nGiven log 2 .3010, log 3 .4771, log 5 .6990, and\\nlog 7 .8451, find:\\n2. log 35. 6. log 147. 10. log 288. 14. log 2205.\\n3. log 30. 7. log 225. 11. log 686. 15. log 7875.\\n4. log 98. 8. log 175. 12. log 504. 16. log 5832.\\n5. log 84. 9. log 420. 13. log 375. 17. log 14112.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0063.jp2"}, "64": {"fulltext": "54 Plane Trigonometry.\\n73. In any system, the logarithm of a fraction is equal to\\nthe logarithm of the num.erator minus the logarithm of the\\ndenominator.\\nAssume the equations\\n\u00e2\u0084\u00a2l; whence, I l\u00c2\u00b0g\u00c2\u00ab\\na^ n (y log\u00e2\u0080\u009e n.\\nDividing the assumed equations,\\na^ m m\\nor a*-^\\na^ n n\\nm\\nWhence, loga\u00e2\u0080\u0094 =x y log,^ m log^ n.\\n74. 1. Given log 2 .3010; find log 5.\\nlog 5 log log 10 log 2 73) 1 .3010 .6990.\\nEXAMPLES.\\nGiven log 2 .3010, log 3 .4771, and log 7 .8451, find\\n2. log^. 5. log33i. 8. log^. 11. log23f\\n3. log I. 6. logg. 9. log4f 12. log^.\\n4. log 45. 7. log 105. 10. log 525. 13. log96f.\\n75. In any system, the logarithm of any power of a quantity\\nis equal to the logarithm of the quantity multiplied by the ex-\\nponent of the power.\\nAssume the equation a m whence, x log\u00e2\u0080\u009e m.\\nRaising both members of the assumed equation to thepth\\npower,\\ngpx _ r^p wixence, log\u00e2\u0080\u009e?/i.^ px p log\u00e2\u0080\u009e m,", "height": "3703", "width": "2370", "jp2-path": "completetrigonom00well_0064.jp2"}, "65": {"fulltext": "Logarithms. 55\\n76. In any system, the logarithm of any root of a quantity\\nis equal to the logarithm of the quantity divided by the index\\nof the root.\\nr 1\\nFor, logaTs/m log\u00e2\u0080\u009e(\u00c2\u00bbr) log\u00e2\u0080\u009e m 75).\\n77. 1. Given log 2 .3010; find log 2^\\nlog 2 3 X log 2 X .3010 5017.\\n3 3\\nNote. To multiply a logarithm by a fraction, multiply first by\\nthe numerator, and divide the result by the denominator.\\n2. Given log 3 A771 find log ^3.\\nlog^ 1^ 1^111 .0596.\\n8 8\\nEXAMPLES.\\nGiven log 2 .3010, log 3 .4771, and log 7 .8451, find\\n3. log2^ 6. log42^ 9. log -^3. 12. log ^28.\\n4. log7l 7. logl5l 10. log ^7. 13. log ^324.\\n5. log 5k 8. log48t 11. log ^5. 14. log -^735.\\n15. Find log (2* x3*).\\nBy 71, log (23 X 3*)= log23 log 3* ilog2 f log3\\n.1003 .5964= .6967.\\nFind the values of the following\\n16. log^i. 18. log||. 20. log:|p. 23. log(Mj*.\\n5 5.\\n17. log5A/2. 19.log-. 21. log 23. log (23 X 2lK\\n5^ V7", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0065.jp2"}, "66": {"fulltext": "56 Plane Trigonometry.\\n78. To prove the relation\\nlog\u00e2\u0080\u009e m\\nlogj m\\nlog\u00e2\u0080\u009e6\\nAssume the equations\\nw m) {x log\u00e2\u0080\u009e m,\\nV whence,\\nb^ m) (y i-Ogj, m.\\nFrom the assumed equations,\\na\\nTaking the ytli root of both members,\\na^ b.\\nTherefore, log\u00e2\u0080\u009e 6 or y\\ny iog\\nThat is, logft m\\nlog\u00e2\u0080\u009e\\n79. jTo ^rove ^/ie relation\\nlogs a X log\u00e2\u0080\u009e h l.\\nPutting m a in the result of 78, we have\\nlogft a 69).\\nlog\u00e2\u0080\u009e6 log\u00e2\u0080\u009e6\\nWhence, logj a x log^ 6 1.\\n80. In the common system., the mantissce of the logarithms\\nof numbers having the same sequence of figures are equal.\\nSuppose, for example, that log 3.053 .4847.\\nThen, log 305.3 log (100 x 3.053) log 100 log 3.053\\n2 .4847 2.4847\\nlog .03053 log (.01 X 3.053) log .01 log 3.053\\n8 10 .4847 8.4847 10 etc.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0066.jp2"}, "67": {"fulltext": "Logarithms. 57\\nIt is evident from the above that, if a number be multi-\\nplied or divided by any integral power of 10, producing\\nanother number with the same sequence of figures, the man-\\ntissse of their logarithms will be equal.\\nThe reason will now be seen for the statement made in\\n65, that only the mantissas are given in a table of loga-\\nrithms of numbers.\\nFor, to find the logarithm of any number, we have only\\nto take from the table the mantissa corresponding to its\\nsequence of figures, and the characteristic may then be\\nprefixed in accordance with the rules of 66 or 67.\\nThus, if log 3.053 .4847, then\\nlog 30.53 1.4847, log .3053 9.4847 10,\\nlog 305.3 2.4847, log .03053 8.4847 10,\\nlog 3053. 3.4847, log .003053 7.4847 10, etc.\\nThis property is only enjoyed by the common system of\\nlogarithms, and constitutes its superiority over others for\\nthe purposes of numerical computation.\\n81. 1. Given log 2 .3010, log 3 .4771 find log .00432.\\nWe have log 432 log (2* x 3^) 4 log 2 3 log 3 2.6353.\\nThen by 80, the mantissa of the result is .6353.\\nWhence by 67, log .00432 7.6353 10.\\nEXAMPLES.\\nGiven log 2 .3010, log 3 .4771, and log 7 .8451, find\\n2. log 2.4. 6. log .00135. 10. Jog .1029.\\n3. log 16.8. 7. log 5880. 11. log 201.6.\\n4. log .81. 8. log .0245. 12. log ^7:5.\\n6. log .0192. 9. log .000486. 13. log (12.6) t-", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0067.jp2"}, "68": {"fulltext": "58 Plane Trigonometry.\\nUSE OF THE TABLE OF LOGARITHMS OF NUMBERS.\\n(For directions as to the use of the Table of Logarithms\\nof Numbers, see pages 1 to 4 of the Introduction to the\\nauthor s New Four Place Logarithmic Tables.)\\nEXAMPLES.\\n82. Find the logarithms of the following numbers\\n1.\\n80.\\n6.\\n.03294.\\n11.\\n.0007178.\\n2\\n6.3.\\n7.\\n.5205.\\n12.\\n5.1809.\\n3.\\n.298.\\n8.\\n20.08.\\n13.\\n1036.5.\\n4.\\n772.3.\\n9.\\n92461.\\n14.\\n.086676.\\n5.\\n1056.\\n10.\\n.0040322.\\n15.\\n.000011507.\\nFind the numbers corresponding to the following loga-\\nrithms\\n16. 1.8055. 21. 8.1646-10. 26. 1.6482.\\n17. 9.4487-10. 22. 7.5209-10. 27. 6.0450-10.\\n18. 0.2165. 23. 2.0095. 28. 4.8016.\\n19. 3.9487. 24. 0.9774. 29. 8.1142-10.\\n20. 2.7371. 25. 9.3178-10. 30. 5.7015-10.\\nAPPLICATIONS.\\n83. The approximate value of an arithmetical quantity,\\nin which the operations indicated involve only multiplica-\\ntion, division, involution, or evolution, may be conveniently\\nfound by logarithms.\\nThe utility of the process consists in the fact that addi-\\ntion takes the place of multiplication, subtraction of divi-\\nsion, multiplication of involution, and division of evolution.\\nNote. In computations with four-place logarithms, the results\\ncannot usually be depended upon to more than four significant figures.", "height": "3704", "width": "2370", "jp2-path": "completetrigonom00well_0068.jp2"}, "69": {"fulltext": "Logarithms. 59\\n84. 1. Find the value of .0631 x 7.208 x .51272.\\nBy 71, log (.0631 x 7.208 x .51272)\\nlog .0631 f log 7.208 log .51272.\\nlog .0631 8.8000-10\\nlog 7.208 0.8578\\nlog. 51272= 9.7099-10\\nAdding, log of result 19.3677 20\\n9.3677 10. (See Note 1.)\\nNumber corresponding to 9.3677 10 .2332.\\nNote 1. If the sum is a negative logarithm, it should be written\\nin such a form that the negative portion of the characteristic may be\\n-10.\\nThus, 19.3677 20 is written in the form 9.3677 10.\\n2. Find the value of\\n7984\\nBy 73, log f 1^ log 336.8 log 7984.\\nlog 336. 8 12. 5273 10 (See Note 2.\\nlog 7984 3.9022\\nSubtracting, log of result 8.6251 10\\nNumber corresponding .04218.\\nNote 2. To subtract a greater logarithm from a less, or to subtract\\na negative logarithm from a positive, increase the characteristic of the\\nminuend by 10, writing 10 after the mantissa to compensate.\\nThus, to subtract 3.9022 from 2.5273, write the minuend in the form\\n12.5273 10 subtracting 3.9022 from this, the result is 8.6251 10.\\n3. Find the value of (.07396)^.\\nBy 75, log (.07396)5 5 x log .07396.\\nlog\\n.07396\\n8.8690-\\n-10\\n5\\n44.3450\\n-50\\n4.3450-\\n10 (See\\nNote 1,\\nlog. 000002213.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0069.jp2"}, "70": {"fulltext": "6o Plane Trigonometry.\\n4. Find the value of V.035063.\\nBy 76, log V.035063 i log .035063.\\nlog .035063 8.5449 -10\\n20. 20 (See Note 3.)\\n3 )28.5449 30\\n9.5150 10 log .3274.\\nNote 3. To divide a negative logarithm, write it in such a form\\nthat the negative portion of the characteristic may be exactly divisible\\nby the divisor, with 10 as the quotient.\\nThus, to divide 8.5449 10 by 3, we write the logarithm in the\\nform 28.5449 30 dividing this by 3, the quotient is 9.5150 10.\\n85. Arithmetical Complement.\\nThe Arithmetical Complement of the logarithm of a num-\\nber, or, briefly, the Cologaritlim of the number, is the loga-\\nrithm of the reciprocal of that number.\\nThus, colog 409 log log 1 log 409.\\nlog 1 10. 10 (Note 2, 84.)\\nlog 409= 2.6117\\ncolog 409= 7.3883-10.\\nAgain, colog .067 log log 1 log .067.\\nlogl= 10. -10\\nlog .067 8.8261 10\\ncolog .067= 1.1739.\\nIt follows from the above that the cologarithm of a number\\nmay be found by subtracting its logarithm from 10 10.\\nNote. The cologarithm may be obtained by subtracting the last\\nsignificant figure of the logarithm from 10, and each of the others\\nfrom 9, 10 being written after the result in the case of a positive\\nlogarithm.", "height": "3679", "width": "2370", "jp2-path": "completetrigonom00well_0070.jp2"}, "71": {"fulltext": "Logarithms. 6i\\n86. Example. Find the value of F;^i7^*\\n8. 709 X. 0946 ^V 8.709 .0946/\\nlog 51384 log -i\u00e2\u0080\u0094 log\\n8.709 .0946\\nlog .51384 colog 8.709 colog .0946.\\nlog.51384 9.7109- 10\\ncolog 8.709 9.0601 -10\\ncolog .0946 1.0241\\n9.7951 10 log. 6239.\\nIt is evident from tlie above example that the logarithm\\nof a fraction either of whose terms is the product of factors,\\nmay be found by the following rule\\nAdd together the logarithms of the factors of the numerator,\\nand the cologarithms of the factors of the denominator.\\nEXAMPLES.\\nNote. A negative number has no common logarithm 62, Note).\\nIf such numbers occur in computation, they should be treated as if\\nthey were positive, and the sign of the result determined irrespective\\nof the logarithmic work.\\nThus, in Ex. 3, 87, the value of 439.2 x 7.1367) is obtained by\\nfinding the value of 439.2 x 7.1367, and putting a negative sign before\\nthe result. See also Ex. 33.\\n87. Find by logarithms the values of the following\\n1. 3.145 X .6839. 4. 9.0654) x .010785).\\n2. 847.6 X .02287. 5. .36552 x .025208.\\n3. 439.2 X 7.1367). 6. .0019036 x 57.143.\\n7 4^6.7 Q .2709 8062.4\\n76.52 .08683 9.5073*\\ng 1.0548 ^Q 6.802 .0001798\\n34.96 .0051264 .033166", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0071.jp2"}, "72": {"fulltext": "62\\nPlane Trigonometry.\\n13.\\n38.961 X .695\\n4994 X .0045\\n15.\\n.87028) X 37\\n(-.0659) X (-42.32)\\nj^ 715 X .02416) ^g .08214 x 73.4)\\n(-.516) X 142.07 .84 X 2808.7\\n17. (7.795)^ 22. (.095129)1\\n18. (.8328)^ 23. (.00010594)^\\n19. (-25.144)3. 24. V5.\\n20. (.01)^. 25. /2.\\n21. (-964.8)^. 26. V^^.\\n2^5\\n27. VIOO.\\n28. 1995.\\n29. V.072563.\\n30. V.0026139.\\n31. a/ -.00095174.\\n32. Find the value of\\n3*\\nBy 86,\\nlog\\nlog 2 log \\\\/5 colog 3 6\\nlog 2 i log 5 1 colog 3.\\nlog2 0.3010\\nlog 5 0.6990 divide by 3 0.2330\\ncolog 3 9.5229 10 multiply by f 9.6024 10\\n0.1364 log 1.369.\\n33. Find the value of\\n3 .032\\n7.962\\n-.03296\\n.962\\nlog\\n03296 J ^03296 ^gg^g _ ^^qq2\\\\\\n7.962 7.962\\nlog .03296 8.5180 -10\\nlog 7.962 =0.9010\\n3)27.6170 30\\n9.2057 -10 log. 1606.\\nKesult, .1606.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0072.jp2"}, "73": {"fulltext": "Logarithms.\\nFind the values of the following\\n34. 4 X 7\\\\\\n63\\n39. (-Mooy.\\n6928/\\n35. t\\n\\\\46\\n40.\\n41.\\n276.9\\n940\\n5^\\n44. -^Sx-^Sxa/T.\\n45.\\n76.1x.0593 \\\\f\\n1.307\\n37.\\n38.\\n(.001)^\\n^7\\nv:o8\\n(-10)*\\n42.\\n-.1\\n-J/iooo\\n46.^\\n47\\n75.44\\n31.4 X .415\\n-t/.0009657\\n43\\n-6?\\n6/3 5/7\\n\\\\5^V8-\\nV.0049784\\n4g^ -(.25693)-^\\n(-.8346)^\\n49. (25.467y\u00c2\u00abx(- .052)12. .7664 x 1.2809\\n50. \\\\/5106.5 X .00003109.\\n51. (837.5 X .0094325)1\\n52. (4.8672)^ x (.17544)^\\nV3:929 X ^65M\\n55.\\n(.00259)^\\n^05287\\n.374 X V.0078359\\n53.\\n^721.33\\n56.\\nV.04142 X .947^)\\n38.014\\n57. .083184 X (.2682)^ x (56.1)2.\\n58.\\n.0005616 X a/424.65\\n59.\\n60.\\n61.\\n(6.73)* X (.03194)^\\n485.7 X .0 7301)^ x ^/MS\\n(9.1273)\u00c2\u00ab X (.7095)*\\n.95048)^ X (8473)3\\n(-2080.9) x^:0572_\\n\u00e2\u0080\u00a2^-.003012x1.955\\n.843)\u00c2\u00ab X -n/17959 x 560.6)*\\n4", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0073.jp2"}, "74": {"fulltext": "64 Plane Trigonometry.\\nEXAMPLES IN THE USE OF TRIGONOMETRIC\\nTABLES.\\n(For directions, see pages 4 to 8 of the Introduction to the\\nauthor s New Four Place Logarithmic Tables.)\\n88. Tables of Logarithmic Sines, Cosines, etc.\\nFind the values of the following\\n1. log tan 35\u00c2\u00b0 39 6. log sin 30\u00c2\u00b0 37.2\\n2. log sin 61\u00c2\u00b0 58 7. log cos 55\u00c2\u00b0 21 48\\n3. log cot 12\u00c2\u00b0 34 8. log cot 48\u00c2\u00b0 3 43\\n4. log cos 26\u00c2\u00b0 56 9. log sec 80\u00c2\u00b0 7\\n5. log tan 82\u00c2\u00b0 3 10. log esc 65\u00c2\u00b0 12\\nFind the angles corresponding in the following\\n11. log tan 0.9164. 16. log cot 0.2154.\\n12. log cos 9.9221 10. 17. log sin 9.1891 10.\\n13. log sin 9.8619 10. 18. log tan 8.9668 10.\\n14. log cot 9.4700 10. 19. log esc 0.1888.\\n15. log cos 9.2204 10. 20. log sec 0.4032.\\nTables of Natural Sines, Cosines, etc.\\nFind the values of the following\\n21. sin 17\u00c2\u00b0 13 23. tan 35\u00c2\u00b0 7\\n22. cos 75\u00c2\u00b0 38 24. cot 68\u00c2\u00b0 46\\nFind the angles corresponding in the following\\n25. sin .7385. 27. tan 1.1897.\\n26. cos .9280. 28. cot 1.8207.", "height": "3704", "width": "2370", "jp2-path": "completetrigonom00well_0074.jp2"}, "75": {"fulltext": "Solution of Right Triangles.\\n6S\\nVI. SOLUTION OF RIGHT TRIANGLES.\\n89. The elements of a triangle are its three sides and its\\nthree angles.\\nWe know by Geometry that a triangle is, in general, com-\\npletely determined when three of its elements are known,\\nprovided one of them is a side.\\nThe solution of a triangle is the process of computing the\\nunknown from the given elements.\\n90. To solve a rigJit triangle, two elements must be given\\nin addition to the right angle, one of which must be a side.\\nThe various cases which can occur may all be solved by\\naid of the following formulae\\nsin^\\nsin B\\nCOS A\\ncos B\\ntan^\\ntan 5\\nWheyi the given elements are a side and an\\n91. Case I.\\nangle.\\nThe formula for computing either of the remaining sides\\nmay be found by the following rule\\nTake that function of the angle which involves the given side\\nand the required side.\\n1. Given c 23, B 21\u00c2\u00b0 33 Find a and h.\\nIn this case, the formulae to be used are\\ncos 5\\nand sm B\\nWhence,\\nc cos B, and h\\nc\\nG sin B.\\n(A)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0075.jp2"}, "76": {"fulltext": "66 Plane Trigonometry.\\nSolution by Natural Functions.\\na 23 X cos 21\u00c2\u00b0 33 23 X .9301 21.39.\\n5 23 X sin 21\u00c2\u00b0 33 23 x .3673 8.448.\\nSolution by Logarithms.\\nTaking the logarithms of both members, in formulae (A),\\nlog a log c log cos B, and log 6 log c log sin B.\\nlogc 1.3617 logc 1.3617\\nlog cos 5 9. 9685 10 log sin B 9. 5651 10\\nlog a 1.3302 log 6 0.9268\\na 21.39. 6 8.448.\\nNote. In examples under Case I. in which the given sides are\\nnumbers of not more than two significant ^figures, and the operations\\nindicated involve only multiplication, it is usually shorter to employ\\nNatural Functions.\\nIn such a case, the results cannot be depended upon to more than\\nfour significant figures.\\n2. Given a .2369, A 67\u00c2\u00b0 18 Find b and c.\\nIn this case, tan A and sin A\\nh G\\nWhence,\\ntan A sin A\\nBy logarithms,\\nlog h \\\\oga log tan A^ and log c log a log sin A.\\nlog a 9.3727 10 log a 9. 3727 10\\nlog tan 0.3785 log sin A 9. 9650 10\\nlog h 8.9942 10 log c 9.4077 10\\nb= .09868. c= .2557.\\n92. Case II. Wlieji both given elements are sides.\\nFirst calculate one of the angles by aid of either formula\\ninvolving the given elements, and then compute the remain-\\ning side by the rule of Case I.", "height": "3705", "width": "2370", "jp2-path": "completetrigonom00well_0076.jp2"}, "77": {"fulltext": "Solution of Right Triangles. 67\\nEx. Given h .1512, c .3081. Find A and a.\\nWe first find A by the formula cos J. and then a by the\\nc\\nformula sin or a c sin\\nc\\nBy logarithms,\\nlog cos J. log 5 log c, and log a log c log sin A.\\nlog h 9.1796 10 log c 9.4887 10\\nlogc= 9.4887 -10 logsin^ 9.9401 -10\\nlog cos A 9.6909 10 log a 9.4288 10\\n60\u00c2\u00b0 36. 4 a= .2684.\\n93. In the Trigonometric solution of an example under\\nCase II., it is necessary to first find one of the angles, and\\nthe remaining side may then be calculated.\\nBut it is possible to compute the third side directly, with-\\nout first finding the angle, by Geometry.\\nThus, in the example of 92, we have\\na2 52 ^2_\\nWhence, a Vc 6^ V(c 6) (c h).\\nBy logarithms, log a [log (c 6) log (c 6)].\\nc 6 .4593 log 9.6621 10\\nc 5 .1569 log 9.1956 10\\n2 18.8577 20\\nloga 9.4289 -10\\na= .2685.\\nIf the given sides are a and h, the expression for c is\\nVa^ which is not adapted to logarithmic computation.\\nIn such a case, it is usually shorter to proceed as in 92.\\nEXAMPLES.\\n94. Solve the following right triangles\\nH. Given A 15\u00c2\u00b0, c 7. 3. Given B 50\u00c2\u00b0, b 20.\\n2. Given B 68\u00c2\u00b0, a 5. 4. Given a .35, c .62.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0077.jp2"}, "78": {"fulltext": "68\\nPlane Trigonometry.\\n5. Given a 21, 5 42. 8. Given 6 586, c= 763.\\n6. Given A 38\u00c2\u00b0, a 8.09. 9. Given A 9\u00c2\u00b0, b= 937.\\n7. Given B 65% c .014. 10. Given a 3.41, b 2.87.\\n11. Given =3 31\u00c2\u00b0 50 a 48.04.\\n12. Given 46\u00c2\u00b0 15 c 5280.\\n13. Given 6 .0469, c .0515.\\n14. Given 5 79\u00c2\u00b0 28 6 842.\\n15. Given B 67\u00c2\u00b0 47 c .00954.\\n16. Given A 43\u00c2\u00b0 30 b 26185.\\n17. Given a 3402, 6 2317.\\n18. Given 5 82\u00c2\u00b0 6 a .08937.\\n19. Given 6 578.9, c 2492.\\n20. Given A 26\u00c2\u00b0 12 c .4694.\\n21. Given B 14\u00c2\u00b0 53 6 1353.\\n22. Given 5 43\u00c2\u00b0 24 a .89658.\\n23. Given a 99.46, c 156.8.\\n24. Given A 62\u00c2\u00b0 44 6 4.2492.\\n25. Given A 74\u00c2\u00b0 17 a .000020386.\\n26. Given B 29\u00c2\u00b0 56 c .00078144.\\n27. Given a 63827, c 92275.\\n28. Given A 58\u00c2\u00b0 39 c 35.733.\\n29. Given 5 35\u00c2\u00b0 8 6 17269.\\n30. Given ct .0067239, 6 .0038453.\\nSolve the following isosceles triangles, in which A and B\\nare the equal angles, and a, 6, and c the sides opposite\\nangles A, B, and C, respectively\\n31. Given A 71\u00c2\u00b0, 6 39.\\n32. Given B 36\u00c2\u00b0 40 c .4688.", "height": "3705", "width": "2370", "jp2-path": "completetrigonom00well_0078.jp2"}, "79": {"fulltext": "Solution of Right Triangles. 69\\n33. Given (7= 83\u00c2\u00b0 52 710.6.\\n34. Given a 6875, c 11318.\\n35. Given B 29\u00c2\u00b0 7 a 2.569.\\n36. Given A 54\u00c2\u00b0 39 c 1.7255.\\n37. Given C 135\u00c2\u00b0 26 c .06377.\\nMISCELLANEOUS PROBLEMS.\\n95. If ^C is the diagonal of rectangle ABCD, and the\\nside AB is horizontal and BC vertical,\\nZ BAC is called the cmgle of elevation\\nof point C from point A, and Z ^(7i\\nthe angle of depression of point A from\\npoint (7.\\n96. 1. From the top of a lighthouse, 150 feet above the\\nsea, the angles of depression of two boats, in line with the\\nlighthouse, are observed to be 12\u00c2\u00b0 and 30\u00c2\u00b0, respectively.\\nFind the distance between the boats.\\nLet A be the position of the first boat, A of the second, B the top\\nof the lighthouse, and C its foot.\\nThen,\\nAA AC- A C BC cot A- BC cot BA C\\n150 (cot 12\u00c2\u00b0 -cot 30\u00c2\u00b0)\\n150(4.7046-1.7321)\\n150 X 2.9725 445.9 ft.\\n2. From the top of a lighthouse, 250 feet above the sea,\\nthe angle of depression of a buoy is observed to be 31\u00c2\u00b0.\\nFind the horizontai distance of the buoy.\\n3. If the radius of a circle is 834, what is the length of a\\nchord which subtends an arc of 46\u00c2\u00b0\\n4. A regular hexagon is circumscribed about a circle\\nwhose diameter is 59. Find the length of its side.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0079.jp2"}, "80": {"fulltext": "yo Plane Trigonometry.\\n5. Find the angle of elevation of a road wMcli rises a\\ndistance of 349 feet in a horizontal distance of five-eighths\\nof a mile.\\n6. A regular polygon of nine sides is inscribed in a circle\\nwhose diameter is 6S. Find the length of its side.\\n7. How far from the top of a tower 121 feet in height\\nmnst an observer stand so that the angle of elevation of its\\ntop may be 23\u00c2\u00b0\\n8. A regular polygon, whose side is 7.6 and angle 144\u00c2\u00b0, is\\ncircumscribed about a circle. Find its radius.\\n9. Find the angle of elevation of the sun when a monu-\\nment whose height is 214.8 feet casts a shadow 167.4 feet\\nin length.\\n10. Find the length of the diagonal of a regular pentagon\\nwhose side is 9.437.\\n11. At a distance of 41.6 feet from the base of a tower,\\nthe angle of elevation of its top is observed to be 59\u00c2\u00b0 36\\nFind its height.\\n12. The middle point of a chord of a circle, 24 units in\\nlength, is distant 7 units from the middle point of its sub-\\ntended arc. How many degrees and minutes are there in\\nthe arc\\n13. If the diameter of a circle is 6374, find the angle at\\nthe centre subtended by an arc whose chord is 2138.\\n14. If the radius of a circle is 9.54, and the distance from\\nthe middle point of a chord to the middle point of its sub-\\ntended arc is 3.87, how many degrees and minutes are there\\nin the arc\\n15. From the top of a tower, the angle of depression of\\nthe extremity of a horizontal base line, 236.1 feet in length\\nmeasured from the foot of the tower, is observed to be\\n29\u00c2\u00b0 48 Find the height of the tower.", "height": "3705", "width": "2370", "jp2-path": "completetrigonom00well_0080.jp2"}, "81": {"fulltext": "Solution of Right Triangles. 71\\n16. A chord of a circle, whose length is 14.95, subtends\\nan arc of 135\u00c2\u00b0 52 What is the distance from the middle\\npoint of the chord to the middle point of the arc\\n17. If a vertical pole casts a shadow which is three-fourths\\nits own length, what is the angle of elevation of the sun\\n18. The radius of the inscribed circle of an equilateral\\ntriangle is .307. Eind its perimeter, and the diameter of\\nthe circumscribed circle.\\n19. The side of a regular octagon is 23.68. Find the\\nradii of its inscribed and circumscribed circles.\\n20. A chord of a circle subtends an arc of 70\u00c2\u00b0 24 If the\\nlength of the chord is 853.4, find the radius of the circle.\\n21. At a point 250 feet from the foot of a cliff surmounted\\nby a lighthouse, the angle of elevation of the top of the\\nlighthouse is 50\u00c2\u00b0, and of its foot 30\u00c2\u00b0. Find the height of\\nthe cliff, and of the lighthouse.\\n22. From the top of a cliff 378 feet above the sea, the\\nangles of depression of two boats, in line with the observer,\\nare observed to be 11\u00c2\u00b0 50 and 29\u00c2\u00b0 20 respectively. Find\\nthe distance between the boats.\\n23. At a distance of 169 feet from the foot of a tower\\nsurmounted by a pole, the tower subtends an angle of 35\u00c2\u00b0,\\nand the pole an angle of 12\u00c2\u00b0. Find the length of the pole.\\n24. How many degrees and minutes are there in the arc\\nincluded between two parallel chords, on the same side of\\nthe centre of a circle, whose distances from the centre are 5\\nand 7, respectively, the radius of the circle being 11\\n25. From the top of a tower, the angle of depression of a\\nstake is 31\u00c2\u00b0 29 What will be the angle of depression of\\nthe stake from a point half way to the top\\n26. The diagonal of a regular pentagon is 43.92. Find\\nthe radius of its inscribed circle.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0081.jp2"}, "82": {"fulltext": "72 Plane Trigonometry,\\n27. A railway runs from AtoB,Si horizontal distance of\\n1250 feet, at an angle of elevation of 8\u00c2\u00b0 12 and then from\\njB to C, a horizontal distance of 375 feet, at an angle of elevar\\ntion of 7\u00c2\u00b0 26 How many feet is C above the plane oi A?\\n28. From a point 200 feet from the foot of a tower sur-\\nmounted by a pole, the angle of elevation of the top of the\\npole is 38\u00c2\u00b0; from a point 150 feet further, the angle of\\nelevation of the foot of the pole is 22\u00c2\u00b0. Find the height\\nof the pole.\\n29. If the radius of the earth is 3956 miles, find the\\nradius in miles of the arctic circle, latitude 66\u00c2\u00b0 32 IST.\\n30. If the diameter of the earth is 7912 miles, what is\\nthe distance of the remotest point of the surface visible\\nfrom the top of a mountain, l-i- miles above the sea\\n31. A flagpole 23 feet long surmounts a tower whose\\nheight is 98 feet. What angle does the flagpole subtend at\\na point on the ground 315 feet from the base of the tower\\n32. An observer notes that a spire bears due north from\\nhim, the angle of elevation of its top being 22\u00c2\u00b0 17 On\\ngoing due east 550 feet, the spire bears 49\u00c2\u00b0 west of north.\\nWhat is the height of the spire\\n33. A regular pyramid stands on a square base, whose\\nside is 50 feet. Each side of the base makes an angle of\\n69\u00c2\u00b0 with the lateral edge. Find the altitude of the pyramid.\\n34. A vessel is sailing due north at a uniform rate of\\nspeed. At 7.30 a.m., a lighthouse is observed to bear 70\u00c2\u00b0\\nwest of north at 8 a.m. it is due west, at a distance of 12\\nmiles. Find the distance and bearing of the lighthouse\\nat 9.30 A.M.\\nFORMULA FOR THE AREA OF A RIGHT TRIANGLE.\\n97. Case I. Given the liyiootenuse and an acute angle.", "height": "3705", "width": "2370", "jp2-path": "completetrigonom00well_0082.jp2"}, "83": {"fulltext": "Solution of Right TriangleSo 73\\nB\\nb\\nDenoting tlie area by K, we have by Geometry,\\n2K=zab.\\nBut by 5, a c sin A, and b c cos A.\\nWhence, 2 K=c^ sin Aco^A \\\\c^ sin 2 A, by (22).\\nThen, 4 /iT c^ sin 2 (32)\\nIn like manner, 4 /r= c^ sin 2 jB. (33)\\nCase II. Given an angle and its opposite side.\\nBy 2, b aGotA.\\nWhence, 2 K=a x a cot A a^ cot J.. (34)\\nIn like manner, 2 K=b^ cot B. (35)\\nCase III. Given an angle and its adjacent side.\\nBy 2, b a tan B.\\nWhence, 2 K=a x a tan 5 a^ tan 5. (36)\\nIn like manner, 2 K=b^ tan A. (37)\\nCase IV. (Twe?i the hypotenuse and another side.\\nSince a^ 5^ c^, we have\\n2 K= ab a^(? a^ a^(c a) (c a). (38)\\nIn like manner, 2K= b-y/ic -\\\\-b){G b). (39)\\nCase V. Given the two sides about the right angle.\\nIn this case, 2 K= ab. (40)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0083.jp2"}, "84": {"fulltext": "74 Plane Trigonometry.\\nEXAMPLES.\\n98. 1. Given c 10.36, B T5\u00c2\u00b0; find the area.\\nBy (33), 4ir=c2sin2^.\\nWhence, log (4 K) 2 log c log sin 2 B.\\nlog c 1.0153 multiply by 2 2.0306\\n2^=150\u00c2\u00b0; log sin 9.6990 -10\\nlog (4^) =1.7296\\n4ir= 53.65, and ^=13.41.\\nNote. To find log sin 150\u00c2\u00b0, take either log cos 60\u00c2\u00b0 or log sin 30\u00c2\u00b0.\\n(See page 7 of the Introduction to the author s New Four Place\\nLogarithmic Tables.)\\nFind the areas of the following right triangles\\n2. Given A 46\u00c2\u00b0, a 2.717.^\\n3. Given B 35\u00c2\u00b0 16 a .557.\\n4. Given a 283.17, b 94.93.\\n5. Given b 4.564, c 7.176.\\n6. Given A 53\u00c2\u00b0 9 c 13.84.\\n7. Given A 20\u00c2\u00b0 57 b .05027.\\n8. Given a .0861, c .4806.\\n9. Given B 67\u00c2\u00b0 48 c 67.409.\\n10. Given B 75\u00c2\u00b0 34 b .0032056.\\n11. Given 81\u00c2\u00b0 23 c 195.84.", "height": "3709", "width": "2468", "jp2-path": "completetrigonom00well_0084.jp2"}, "85": {"fulltext": "General Properties of Triangles.\\nIS\\nVII. GENERAL PROPERTIES OF\\nTRIANGLES.\\n99. In any triangle, the sides are proportional to the sines\\nof their opposite angles.\\nI. To prove a 6 sin sin B. (41)\\nThere will be two cases, according as angles A and B are\\nboth, acute (Fig. 1), or one of them obtuse (Fig. 2).\\nIn each case, draw line CD perpendicular to AB.\\nThen in each figure,\\nAlso in Fig. 1,\\nAnd in Fig. 2,\\nCD b sin A 5).\\nCD a sin B.\\nCD a sin CBD\\na sin (180\u00c2\u00b0 -B) a sin B\\n32).\\nThen in either case, b sin a sin B.\\nWhence by the theory of proportion,\\na 6 sin sin B.\\nIn like manner, b c sin B sin C,\\nand c a sin (7 sin A.\\n(42)\\n(43)\\n100. In any triangle, the sum of any tico sides is to their\\ndifference as the tangent of half the sum of the opposite angles\\nis to the tangent of half their difference.\\nBy (41). a 6 sin J. sin B.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0085.jp2"}, "86": {"fulltext": "76 Plane Trigonometry.\\nWhence by composition and division,\\na-{-b a b sin^H-sin_B sin^\u00e2\u0080\u0094 sinjB.\\na -\\\\-b sin A sin B\\nOr,\\na b sin A sin B\\nsin sin _ tani(^ (21).\\nsin^-sinjB tan-i-(^-5)\\nWhence,\\na b t2,i\\\\\\\\ {A B)\\na b tan ^{A B)\\nT n b^C tani(5+C)\\nIn like manner, 2_a z\\nb-c tan i (J3 G)\\nand\\nc4- a _ tani((7 J.)\\nc a tan-i-((7\\n(44)\\n(45)\\n(46)\\n101. In any triangle, the square of any side is equal to the\\nsum of the squares of the other two sides, minus twice their\\nproduct into the cosine of their included angle.\\nI. To x^rove a^ W c^ 2bc cos A.\\nCase I. When the included angle A is acute.\\nC\\n(47)\\nThere will be two cases, according as angle B is acute\\n(Fig. 1), or obtuse (Fig. 2).\\nIn each case, draw line CD perpendicular to AB.\\nIn Fig. 1, BD c- AD, and in Fig. 2, BD AD c.\\nSquaring, we have in either case,\\nBD^ AD^ c --2cx AD.", "height": "3681", "width": "2449", "jp2-path": "completetrigonom00well_0086.jp2"}, "87": {"fulltext": "General Properties of Triangles.\\nAdding CD to both members,\\nBlf -hCI? ad +CD -\\\\-c -2cx ad.\\nBut, Bff +Clf o?, and Aff Ud h\\\\\\nAlso, by 5, AD h cos A.\\nWhence, o? h^ c^ 2bc cos A.\\nCase II. When the included angle A is obtuse.\\nQ\\n77\\nDraw line CD perpendicular to AB.\\nWe have BD AD c.\\nSquaring, and adding CD to both members,\\nBD -hCD ad +CD^ c^ 2cxAD.\\nBut, BD -{-CD a and AD CD h\\\\\\nAnd by 5,\\nAD h cos CAD h cos (180\u00c2\u00b0 6 cos 32).\\nWhence, a^ b^ c^ 2 be cos A.\\nIn like manner, b^ c^ a^ 2ca cos B, (48)\\nand c^ a^ -\\\\-b^ 2 ab cos C (49)\\n102. To ex^jress the cosines of the angles of a triangle in\\nterms of the sides of the triangle.\\nBy (47), a^==W c -2bc cos A.\\nTransposing, 2 be cos A V^ c^ a^.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0087.jp2"}, "88": {"fulltext": "78 Plane Trigonometry.\\nWhence, cos A /gg)\\n2 be\\nIn like manner, cos B ^^^I (51)\\n2ca\\nand cos C +J ^l (52)\\n2ab\\n103. jPo express the sines, cosines, and tangents of the half\\nangles of a triangle in terms of the sides of the triangle.\\nBy (50),\\n2 5c 2 he\\nWhence by (28),\\no 1 A a^ (b cY\\n2 sin^ ^A=\\n2bc\\nOr, sinH^-\\nAbe\\nDenoting the sum of the sides, a 6 c, by 2 s, we have\\na b c (a b c) 2b 2 s 2b 2(s b),\\nand a 6 c (a 5 c) 2c 2s 2c 2(s c).\\nWhence, sin^ i 4(.-6)(.-c)\\nOr,\\nsinl^=.J5^ME5. (53)\\nIn like manner, sin i B =\\\\h^ (54)\\nca\\nsmiC=^J (55)\\nand _\\nab", "height": "3681", "width": "2448", "jp2-path": "completetrigonom00well_0088.jp2"}, "89": {"fulltext": "General Properties of Triangles. 79\\nAgain, by (50),\\n1 cos 1 H\\n2 be 2 be\\nWhence by (29),\\n2cosH^\\n(P cf 0}\\n9\\nbe\\nOr, cos^^-\\nBut, 6 c a 2s,\\nand b e a (b G a) 2 a 2 {s a).\\nWhence,\\n30S- 1 A\\n_46\\n4. be\\nCOS i J.\\ns(s\\nbe\\na).\\nOr, cos i A A/H^^^. (56)\\nIn like manner, cos -J--B h (57)\\nea\\nand cos i (7 ^^ii\u00e2\u0080\u0094 (58)\\nDividing (53) by (56), we have.\\nsin^A l (s-b)(s-e) I be\\ncos i J. be ^s{s a)\\nWhence by (4), tan i J ^)(s-e)\\ns is Ct)\\nIn like manner, tani^^^ r (60)\\n2 s(s_5)\\nand UniC=J (61)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0089.jp2"}, "90": {"fulltext": "8o\\nPlane Trigonometry.\\nNote. Since each angle of a triangle is less than 180\u00c2\u00b0, its half is\\nless than 90\u00c2\u00b0 hence, the positive sign must be taken before the radi-\\ncal in each formula of 103.\\nFORMULA FOR THE AREA OF AN OBLIQUE TRIANGLE.\\n104. Case I. Given two sides and their included angle.\\nI. When the given parts are h, c, and A.\\nA\\nThere will be two cases, according as A is acute (Fig. 1),\\nor obtuse (Fig. 2).\\nIn each case, draw line CD perpendicular to AB.\\nThen denoting the area by K, we have by Geometry,\\n2K=cxCD.\\nBut in Fig. 1, CD b sin A 5).\\nAnd in Fig. 2, CD b sin CAD\\nh sin (180\u00c2\u00b0 6 sin 32).\\nThen in either case,\\n2 K= hc^xuA.\\nIn like manner, 2 K= ca sin B,\\nand 2 K= ah sin C.\\nCase II. Given a side and all the angles.\\nI. When the given parts are a. A, B, and G.\\nBy (64), 2K=absmC\\n(62)\\n(63)\\n(64)", "height": "3705", "width": "2457", "jp2-path": "completetrigonom00well_0090.jp2"}, "91": {"fulltext": "General Properties of Triangles. 8i\\n-D i. u /y,-,\\\\ b sin B J a sin B\\nBut by (41), or b=\u00e2\u0080\u0094.\\na smA sin A\\nWhence, 2K=ax ^L$^ x sin G\\nsmA\\na^ sin B sin O\\nsin^\\n(65)\\nT Ti o T^ b^ sin C sin A\\nIn like manner, 2K= (66)\\nsm^\\nand ^j^^l^nA^^\\nsm C\\nCase III. Given the three\\nBy (62), 2 6c sin 2 6c sin i cos i A, by (22).\\nDividing by 2, and substituting the values of sin i A and\\ncos I -4 from (53) and (56), we have.\\nhe, he\\nVs(s-a)(s-b)(s-c). (68)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0091.jp2"}, "92": {"fulltext": "82 Plane Trigonometry.\\nVIII. SOLUTION OF OBLIQUE TRIANGLES\\nIn the solution of oblique triangles^ we may distinguish\\nfour cases.\\n105. Case I. Given a side and any two angles.\\nThe third angle may be found by Geometry, and then by\\naid of 99 the remaining sides may be calculated.\\nThe triangle is possible for any values of the given ele-\\nments, provided the sum of the given angles is 180\u00c2\u00b0.\\n1. Given 20, A 104\u00c2\u00b0, 19\u00c2\u00b0 find C, a, and c\u00c2\u00bb\\nWe have C 180\u00c2\u00b0 -(A-{-B)= 180\u00c2\u00b0 123\u00c2\u00b0 57\u00c2\u00b0.\\nBy 99,\\na _ sin\\nh sin B\\nand\\n_ sin C\\nsin^\\nThen,\\na sin\\nA CSC B,\\nand c 6 sin esc B.\\nWhence,\\nlog a log 6\\nlog sin\\nA log CSC B,\\nd\\nlog c log 6\\nlog sin\\nC log CSC B.\\nlog 6:\\n1.3010\\nlog 1.3010\\nlog sin A\\n9.9869 10\\nlog sin a =9.9236\\n-10\\nlog CSC B\\n0.4874\\nlog CSC 0.4874\\nloga\\n:z 1.7753\\nlogc 1.7120\\na\\nz 59.61.\\nc 51.52.\\nNote. To find the log cosecant of an angle, subtract the log sine\\nfrom 10 10. To find log sin 104\u00c2\u00b0, take either log cos 14\u00c2\u00b0 or log sin\\n76\u00c2\u00b0. (See page 7 of the author s New Four Place Logarithmic\\nTables.)\\nEXAMPLES.\\nSolve the following triangles\\n2. Given a 180, A 38\u00c2\u00b0, B 75\u00c2\u00b0 20\\n3. Given b 8.19, B 52\u00c2\u00b0, C 109\u00c2\u00b0.\\n4. Given c .0246, 83\u00c2\u00b0 30 5 38\u00c2\u00b0 50", "height": "3681", "width": "2472", "jp2-path": "completetrigonom00well_0092.jp2"}, "93": {"fulltext": "Solution of Oblique Triangles. 83\\n5. Given 6 67.13, A 26\u00c2\u00b0 18 C 44\u00c2\u00b0 35\\n6. Given c .45924, ^=74\u00c2\u00b0 43 C=61\u00c2\u00b029\\n7. Given a 3024, B 133\u00c2\u00b0 34 C 22\u00c2\u00b0 57\\n(For additional examples under Case I, see 112.)\\n106. Case II. Given two sides and their included angle.\\nSince one angle is known, the sum of the remaining angles\\nmay be found, and then their difference may be calculated\\nby aid of 100.\\nKnowing the sum and difference of the angles, the angles\\nthemselves may be found, and then the remaining side may\\nbe computed as in Case I.\\nThe triangle is possible for any values of the data.\\n1. Given a S2, c 167, 5 98\u00c2\u00b0; find A, C, and b.\\nBy Geometry, 1 80\u00c2\u00b0 -5 82\u00c2\u00b0.\\nBy 100,\\nc a _ tan-i (C+\\nc a tan (^C A)\\nOr, tani((7-^):=^^^tani((7\\nThen,\\nlog tan C J.) log (c a) colog (c a) log tan 1 A).\\nc-a Sb. log =1.9294\\nc a 249. colog 7.6038 10\\n1(0 41\u00c2\u00b0. log tan 9.9392 10\\nlog tan ^(C-A)= 9.4724 10\\ni(C-^)=16\u00c2\u00b031.7\\nThen, (7 K -4)+ K^- 67\u00c2\u00b031.7\\nand A l(C A)- ^(C- A) 24:\u00c2\u00b0 28.SL\\nTo find the remaining side, we have by 99,\\nI a sin B n a\\nb a sm J5 esc A\\nsinA", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0093.jp2"}, "94": {"fulltext": "84 Plane Trigonometry.\\nWhence, log 6 log a log sin B log esc A.\\nloga 1.9138\\nlogsin5 9.9958 -10\\nlog CSC 0.3828\\nlog 2.2924\\n6 196.1.\\nEXAMPLES.\\nSolve the following triangles\\n2. Given a 67, c 33, B 36\u00c2\u00b0.\\n3. Given a 986, 6 544, 0=134\u00c2\u00b0.\\n4. Given b .149, c .427, A 71\u00c2\u00b0.\\n5. Given a 3.95, b 6.64, 68\u00c2\u00b0 30\\n6. Given a 2937, c 6185, B 55\u00c2\u00b0 46\\n7. Given 6 .01292, c .00286, 26\u00c2\u00b0 32\\n(For additional examples under Case II., see 112.)\\n107. Case III. Given the three sides.\\nThe angles might be calculated by the formulae of 102\\nbut as these are not adapted to logarithmic computation, it\\nis usually more convenient to use the formulae of 103.\\nEach angle should be computed trigonometrically for we\\nthen have a check on the work, since their sum should be\\n180\u00c2\u00b0.\\nIf all the angles are to be computed, the tangent formulae\\nare the most convenient, since only four different numbers\\noccur in the second members.\\nIf but one angle is required, the cosine formula involves\\nthe least work.\\nThe triangle is possible for any values of the data, pro-\\nvided no side is greater than the sum of the other two.", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0094.jp2"}, "95": {"fulltext": "Solution of Oblique Triangles. 85\\nIf all the angles are required, and the tangent formulae\\nare used, it is convenient to modify them as follows by (59),\\ntan 1^ V(--^ ^r t J-\\nS (S Clj S Ct s\\nDenoting by r, we have\\ntani J[\\na\\nIn like manner, tan \\\\B and tan 1 G\\ns c\\n1. Given a 2.5, b 2.S, 2.2-, find A, B, and O.\\nHere, 2 s a b c 7.5, and s 3.75.\\nThen, s a 1.25, s b .95, s c 1.55.\\nBy logarithms,\\nlog r 1 [log (s a) log (s 5) log (s c) colog s].\\nAlso, log tan 1 log r log (s a),\\nlog tan ^B logr log (s 6)\\nlog tan 1 C log r log (s c).\\nlog (s a) 0.0969 log r 9.8455 10\\nlog (s 9.9777 10 log (s-b)= 9.9777 10\\nlog (s c) 0. 1903 log tan 1 5 9.8678 10\\ncolog 5 9.4260- 10\\nJ5 36\u00c2\u00b024.6\\n5=72\u00c2\u00b0 49.2\\n2 )19.6909 20\\nlog r 9.8455 10\\nlog (s- a) 0.0969 log r 9.8455 10\\ntani^ 9.7486- 10\\n(s-c) 0.1903\\nJ^ 29\u00c2\u00b016.3\\n58\u00c2\u00b0 32.6 JC =24\u00c2\u00b019.7 o\\nCheck, A B+ C= 180\u00c2\u00b0 1.2\\ntan ^0=9.6552- 10\\nC =24\u00c2\u00b019.7\\n48\u00c2\u00b0 39.4", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0095.jp2"}, "96": {"fulltext": "86 Plane Trigonometry.\\n2. Given a 7, b ll, c 9.6 find B.\\nBy \u00c2\u00a7103, cos^B\\nCO,\\nOr, log cos 1 5 i[log s log (s 5) colog c colog a].\\nHere, 2s 27.6 whence, s 13.8, s 2.8.\\nlogs 1.1399\\nlog (s- 6) 0.4472\\ncolog c 9.0177 -10\\ncolog a 9.1549 -10\\n2 )19.7597 -20\\nlog cos 1^ 9.8799 10\\n|5 40\u00c2\u00b040.9 and 5 81 21.8\\nEXAMPLES.\\nSolve tlie following triangles\\n3. Given a 5, h l, c=zQ\\n4. Given a 10, 6 9, c 8.\\n5. Given a M, b A3, c .89.\\n6. Given a 70.5, b 56.2, c 63.9 find A.\\n7. Given a .0292, .0185, c .0357; find 5.\\n8. Given a 302, b 427, c 674 find C.\\n(For additional examples under Case III., see 112.)\\n108. Case IV. Given two sides, and the angle opposite to\\none of them.\\nIt was stated in 89 that a triangle is in general com-\\npletely determined when three of its elements are known,\\nprovided one of them is a side. The only exceptions occur\\nin Case IV.", "height": "3702", "width": "2478", "jp2-path": "completetrigonom00well_0096.jp2"}, "97": {"fulltext": "Solution of Oblique Triangles.\\n87\\nTo illustrate, let us consider the following example\\nGiven a 52.1, 61.2, A 31 26 find B, C, and c.\\nBy \u00c2\u00a799, sin^^6^ sin 5 ^il^.\\nsin A a a\\nWhence, log sin B \\\\ogb colog a log sin A.\\nlog6 1.7868\\ncologa 8.2832 10\\nlogsin^ 9.7173 -10\\nlog sin^ 9.7873- 10\\nB 37\u00c2\u00b0 47.5 from the table.\\nBut in finding tlie angle corresponding, attention must be paid to\\nthe fact that an angle and its supplement have the same sine 32).\\nTherefore another value of B will be 180\u00c2\u00b0 37\u00c2\u00b0 47.5 or 142\u00c2\u00b0 12.5\\nand calling these values ^i and B2, we have\\nBi 37\u00c2\u00b0 47.5 and B2 142\u00c2\u00b0 12.5\\nThe reason for the ambiguity is at once apparent when we attempt\\nto construct the triangle from the data.\\nBi D\\nWe first lay off angle DAF=Sr26 and on AF take AC =61.2.\\nWith O as a centre, and a radius equal to 52.1, describe an arc cutting\\nAD at Bi and B2. Then either of the triangles ABiC or AB^C\\nsatisfies the given conditions.\\nThe two values of B which were obtained are the values of angles\\nAB\\\\C and AB^G^ respectively; and it is evident geometrically that\\nthese angles are supplementary.\\nTo complete the solution, denote angles ACBi and ACBq, by C\\\\ and\\nC2, and sides ABi and AB2 by C\\\\ and C2, respectively.\\nThen, Ci 180\u00c2\u00b0 ^1) 180\u00c2\u00b0 69\u00c2\u00b0 13.5 110\u00c2\u00b0 46.5\\nand (72 180\u00c2\u00b0 (J. B2) 180\u00c2\u00b0 173\u00c2\u00b0 38.5 6\u00c2\u00b0 21.5", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0097.jp2"}, "98": {"fulltext": "88 Plane Trigonometry.\\nAgain, by 99, ^-H^, and ^J^^.\\na sin^ a sin A\\nWhence, Ci a sin Ci esc A^ and C2 a sin C2 esc A\\nloga 1.7168 loga 1.7168\\nlog sin Ci 9.9708 10 log sin C2 9.0443 10\\nlog esc JL 0.2827 log esc 0.2827\\nlog ci 1.9703 log C2 1.0438\\nci 93.40. C2 11.06.\\n109. Whenever an angle of an oblique triangle is deter-\\nmined from its sine, both the acute and obtuse values must\\nbe retained, unless one or both can be shown to be inadmis-\\nsible hence there may sometimes be two solutions, some-\\ntimes one, and sometimes none, in an example under Case IV.\\n1. Let the data be a, h, and A, and suppose h a.\\nBy Geometry, B must be hence, only the acvte value\\nof B can be taken; in this case there is but one solution.\\n2. Let the data be a, b, and A, and suppose b a.\\nSince B must be A, the triangle is impossible unless A\\nis acute.\\nAgain, since and 6 is a, sinB is sin A\\nsm^ a\\nHence, both the acute and obtuse values of B are A,\\nand there are two solutions, except in the following cases\\nIf log sin B 0, then smB=l 68), and B 90\u00c2\u00b0, and\\nthe triangle is a rigJit triangle if log sin B is positive, then\\nsin jB is 1, and the triangle is impossible.\\nThe above results may be stated as follows\\nIf, of the given sides, that adjacent to the given angle is\\nthe less, there is but one solution, which corresponds to the\\nacute value of the opposite angle.\\nIf the side adjacent to the given angle is the greater, there\\nare two solutions, unless the log sine of the opposite angle is\\nor positive in which cases there are one solution (a right\\ntriangle), and no solution, respectively.", "height": "3681", "width": "2474", "jp2-path": "completetrigonom00well_0098.jp2"}, "99": {"fulltext": "Solution of Oblique Triangles. 89\\n110. We will illustrate these points by examples\\n1. Given a 7.42, h 3.39, A 105\u00c2\u00b0 13 find B.\\nSince 6 is a, there is but one solution, corresponding to the acute\\nvalue of B.\\nBy 99, sin B P^ILA.\\na\\nlog 0.5302\\ncolog\u00c2\u00ab=: 9.1296- 10\\nlogsin^=: 9.9845- 10\\nlog sin 5 9.6443 10\\n5 26\u00c2\u00b0 9.6\\n2. Given b S, c 2, C= 100\u00c2\u00b0 find B.\\nSince is c, and C is obtuse, the triangle is impossible.\\n3. Given a 22.764, c 50, A 27\u00c2\u00b0 4.8 5 find C.\\nWe have, sin C ^^HL^.\\na\\nlogc 1.6990\\ncolog\u00c2\u00ab 8.6428 -10\\nlog sin^ 9.6582 -10\\nlogsin (7= 0.0000\\nTherefore, sin (7 1, and C= 90\u00c2\u00b0.\\nHere there is but one solution a right triangle.\\n4. Given a =.83, 6 .715, i? 6r47 find A\\nWe have, sin^\\nb\\nlog a 9.9191 -10\\ncolog 0. 1457\\nlog sin 9.9451 10\\nlog sin vl 0.0099\\nSince logsin J. is positive, the triangle is impossible.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0099.jp2"}, "100": {"fulltext": "90 Plane Trigonometry.\\nEXAMPLES.\\n111. Solve the following triangles\\n1.\\nGiven a 7.3,\\nb 6.6,\\n56\u00c2\u00b0.\\n2.\\nGiven b 86,\\nc 159,\\n0=115\u00c2\u00b0.\\n3.\\nGiven 60.93,\\nc 76.09,\\n5 133\u00c2\u00b0 41.\\n4.\\nGiven b 3S,\\nc 48.\\nB 34\u00c2\u00b0.\\n5.\\nGiven a .279,\\nc .227,\\n(7 =65\u00c2\u00b0 45\\n6\\nGiven a 3215,\\nc 6754,\\n28\u00c2\u00b0 26\\n7.\\nGiven a .06358,\\nc .08604,\\n0=19\u00c2\u00b0 14\\n8.\\nGiven a 186.7,\\nb 394.2,\\nJ5 114\u00c2\u00b0 28\\n9.\\nGiven a .462,\\nc .647,\\nA 31\u00c2\u00b0 7\\n(For additional examples under Case IV., see 112.)\\nMISCELLANEOUS EXAMPLES.\\n112. Solve the following triangles\\n1.\\nGiven\\na 934,\\nb 756,\\n0=73\u00c2\u00b0 16\\n2.\\nGiven\\nc 8.706,\\n5 38\u00c2\u00b0 45\\n0=31\u00c2\u00b0 59\\n3.\\nGiven\\na 61,\\n6 85,\\nc 48.\\n4.\\nGiven\\na .425,\\nc .454,\\n0=37\u00c2\u00b0 9\\n5.\\nGiven\\nb .0479,\\nc .0144,\\n121\u00c2\u00b0 28\\n6.\\nGiven\\na 7824,\\nc 3202,\\nA 140\u00c2\u00b0 53\\n7.\\nGiven\\nb .0005639,\\n44\u00c2\u00b0 24\\n5 116\u00c2\u00b0 9\\n8.\\nGiven\\na 1.5,\\nb 1.3,\\nc 1.9.\\n9.\\nGiven\\na 576,\\nb 813,\\nA 23\u00c2\u00b0 25\\n10.\\nGiven\\nb 2615,\\nc 6086,\\nA 115\u00c2\u00b0 10\\n11.\\nGiven\\nb 9.874,\\nc 7.486,\\n5 81\u00c2\u00b0 47\\n12.\\nGiven\\na 71387,\\nB 42\u00c2\u00b0 56\\n0=76\u00c2\u00b0 7", "height": "3681", "width": "2456", "jp2-path": "completetrigonom00well_0100.jp2"}, "101": {"fulltext": "Solution of Oblique Triangles. 91\\n13. Given b 51.434, c 47.955. C 72\u00c2\u00b0 54\\n14. Given a .008727, c .007065, B 84\u00c2\u00b0 56\\n15. Given a .031, 6 .024, c .028.\\n16. Given a .19597, 6 .13927, 45\u00c2\u00b0 17\\n17. Given a 3.5374, 5 9.6036, A 97\u00c2\u00b0 46\\n18. Given a .40932, 53\u00c2\u00b0 13 C=67\u00c2\u00b032\\n19. Given a 31.06, b 51.49, C 47\u00c2\u00b0 43\\n20. Given a .019186, b .033728, 125\u00c2\u00b0 33\\n21. Given a 353.85, c 579.42, 5 19\u00c2\u00b0 37\\n22. Given 6 24883, c 20609, C 48\u00c2\u00b06\\nAREA OF AN OBLIQUE TRIANGLE.\\n113. 1. Given a 18.063, A 96\u00c2\u00b0 30 B Sd\u00c2\u00b0; find Kl\\nBy 104, 2 K= sm -B sm C ^2 gin sin esc\\nsill J.\\nWhence,\\nlog (2 K) 2 log a log sin 5 log sin C log esc A.\\nHere, C 180\u00c2\u00b0 5) 48= 30\\nlog a 1.2568 multiply by 2 2.5136\\nlogsin5 9.7586 10\\nlog sin C= 9.8745 -10\\nlogescJ. 0.0028\\nlog (2 IT) 2.1495\\n2/f= 141.1, and ir= 70.55.\\nEXAMPLES.\\nFind the areas of the following triangles\\n2. Given a 26 A, c 47.9, B 67\u00c2\u00b0.\\n3. Given a 8.05, B 65\u00c2\u00b0 30 C 81\u00c2\u00b0 40\\n4. Given a 7, 6 9, c 6.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0101.jp2"}, "102": {"fulltext": "9^ Plane Trigonometry.\\n5. Given c .518, ^==67\u00c2\u00b0 45 B 37\u00c2\u00b019\\n6. Given b 15.32, c 36.78, A 105\u00c2\u00b0 43\\n7. Given 5-210.6, 32\u00c2\u00b0 21 O 108\u00c2\u00b0 56\\n8. Given c .004096, A 17\u00c2\u00b0 4.5 46\u00c2\u00b0 8\\n9. Given a .73, b .55, c .63.\\n10. Given a .0006854, b .0009743, C 61\u00c2\u00b0 44\\n11. Given a -3 7.219, 23\u00c2\u00b0 33 JB 124\u00c2\u00b0 12\\n12. Given a 5.321, c 8.467, 5 152\u00c2\u00b0 51\\n13. Given a 39.5, b 47.3, c 50.8.\\n14. Given b 250.8, A 77\u00c2\u00b0 53 O 56\u00c2\u00b0 29\\n15. Given 19146, c .42829, 59\u00c2\u00b0 7\\n16. Given a .078, b .091, c .084.\\n17. Given 109.41, 77\u00c2\u00b0 46 5 43\u00c2\u00b0 32\\n18. Given a 5.7434, b 8.6326, C 129\u00c2\u00b0 17\\n19. Given a 307.4, b 351.9, c 335.7.\\n20. Given a .0083214, A S^\u00c2\u00b0U 105\u00c2\u00b0 23\\n21. Given a .064325, c .033777, 5 141\u00c2\u00b0 38\\nMISCELLANEOUS PROBLEMS.\\n114. 1. To find the distance of an inaccessible object A\\nfrom a position B, I measure a base-line BG 675 feet in\\nlength, and observe the angles ABO and ACB to be 101\u00c2\u00b0 17\\nand 36\u00c2\u00b0 55 respectively. Find the distance AB.\\n2. In a field ABCD, the sides AB, BC, CD, and DA\\nare 16, 23, 18, and 29 rods, respectively, and the diagonal\\nAC is 34 rods. Find the area of the field.\\n3. From the top of a cliff the angles of depression of\\ntwo stakes in the plain beloAv, in line with the observer,\\nand 725 feet apart, are found to be 35\u00c2\u00b0 10 and 19\u00c2\u00b0 40 respec-\\ntively. Find the height of the cliff above the plain.", "height": "3701", "width": "2470", "jp2-path": "completetrigonom00well_0102.jp2"}, "103": {"fulltext": "Solution of Oblique Triangles. 93\\n4. The area of a triangle is 437, and two of its sides are\\n36 and 43. Find the angle between them.\\n5. From a point in the same horizontal plane with the\\nbase of a tower, the angle of elevation of its top is 42\u00c2\u00b0, and\\nfrom a point 200 feet farther away, it is 26\u00c2\u00b0. Find the\\nheight of the tower, and the distance of its base from each\\npoint of observation.\\n6. Two vessels start at the same point, at the rates of\\n9.7 and 5.5 miles an hour, respectively, the first due east, and\\nthe second due southwest. Find the distance between them\\nat the end of an hour and a half, and the bearing of each\\nfrom the other.\\n7. Two sides of a triangle are .85 and .74, and the differ-\\nence between their opposite angles is 18\u00c2\u00b0 27 Solve the\\ntriangle.\\n8. The area of a triangle ABC is 980, its angle A is\\n56\u00c2\u00b0 20 and its side b is 44. Find B, c, and a.\\n9. The sides of a triangle are 5, 7, and 9, respectively.\\nFind the radius of the inscribed circle.\\n(By Geometry, the area of a triangle is equal to one-half its perim-\\neter multiplied by the radius of the inscribed circle.)\\n10. Two sides of a parallelogram are 8 and 5, and include\\nan angle of 61\u00c2\u00b0. Find the diagonals.\\n11. The diagonals of a field ABCD intersect at E at an\\nangle of 78\u00c2\u00b0. If AE, BE, CE, and DE are 27, 31, 59, and\\n64 feet, respectively, find the area of the field.\\n12. The bases of a trapezoid are 49 and 95, and the angles\\nat the extremities of the latter are 64\u00c2\u00b0 and 71\u00c2\u00b0. Find the\\nnon-parallel sides.\\n13. Two vessels, A and B, are sailing due northeast. At\\na certain time, B lies 8 miles due south of A, and at the\\nexpiration of an hour 75\u00c2\u00b0 east of south. If the rate of A\\nis 6 miles an hour, find the rate of B.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0103.jp2"}, "104": {"fulltext": "94 Plane Trigonometry.\\n14. From a point in the same horizontal plane with the\\nbase of a tower, the angle of elevation of its top is 39\u00c2\u00b0 and\\nfrom a point 150 feet vertically above the first, the angle of\\ndepression of the top is 43\u00c2\u00b0. Find the height of the tower,\\nand its distance from the first point of observation.\\n15. From two points on either side of, and in line with, a\\ntower, 300 feet apart, the angles of elevation of its top are\\nobserved to be 31\u00c2\u00b0 and 27\u00c2\u00b0, respectively. Find the height\\nof the tower.\\n16. From a point in the same horizontal plane with the\\nbase of a tower, the angle of elevation of its top is 22\u00c2\u00b0, and\\nits bearing 31\u00c2\u00b0 west of north. From another point 400 feet\\nwest of the first, the bearing is 26\u00c2\u00b0 east of north. Find the\\nheight of the tower.\\n17. One of the non-parallel sides of a trapezoid is 15, the\\nangle between it and the longer base is 78\u00c2\u00b0, the angle at the\\nother extremity of the longer base is 62\u00c2\u00b0, and the shorter\\nbase is 9. Find the other two sides.\\n18. Two sides of a parallelogram are 103 and 54, and one\\nof the diagonals is 137. Find the angles of the parallelo-\\ngram, and the other diagonal.\\n19. From a ship, two lighthouses bear due northwest.\\nAfter sailing 18 miles in a direction 35\u00c2\u00b0 west of south, the\\nlighthouses bear 6\u00c2\u00b0 west of north and 9\u00c2\u00b0 east of north, re-\\nspectively. Find the distance between the lighthouses.\\n20. The sides AB and BC, of quadrilateral ABCD, are 9\\nand 5, respectively, and the angles. A, B, and C are 84\u00c2\u00b0,\\n109\u00c2\u00b0, and 96\u00c2\u00b0, respectively. Find the sides AD and CD.\\n21. From a position at the foot of a hill surmounted by a\\ntower, the angle of elevation of the top of the tower is 56\u00c2\u00b0.\\nAfter walking 1260 feet toward the foot of the tower, up a\\nslope whose angle with a horizontal plane is 29\u00c2\u00b0, the tower\\nsubtends an angle of 65\u00c2\u00b0. How far is the top of the tower\\nabove the horizontal plane of the foot of the hill", "height": "3681", "width": "2370", "jp2-path": "completetrigonom00well_0104.jp2"}, "105": {"fulltext": "Solution of Oblique Triangles. 95\\n22. The sides AB, BC, and CD, of quadrilateral ABCD,\\nare 23, 41, and 36, respectively, and the angles B and C are\\n116\u00c2\u00b0 and 131\u00c2\u00b0, respectively. Find the side AD and the\\nangles A and D.\\n23. The diagonals of a parallelogram are 58 and 92,\\nrespectively, and intersect at an angle of 55\u00c2\u00b0. Find the\\nsides and angles of the parallelogram.\\n24. From two points on the slope of a hill, in the same\\nvertical plane with the summit, the angles of elevation of\\nthe top are 11\u00c2\u00b0 and 18\u00c2\u00b0, respectively. The points are 300 feet\\napart, and the second 40 feet above the horizontal plane of\\nthe first. How far is the top of the hill above the hori-\\nzontal plane of the first point\\n25. To find the distance between two inaccessible buoys,\\nA and B, a line CD, 150 feet in length, is measured on the\\nshore. At C the angles ACD and BCD are observed to be\\n83\u00c2\u00b0 and 69\u00c2\u00b0, respectively, and at D the angles ADC and\\nBDG are observed to be 74\u00c2\u00b0 and 97\u00c2\u00b0, respectively. Find\\nthe distance AB.\\n26. A bluff, with a lighthouse on its edge, is observed\\nfrom a boat, the angle of elevation of the top of the light-\\nhouse being 25\u00c2\u00b0. After rowing 1000 feet directly toward the\\nlighthouse, the angles of elevation of its top and bottom are\\nfound to be 53\u00c2\u00b0 and 39\u00c2\u00b0, respectively. Find the height of\\nthe bluff, and of the lighthouse.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0105.jp2"}, "106": {"fulltext": "", "height": "3681", "width": "2438", "jp2-path": "completetrigonom00well_0106.jp2"}, "107": {"fulltext": "SPHERICAL TRIGONOMETRY.\\nD C\\nIX. GEOMETRICAL PRINCIPLES.\\n115. If a triedral angle be formed with its vertex at the\\ncentre of a sphere, it intercepts on the surface a spherical\\ntriangle.\\nThe triangle is bounded by three arcs of great circles,\\ncalled its sides, which measure the face angles of the tri-\\nedral angle.\\nThe angles of the spherical triangle are the spherical\\nangles formed by the adjacent sides and each is equal to\\nthe angle between two straight lines drawn, one in the plane\\nof each of its sides, perpendicular to the intersection of these\\nplanes at the same point.\\nThe sides of a spherical triangle are usually expressed in\\ndegrees.\\n116. A spherical triangle is called right when it has a\\nright angle quadrantal when it has one side a quadrant.\\n117. Spherical Trigonometry treats of the trigonometric\\nrelations between the sides and angles of a spherical tri-\\nangle.\\nThe face and diedral angles of the triedral angle are not\\naltered by varying the radius of the sphere and hence the\\nrelations between the sides and angles of a spherical triangle\\nare independent of the length of the radius.\\n97", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0107.jp2"}, "108": {"fulltext": "98 spherical Trigonometry.\\n118. We shall limit ourselves in the present work to such\\ntriangles as are considered in Geometry, where each angle\\nis less than two right angles, and each side less than the\\nsemi-circumference of a great circle; that is, where each\\nelement is less than 180\u00c2\u00b0.\\n119. The proofs of the following properties of spherical\\ntriangles may be found in any treatise on Solid Geometry\\n1. Any side of a spherical triangle is less than the sum\\nof the other two sides.\\n2. The sum of the sides of a spherical triangle is less\\nthan 360\u00c2\u00b0.\\n3. The sura of the angles of a spherical triangle is greater\\nthan 180\u00c2\u00b0, and less than 540\u00c2\u00b0.\\n4. If A B C is the polar triangle of spherical triangle\\nABO, that is, if A, B, and C are poles of sides B C, CA\\nand A B respectively, then ABC is the polar triangle of\\nspherical triangle A B C.\\n5. In two polar triangles, each angle of one is measured\\nby the supplement of that side of the other of which it is\\nthe pole that is,\\na 180\u00c2\u00b0 -A. b 180\u00c2\u00b0 B. c 180\u00c2\u00b0 C.\\nA 180\u00c2\u00b0 -a. B 180\u00c2\u00b0 b. C 180\u00c2\u00b0 c.\\n6. If two angles of a spherical triangle are unequal, the\\nsides opposite are unequal, and the greater side lies opposite\\nthe greater angle conversely, if two sides of a spherical\\ntriangle are unequal, the angles opposite are unequal, and\\nthe greater angle lies opposite the greater side.", "height": "3703", "width": "2454", "jp2-path": "completetrigonom00well_0108.jp2"}, "109": {"fulltext": "Geometrical Principles. 99\\n120. A spherical triangle is called tri -rectangular wlien it\\nhas three right angles each side is a quadrant, and each\\nvertex is the pole of the opposite side.\\n121. I. Let O be the right angle of right spherical tri-\\nangle ABC, and suppose a 90\u00c2\u00b0 and b 90\u00c2\u00b0.\\nE\\nComplete the tri-rectangular triangle A B C; also, since\\nB is the pole of AC, and A of BC, construct the tri-\\nrectangular triangles AB B and A BE.\\nThen since B lies within triangle AB B, AB or c is 90\u00c2\u00b0.\\nSince BC is B C, ZAis Z B AD, or 90\u00c2\u00b0.\\nSince ACis A C, Z B is Z A BE, ot KdO\\nII. Suppose a 90\u00c2\u00b0 and b 90\u00c2\u00b0.\\nh\\nComplete the lune ABA C.\\nThen in right triangle A BC, A C= 180\u00c2\u00b0 b.\\nThat is, sides a and A C of triangle A BC are each 90\u00c2\u00b0\\nand by I., A B and angles A and A BC are each 90\u00c2\u00b0.\\nBut, c 180\u00c2\u00b0 A B, A A and B 180\u00c2\u00b0 A BC.\\nWhence, c is 90\u00c2\u00b0, A 90\u00c2\u00b0, and B 90\u00c2\u00b0.\\nL.ofC.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0109.jp2"}, "110": {"fulltext": "loo spherical Trigonometry.\\nSimilarly, if a is 90\u00c2\u00b0 and h 90\u00c2\u00b0, then c is 90\u00c2\u00b0,\\nA 90\u00c2\u00b0, and B 90\u00c2\u00b0.\\nIII. Suppose a 90 and b 90\u00c2\u00b0.\\nComplete the lune ACBO.\\nThen in right triangle ABO,\\nAC 180\u00c2\u00b0 b, and BC 180\u00c2\u00b0 -a.\\nThat is, sides AC and BC of triangle ABC are each\\n90\u00c2\u00b0 and by I., AB and angles BAC and ABC are\\neach 90\u00c2\u00b0.\\nBut, A 180\u00c2\u00b0 BAC, and 5 180\u00c2\u00b0 ABC.\\nWhence, c is 90\u00c2\u00b0, A 90\u00c2\u00b0, and B 90\u00c2\u00b0.\\nHence, in any right spherical triangle\\n1. If the sides about the right angle are in the saine quad-\\n7 ant, the hypotemise is 90\u00c2\u00b0 if they are in differeiit quad-\\nrants, the hypotenuse is 90\u00c2\u00b0.\\n2. An angle is in the same quadrant as its opposite side.\\n122. In the figure of 119, we have, by 119, 1,\\na b c\\nPutting for a b and c the values given in 119, 5, we\\nhave\\n180\u00c2\u00b0-^ 180\u00c2\u00b0-5 180\u00c2\u00b0- O, or 5+ O-^ 180\u00c2\u00b0.\\nAgain, by 118, B C 180\u00c2\u00b0 A.\\nWhence, B+C-Ai8 180\u00c2\u00b0.\\nTherefore, C is between 180\u00c2\u00b0 and 180\u00c2\u00b0.\\nSimilarly, C+A B and C are between 180\u00c2\u00b0\\nand 180\u00c2\u00b0.", "height": "3681", "width": "2454", "jp2-path": "completetrigonom00well_0110.jp2"}, "111": {"fulltext": "Right Spherical Triangles.\\nlOI\\nX. RIGHT SPHERICAL TRIANGLES.\\n123. Let G be the right angle of right spherical triangle\\nABC, and the centre of the sphere.\\nDraw radii OA, OB, and OC.\\nAt any point A of OA, draw lines J. 5 and A C in\\nplanes GAB and OAC, respectively, perpendicular to OA,\\nmeeting OB and OC at J5 and C\\\\ respectively also, draw\\nline B C.\\nThen, OA is perpendicular to plane A B C.\\nWhence, each of the planes A B C and OBC is perpen-\\ndicular to plane OAC, and hence B C is perpendicular\\nto OAC.\\nTherefore, B C is perpendicular to A C and OC\\nBy 115, sides a, b, and c measure angles BOC, COA,\\nand AOB, respectively, and the angle A of the spherical\\ntriangle is equal to angle B A C.\\nIn right triangle OA B we have\\n.,r,r\u00e2\u0080\u009e OA OC OA\\n.osc .osA OB \u00e2\u0080\u0094X^r\\nBut in right triangles OB C and OCA\\nWhence,\\nOC OA\\ncos a, and cos o.\\nOB OC\\ncos c COS a cos b.\\n(69)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0111.jp2"}, "112": {"fulltext": "I02 Spherical Trigonometry.\\nB C\\nAnd, cos^ cos5 0 =-jy^, ^y^ (71)\\nIn like manner, sin B (72)\\nand cos 5 (73)\\ntan c\\nB C\\nOB\\nA B\\nsin a\\nA B\\nsm c\\nOB\\nA C\\nA C\\nA B\\nOA\\nA B\\nOA\\ntan\\ntan c\\nsin 5\\nsmc\\ntana\\n(70)\\n124. From (70) and (71), we obtain\\nJ sin A sin a tan c sin a\\ntan X\\ncos A sin c tan b cos c tan b\\nWhence by (69), tan A ^^L^^ (74)\\ncos a cos b tan o sm b\\nIn like manner, tan 5 (75)\\nsm a\\n125. By (4), sin a cos a tan a then (70) may be written\\ntana\\ncos a tan a tan c\\nsm\\ncos c tan c cos c\\ncos a\\nWhence by (69) and (73),\\nA COS 5 /_-s\\nsm (76)\\ncos\\nIn like manner, sin B (77)\\ncos a", "height": "3710", "width": "2457", "jp2-path": "completetrigonom00well_0112.jp2"}, "113": {"fulltext": "Right Spherical Triangles. 103\\n126. From (69), (76), and (77), we have\\n7 cos A cos B 1. A 1. T /\u00e2\u0080\u009ert\\\\\\ncos c cos a cos X cot A cot B. (78)\\nsm sm\\n127. The proofs of 123 cannot be regarded as general,\\nfor in the construction of the figure we have assumed a and\\n6, and therefore c and A 121), to be less than 90\u00c2\u00b0.\\nTo prove formulae (69) to (73) universally, we must con-\\nsider two additional cases\\nCase I. WJien 07ie of the sides a and b is 90\u00c2\u00b0, and the\\nother 90\u00c2\u00b0.\\nc^ B\\nIn right spherical triangle ABC, let a be 90\u00c2\u00b0 and b 90\u00c2\u00b0.\\nComplete the lune ABA C; then, in spherical triangle\\nA BC,\\nA B=lSO\u00c2\u00b0-c, A C=lSO\u00c2\u00b0-b, A =A, and A BC=1S0\u00c2\u00b0-B.\\nBut by 121, c is 90\u00c2\u00b0, A 90\u00c2\u00b0, and B 90\u00c2\u00b0.\\nHence, each element, except the right angle, of right\\nspherical triangle A BC is 90\u00c2\u00b0 and we have by 123,\\ncos A B cos a cos A C,\\n,inA sin A BC ^4^,\\nsm A B smA B\\ntSLnA B tan 5\\nPutting for A B, A C, A and A BC their values, we have\\ncos (180\u00c2\u00b0 c) cos a cos (180\u00c2\u00b0 6),", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0113.jp2"}, "114": {"fulltext": "I04 Spherical Trigonometry.\\nsin A sin (180\u00c2\u00b0 B)= sin (180\u00c2\u00b0 -6)\\nsin(180\u00c2\u00b0-cy sin (180\u00c2\u00b0 -cy\\ncos A ^^^(180\u00c2\u00b0-^) ^^g .^g()o _ ^x tana\\ntan (180\u00c2\u00b0 -c) tan (180\u00c2\u00b0 -c)\\nWhence, by 32, cos c cos a cos b),\\nsin\\nsma\\nsin c\\ncos A\\ntan 6\\ntan c\\ncosB\\nsm c\\ntan g\\ntanc\\nand we obtain formiilse (69) to (73) as before.\\nIn like manner, the formulae may be proved to hold when\\na is 90\u00c2\u00b0 and b 90\u00c2\u00b0.\\nCase II. When both a and b are 90\u00c2\u00b0.\\nIn right spherical triangle ABO, let a and 6 be 90\u00c2\u00b0.\\nComplete the lune ACBC.\\nBy 121, c is 90\u00c2\u00b0, A 90\u00c2\u00b0, and B 90\u00c2\u00b0.\\nHence, each element, except the right angle, of right\\nspherical triangle ABG is 90\u00c2\u00b0; and we have by 123,\\ncosc\\ncos\\nAC\\ncosBC,\\nsmc\\nsin ABC\\nsin AC\\nsmc\\ntanc\\nGos ABO\\ntan 5(7\\ntanc", "height": "3681", "width": "2451", "jp2-path": "completetrigonom00well_0114.jp2"}, "115": {"fulltext": "Right Spherical Triangles. 105\\nPutting for AC, BC, BAG and ABC their values,\\ncos e cos (180\u00c2\u00b0 a) cos (180\u00c2\u00b0 b),\\nsin(180\u00c2\u00b0-^) ^HL(l\u00c2\u00a7l^^^^,\\nsmc\\ntano\\nsin(180\u00c2\u00b0-\\nB)\\nsin (180\u00c2\u00b0-\\nsine\\ncos(180\u00c2\u00b0-\\n.B)\\ntan (180\u00c2\u00b0-\\ntanc\\n-a).\\nence, by 32, cos c\\ncos a)\\ncos\\nb),\\nA sin a\\nsm A\\nsmc\\nsin\\nL^:.\\n_ sin 6\\nsmc\\ntan b\\ncos\\ntanc\\ncos\\nB\\ntana\\ntan c\\nand we obtain formulae (69) to (73), as before.\\n128. The formulge of 123 to 126 are collected below\\nfor convenience of reference\\ncos c cos a cos b.\\nsin^-^i\\nsinB-^!^\\nsmc\\nsmc\\nA tan 5\\ncos^\\ntanc\\n-p tan a\\ncos B\\ntanc\\ntan^-*^\\nJ. tan 6\\ntanjt\\nsm\\nsma\\nsi^4_cosB\\nginJJ_COS^,\\ncoso\\nCOS a\\ncosc\\ncot A cot B.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0115.jp2"}, "116": {"fulltext": "io6 Spherical Trigonometry.\\nThe student should compare the formulae for the sines,\\ncosines, and tangents of A and B with the corresponding\\nformulae in 2 and 4.\\n129. liTapier s Rules of Circular Parts.\\nThese are two rules which include all the formulae of\\n128.\\nCO. B\\nCO. A\\nIn any right spherical triangle, the elements a and b, and\\nthe complements of elements A, B, and c (written in abbre-\\nviated form, CO. A, co. B, and co. c), are called the circular\\nparts.\\nIf we suppose them arranged in the order in which the\\nletters occur in the triangle, any one of the five may be\\ntaken and called the middle part; the two immediately\\nadjacent are called the adjacent parts, and the remaining\\ntwo the opposite parts.\\nThen Napier s rules are\\nI. The sine of the middle part is equal to the product of\\nthe tangents of the adjacent p)arts.\\nXI. The sine of the middle part is equal to the product of\\nthe cosines of the opposite parts.\\n130. Napier s rules may be proved by taking each cir-\\ncular part in succession as the middle part, and showing\\nthat the results agree with the formulae of 128.\\n1. If a be taken as the middle part, h and co. B are the\\nadjacent parts, and co. c and co. A the opposite parts.\\nThen the rules give\\nsin a tan h tan (co. B), and sin a cos (co. c) cos (co. A),", "height": "3681", "width": "2439", "jp2-path": "completetrigonom00well_0116.jp2"}, "117": {"fulltext": "Right Spherical Triangles. 107\\nOr by 31, sin a tan h cot B, and sin a sin c sin A\\nwhich are equivalent to (75) and (70).\\n2. If h be taken as the middle part, a and co. A are the\\nadjacent parts, and co. c and co. B the opposite parts.\\nThen, sin b tan a tan (co. A) tan cot A,\\nand sin 6 cos (co. c) cos (co. B) sin c sin B\\nwhich are equivalent to (74) and (72).\\n3. If CO. c be taken as the middle part, co. A and co. B\\nare the adjacent parts, and a and b the opposite parts.\\nThen,\\nsin (co. c) tan (co. A) tan (co. B), and sin (co. c) cos a cos 6.\\nOr, cos c cot A cot 5, and cos c cos a cos b\\nwhich agree with (78) and (69).\\n4. If CO. A be taken as the middle part, b and co. c are\\nthe adjacent parts, and a and co. B the opposite parts.\\nThen,\\nsin (co. tan b tan (co. c), and sin (co. A) cos a cos (co. B).\\nOr, cos tan b cot c, and cos A cos a sin jB\\nwhich are equivalent to (71) and (77).\\n5. If CO. B be taken as the middle part, a and co. c are\\nthe adjacent parts, and b and co. A the opposite parts.\\nThen,\\nsin(co. 5) tan a tan (co. c), and sin(co. 5) cos b cos (co. A).\\nOr, cos -B tan a cot c, and cos B cos 5 sin A\\nwhich are equivalent to (73) and (76).\\nWriters on Trigonometry differ as to the practical value of\\nKapier s rules but in the opinion of the highest authorities,\\nit seems to be regarded as preferable to attempt to remem-\\nber the formulae by comparing them with the analogous\\nformulae for plane right triangles, as stated in 128.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0117.jp2"}, "118": {"fulltext": "io8 Spherical Trigonometry.\\nSOLUTION OF RIGHT SPHERICAL TRIANGLES.\\n131. To solve a right spherical triangle, two elements\\nmust be given in addition to the right angle.\\nThere may be six cases\\n1. Given the hypotenuse and an adjaceyit angle.\\n2. Given an ajigle and its opposite side.\\n3. Given an angle and its adjacent side.\\n4. Given the hypotenuse and another side.\\n5. Given the tivo sides a and b.\\n6. Given the two angles A and B.\\n132. Either of these cases may be solved by aid of 128.\\nIf any two elements are given, the formula for computing\\neither remaining element may be found as follows\\nTake the formula which involves the given parts and the\\nrequired part.\\nIf all the remaining elements are required, the following\\nrule will be found convenient\\nTake the three forniulm ivhich involve the given parts.\\n133. It is convenient to have a check on the logarithmic\\nwork, which may always be done without the necessity of\\nlooking out any new logarithms.\\nExamples of this will be found in 136.\\nThe check formula for any particular case may be selected\\nfrom the set in 128 by the following rule\\nTake the formida which involves the three required parts.\\nNote. If Napier s rules are used, the following rule will indicate\\nwhich of the circular parts corresponding to the given elements and\\nany required element is to be regarded as the middle part.", "height": "3681", "width": "2569", "jp2-path": "completetrigonom00well_0118.jp2"}, "119": {"fulltext": "Right Spherical Triangles. 109\\nIf these three circular parts are adjacent, take the middle one as the\\nmiddle part, and the others are then adjacent parts.\\nIf they are not adjacent, take the part which is not adjacent to\\neither of the others as the middle part, and the others are then oppo-\\nsite parts.\\nFor the check formula, proceed as above with the circular parts\\ncorresponding to the three required elements.\\nThus, if c and A are the given elements,\\n1. To find a, consider the circular parts a, co. c, and co. A of\\nthese, a is the middle part, and co. c and co. A are opposite parts.\\nThen, by Napier s rules,\\nsin a cos (co. c) cos (co. A) sin c sin A.\\n2. To find b, the circular parts are b, co. c, and co. A in this case\\nCO. A is the middle part, and b and co. c are adjacent parts. Then,\\nsin (co. A) tan b tan (co. c), or cos A tan b cot c.\\n3. To find B, the circular parts are co. B, co. c, and co. A; co. c is\\nthe middle part, and co. A and co. B are adjacent parts. Then,\\nsin (co. c) tan (co. A) tan (co. B), or cos c cot A cot B.\\n4. For the check formula, the circular parts are a, b, and co. B\\na is the middle part, and b and co. B are adjacent parts. Then,\\nsin a tan b tan (co. B) tan b cot B.\\n134. In solving spherical triangles, careful attention\\nmust be paid to the algebraic signs of the functions the\\ncosines, tangents, and cotangents of angles between 90\u00c2\u00b0 and\\n180\u00c2\u00b0 being taken negative 21).\\nIt is convenient to place the sign of each function just\\nabove or below it, as shown in the examples of 136 the\\nsign of the function in the first member being then de-\\ntermined in accordance Avith the principle that, in multipli-\\ncation or division^ like signs produce and unlike signs\\nproduce\\nNote. In the examples after the first of 136, the signs are\\nomitted in every case where both functions in the second member\\nare positive.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0119.jp2"}, "120": {"fulltext": "no spherical Trigonometry.\\n135. In finding angles corresponding, if the function is a\\ncosine, tangent, or cotangent, its sign determines whetlier\\nthe angle is acute or obtuse that is, if it is the angle is\\nacute; and if it is the angle is obtuse, and the supple-\\nment of the acute angle obtained from the tables must be\\ntaken 32).\\nIf the function is a sine, since the sine of an angle is\\nequal to the sine of its supplement 32), both the acute\\nangle obtained from the tables and its supplement must\\nbe retained as solutions, unless the ambiguity can be\\nremoved by the principles of 121.\\nEXAMPLES.\\n136. 1. Given 5=z33\u00c2\u00b050 a ==108\u00c2\u00b0; find A, b, and c.\\nBy the rule of 132, the formulae from 128 are,\\nT) cos J. -D tan h t tan a\\nsm B tan B cos B\\ncos a sin a tan c\\n+Q rj a\\nThat is, cos A cos a sin B^ tan 6 sin a tan 5, tan c\\ncos^\\nHence, log cos A log cos a log sin B.\\nlog tan h log sin a log tan B.\\nlog tan c log tan a log cos B.\\nSince cos A and tan c are negative, the supplements of the acute\\nangles obtained from the tables must be taken 135).\\nNote 1. When the supplement of the angle obtained from the\\ntables is to be taken, it is convenient to v^rite 180\u00c2\u00b0 minus the element\\nin the first member, as shown below in the cases of A and c.\\nBy the rule of 133, the check formula for this case is\\ncos A or log cos A log tan h log tan c.\\ntan c\\nThe values of log tan h and log tan c may be taken from the first\\npart of the work, and their difference should be equal to the result\\npreviously found for log cos A.", "height": "3681", "width": "2451", "jp2-path": "completetrigonom00well_0120.jp2"}, "121": {"fulltext": "Right Spherical Triangles. 1 1 1\\nlog COS a 9.4900 10 log tan a 0.4882\\nlog sin 5 9. 7457 10 log cos 5 9. 9194 10\\nlogcos^ 9.2357 10 log tan c 0.5688\\n180\u00c2\u00b0 80\u00c2\u00b0 5.5 180\u00c2\u00b0 c 74\u00c2\u00b0 53.8\\n99\u00c2\u00b0 54.5 0=105\u00c2\u00b0 6.2\\nlog sin a 9.9782 10 Check.\\nlog tan B 9.8263 10 log tan h 9.8045 10\\nlog tan h 9.8045 10 log tan c 0.5688\\nb 32\u00c2\u00b0 31. 1 log cos A 9.2357 10\\n2. Given c 70\u00c2\u00b0 30 100\u00c2\u00b0; find a, h, and\\nIn this case, the three formulae are,\\nsin^=^HIi!, cos^=^^^, cos c cot cot 5.\\nsin c tan c\\nThat is, sin a sin c sin A^ tan h tan c cos A^ cot B cos c tan A.\\nHere, the side a is determined from its sine but the ambiguity is\\nremoved by the principles of 121 for a and A must be in the same\\nquadrant. Therefore a is obtuse, and the supplement of the angle\\nobtained from the table must be taken.\\nBy 133, the check formula is\\ntan B or sin a tan b cot B.\\nsin a\\nNote 2. The check formula should always be expressed in terms\\nof the functions used in determining the required parts thus, in the\\ncase above, the check formula is transformed so as to involve cot B\\ninstead of tan B.\\nlog sin c 9.9743 10 log cos c 9.5235 10\\nlog sin 9. 9934 1 log tan A 0.7537\\nlog sin a 9.9677 10 log cot B 0.2772\\n180\u00c2\u00b0 a 68\u00c2\u00b0 10 180\u00c2\u00b0 B 27\u00c2\u00b0 50.6\\na 111\u00c2\u00b0 50 B 152\u00c2\u00b0 9.4\\nlog tan c =0.4509\\nlog cos A 9.2397 10 Check.\\nlog tan b 9.6906 10 log tan 6 9.6906 10\\n180\u00c2\u00b0 6 26\u00c2\u00b0 7.5 log cot B 0.2772\\n153\u00c2\u00b0 52.5 logsin a =9.9678-10", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0121.jp2"}, "122": {"fulltext": "112 Spherical Trigonometry.\\nNote 3. We observe here a difference of .0001 in the two values\\nof log sin a. This does not necessarily indicate an error in the work,\\nfor such a small difference might easily be due to the fact that the\\nlogarithms are only approximately correct to the fourth decimal place.\\n3. Given a 132\u00c2\u00b0 6 h 77\u00c2\u00b0 51 find A, B, and c.\\nIn this case, the three formulse are,\\nA tan a -r, tan b\\ntan A tan B cos c cos a cos b.\\nsin b sin a\\nThe check formula is\\ncos c cot A cot B, or cos c tan A tan B 1.\\nThat is, log cos c log tan A log tan B log 1 0. _\\nlog tan a 0.0440 log cos a 9.8263 10\\nlog sin b 9.9901 10 log cos b 9.3232 10\\nlog tan 0. 0539 log cos c 9. 1495 10\\n180\u00c2\u00b0 48\u00c2\u00b0 32.8 180\u00c2\u00b0 c 81\u00c2\u00b0 53.4\\nA 131\u00c2\u00b0 27.2 c 98\u00c2\u00b0 6.Q\\nlog tan b =0.6670 Check.\\nlog sin a 9.8704 10 log cos c 9.1495 10\\nlog tan 0. 7966 log tan A 0.0539\\nB 80\u00c2\u00b0 55.4 log tan 0. 7966\\nlog 1 0.0000\\n4. Given A 105\u00c2\u00b0 59 a 128\u00c2\u00b0 33 find b, B, and c.\\nThe formulae are,\\nsmb=^^^, g.+^_cos^\\ntan A cos a sin A\\nThe check formula is sin B\\nsin c\\nIn this example, each required part is determined from its sine\\nand as the ambiguity cannot be removed by 121, both the acute\\nangle obtained from the tables and its supplement must be retained in\\neach case.", "height": "3681", "width": "2456", "jp2-path": "completetrigonom00well_0122.jp2"}, "123": {"fulltext": "Right Spherical Triangles. 113\\nlog tan a 0.0986 log sin a 9.8932 10\\nlog tan 0.5430 log sin 9.9828 10\\nlog sin b 9.5556 10 log sin c 9.9104 10\\nb 21\u00c2\u00b0 3.9 c =54\u00c2\u00b0 26.7\\nor 158\u00c2\u00b0 56.1 or 125\u00c2\u00b0 33.3\\niogcosJ[ 9.4399 -10\\nlog cos a 9.7946 10 ^^^^g _\\nlog sin B 9.6453 10 9.9104-10\\nB 26\u00c2\u00b0 13. 5 j^g g.^ g g^g^ _ ^Q\\nor 153\u00c2\u00b0 46.5\\nIt does not follow, however, that these values can be combined pro-\\nmiscuously for by 121, since a is 90\u00c2\u00b0, with the value of b less\\nthan 90\u00c2\u00b0 must be taken the value of c greater than 90\u00c2\u00b0, and the value\\nof B less than 90\u00c2\u00b0 while with the value of b greater than 90\u00c2\u00b0 must be\\ntaken the value of c less than 90\u00c2\u00b0, and the value of B greater than 90\u00c2\u00b0.\\nThus the only solutions are\\n1. 6 21\u00c2\u00b0 3.9 c 125\u00c2\u00b0 33.3 _B 26\u00c2\u00b0 13.5\\n2. 6 158\u00c2\u00b0 56.1 c 54\u00c2\u00b0 26.7 5 15-3\u00c2\u00b0 46.5\\nNote 4. The figure shows geometrically why there are two solu-\\ntions in this case.\\nB....\\nFor if AB and ^0 be produced to A forming lune ABA C, tri-\\nangle A BC has side a and angle A equal, respectively, to side a and\\nangle A of triangle ABC^ and both triangles are right-angled at C.\\nIt is evident that sides A B and A C and angle A BC are the sup-\\nplements of sides c and b and angle ABC, respectively.\\nSolve the following right spherical triangles\\n5. Given a 31\u00c2\u00b0, c 60\u00c2\u00b0.\\n6. Given A 27\u00c2\u00b0, B 73\u00c2\u00b0.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0123.jp2"}, "124": {"fulltext": "114 Spherical Trigonometry.\\n7. Given a S% 6 22\u00c2\u00b0.\\n8. Given B 25\u00c2\u00b0, c 34\u00c2\u00b0.\\n9. Given A 40\u00c2\u00b0, a 26\u00c2\u00b0.\\n10. Given 5 127\u00c2\u00b0 20 a 82\u00c2\u00b0.\\n11. Given b 18\u00c2\u00b0, c 112\u00c2\u00b0 10\\n12. Given A 120\u00c2\u00b0, c 161\u00c2\u00b0 50\\n13. Given A 159\u00c2\u00b0 40 b 135\u00c2\u00b0.\\n14. Given 5 110\u00c2\u00b0 50 5 118\u00c2\u00b0 30\\n15. Given a 49\u00c2\u00b0 10 6 100\u00c2\u00b0.\\n16. Given A 170\u00c2\u00b0 50 b 55\u00c2\u00b0.\\n17. Given A 28\u00c2\u00b0 20 c 108\u00c2\u00b0 40\\n18. Given A 104\u00c2\u00b0 50 B 156\u00c2\u00b0 30\\n19. Given a 164\u00c2\u00b0 10 c 133\u00c2\u00b050\\n20. Given 5 99\u00c2\u00b0 40 c 50\u00c2\u00b030\\n21. Given b 130\u00c2\u00b0 40 c 70\u00c2\u00b0 10\\n22. Given a 129\u00c2\u00b0 30 6 166\u00c2\u00b0 50\\n23. Given 24\u00c2\u00b0 31 6 19\u00c2\u00b0 9\\n24. Given A 83\u00c2\u00b0 15 a 76\u00c2\u00b0 46\\n25. Given B 115\u00c2\u00b0 22 a 145\u00c2\u00b0 39\\n26. Given 6 43\u00c2\u00b0 57 c 62\u00c2\u00b05\\n27. Given A 81\u00c2\u00b0 29 B 131\u00c2\u00b0 51\\n28. Given a 147\u00c2\u00b0 35 c 52\u00c2\u00b013\\n29. Given A 139\u00c2\u00b0 4 c 63\u00c2\u00b0 47\\n30. Given B 39\u00c2\u00b0 43 a 54\u00c2\u00b0 26\\n31. Given b 153\u00c2\u00b0 18 c 121\u00c2\u00b0 54\\n32. Given A 37\u00c2\u00b0 m\\\\ b 157\u00c2\u00b0 12\\n33. Given B 114\u00c2\u00b0 38 c 168\u00c2\u00b0 23", "height": "3702", "width": "2451", "jp2-path": "completetrigonom00well_0124.jp2"}, "125": {"fulltext": "Right Spherical Triangles. 115\\n34. Given a 66\u00c2\u00b0 6 c- 109\u00c2\u00b0 44\\n35. Given A 30\u00c2\u00b0 48 c 13\u00c2\u00b0 27\\n36. Given B 69\u00c2\u00b0 16 a 160\u00c2\u00b0 55\\n37. Given a 142\u00c2\u00b0 42 5 78\u00c2\u00b0 6\\n38. Given A 126\u00c2\u00b0 53 B 47\u00c2\u00b0 34\\n39. Given B 16\u00c2\u00b0 24 c 140\u00c2\u00b0 37\\n40. Given B 98\u00c2\u00b0 17 b 143\u00c2\u00b0 8\\n137. Quadrantal Triangles.\\nBy 119, 5, tlie polar triangle of a quadrantal triangle is\\na right spherical triangle.\\nHence, to solve a quadrantal triangle, we have only to\\nsolve its polar triangle, and take the siqjplements of the\\nresults.\\n1. Given c 90\u00c2\u00b0, a 67\u00c2\u00b0 38 6 48\u00c2\u00b0 50 find A, B,\\nand C.\\nDenoting the polar triangle by A B C, we have by 119, 5,\\nC 90\u00c2\u00b0, A 112\u00c2\u00b0 22 B 131\u00c2\u00b0 10 to find a 6 and c\\nBy 132, the formulae for the solution are\\ncos a ^2^, cos b \u00c2\u00a32^, and cos c co t A co t B\\nsin B sin A\\nThe check formula is cose cos a cos 6\\nlog cos A 9.5804 10 log cot A 9.6143 10\\nlog sin B 9.8767 10 log cot B 9.9417 10\\nlog cos a 9.7037 10 log cos c 9.5560 10\\n180\u00c2\u00b0 -a 59\u00c2\u00b0 38.2 c 68\u00c2\u00b0 54.8\\nlogcos^ 9.8184 -10 Check.\\nlog sin A 9.9660 10 log cos a 9.7037 10\\nlog cos b 9.8524-10\\n180\u00c2\u00b0 b 44\u00c2\u00b0 36.7 log cos c 9.5561 10", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0125.jp2"}, "126": {"fulltext": "ii6 Spherical Trigonometry.\\nThen in the given quadrantal triangle, we have\\nA 180\u00c2\u00b0-a 59\u00c2\u00b0 38.2\\n180\u00c2\u00b0 -6 44\u00c2\u00b0 36.7\\n(7=180\u00c2\u00b0-c lll\u00c2\u00b0 6.2\\nEXAMPLES.\\nSolve the following quadrantal triangles\\n2. Given 157\u00c2\u00b0, (7=121\u00c2\u00b0.\\n3. Given a 117\u00c2\u00b0, b 142\u00c2\u00b0 50\\n4. Given 43\u00c2\u00b0, J5 106\u00c2\u00b0.\\n5. Given 162\u00c2\u00b020 (7= 64\u00c2\u00b0 40\\n6. Given 30\u00c2\u00b0 10 a =72\u00c2\u00b0 30\\n7. Given 118\u00c2\u00b0 16 6 =137\u00c2\u00b0 57\\n8. Given a 51\u00c2\u00b0 34 C=25\u00c2\u00b049\\n9. Given 5 141\u00c2\u00b0 13 0=49\u00c2\u00b0 35\\n10. Given a 17\u00c2\u00b0 41 B 38\u00c2\u00b0 24\\n11. Given B 159\u00c2\u00b0 2 b 136\u00c2\u00b0 28\\n138. Isosceles Spherical Triangles.\\nWe know, by Geometry, that if an arc of a great circle be\\ndrawn from the vertex of an isosceles spherical triangle to\\nthe middle point of the base, it is perpendicular to the base,\\nbisects the vertical angle, and divides the triangle into two\\nsymmetrical right spherical triangles.\\nBy solving one of these, we can find the required parts\\nof the given triangle.\\n1. Given a 115\u00c2\u00b0, 6 115\u00c2\u00b0, C=71\u00c2\u00b040 find A, B,\\nand a\\nDenoting the elements of one of the right triangles by A B C,\\na 6 and c where C is the right angle, we have\\nc a 115\u00c2\u00b0, and A iC 35\u00c2\u00b0 50", "height": "3710", "width": "2457", "jp2-path": "completetrigonom00well_0126.jp2"}, "127": {"fulltext": "Right Spherical Triangles. 117\\nWe have then to find the parts a and B in this triangle.\\nBy 128, sin A^ ?HL\u00c2\u00ab!, and cos d cot A cot B\\nsin c\\nOr, sin a sin c sin A and cot B cos c tan A\\nlog sin c 9.9573 10 log cos c 9.6259 10\\nlog sin A 9.7675 10 log tan A 9.8586 10\\nlog sin a 9. 7248 10 log cot B 9.4845 10\\na 32\u00c2\u00b0 3.0 180\u00c2\u00b0 B 73\u00c2\u00b0 1.8\\nB 106\u00c2\u00b0 58.2\\nThen in the given isosceles triangle,\\nA B=B 106\u00c2\u00b0 58.2 and c 2 a 64\u00c2\u00b0 6.0\\nEXAMPLES.\\nSolve the following isosceles spherical triangles\\n2. Given A 27\u00c2\u00b0, B 27\u00c2\u00b0, c 135\u00c2\u00b0.\\n3. Given a 152\u00c2\u00b0, 6=152\u00c2\u00b0, 0= 68\u00c2\u00b0.\\n4. Given a 112\u00c2\u00b0 36 b 112\u00c2\u00b0 36 c 123\u00c2\u00b0 50\\n5. Given 159\u00c2\u00b0 14 5 159\u00c2\u00b0 14 a =137\u00c2\u00b0 47", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0127.jp2"}, "128": {"fulltext": "ii8\\nSpherical Trigonometry.\\nXL OBLIQUE SPHERICAL TRIANGLES.\\nGENERAL PROPERTIES OF SPHERICAL TRIANGLES.\\n139. In any spherical triangle, the sines of the sides are\\nproportional to the sines of their opposite angles.\\nD\\nLet ABC be any spherical triangle, and draw arc CD per-\\npendicular to AB.\\nThere will be two cases according as CD falls upon AB\\n(Fig. 1), or AB produced (Fig. 2).\\nIn right spherical triangle ACD, in either figure, we have\\nby (70),\\nsinCi)\\nAlso, in Fig. 1,\\nAnd in Fig. 2,\\nsin J.\\nsin 5\\nsin 6\\nsin(7i)\\nsma\\nsin5=sin(180\u00c2\u00b0-(75i))\\nsin CBD 32)\\nsin CD\\nsma\\nDividing these equations, we have in either case\\nsin (72)\\nsin A sin h sin a\\nsin B sin CD sin h\\nsin a\\n(79)", "height": "3710", "width": "2460", "jp2-path": "completetrigonom00well_0128.jp2"}, "129": {"fulltext": "Oblique Spherical Triangles. 119\\nT T1 sin^ sin 5\\nIn like manner, 77 j (80)\\nsm C sm c\\nsin^ sina\\nand 7^ (81)\\nsm C sm c\\n140. In any spherical triangle, the cosine of any side is\\nequal to the product of the cosines of the other tivo sides, plus\\nthe continued product of their sines and the cosine of their\\nincluded angle.\\nIn right spherical triangle BCD, in Fig. 1, 139, we have,\\nby (69),\\ncos a cos BD cos CD cos (c AD) cos CD.\\nAnd in Fig. 2,\\ncos a cos BD cos CD cos (AD c) cos CD.\\nThen in either case, by (12),\\ncos a cos c cos AD cos CD sin c sin AD cos CD.\\nBut in right spherical triangle ACD, by (69),\\ncos AD cos CD cos h.\\ncos\\nAlso, sin AD cos CD sin AD 7^ cos h tan J.i).\\ncos AD\\n-D 4- T. sin5 sin 5\\nBut, smce tan cos b\\ncos b tan b\\ntan AD\\nThen, sin AD cos Oi) sin b r\u00e2\u0080\u0094 sin b cos by (71).\\ntan o\\nWhence, cos a cos cos c sin 6 sin c cos A. (82)\\nIn like manner,\\ncos b cos c cos a sin c sin a cos B, (83)\\nand cos c cos a cos b sin a sin b cos C (84)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0129.jp2"}, "130": {"fulltext": "I20 Spherical Trigonometry.\\n141. Let ABC and A B C be a pair of polar triangles.\\nApplying formula (82) to triangle A B C, we obtain\\ncos a cos b cos c sin b sin c cos A\\nPutting for a b c and A the values given in 119, 5,\\ncos (180\u00c2\u00b0 -A) cos (180\u00c2\u00b0 B) cos (180\u00c2\u00b0 C)\\n-f sin (180\u00c2\u00b0 B) sin (180\u00c2\u00b0 C) cos (180\u00c2\u00b0 a).\\nWhence, by 32,\\ncos A cos B) cos (7) sin B sin C(\u00e2\u0080\u0094 cos a).\\nThat is,\\ncos A cos 5 cos sin B sin (7 cos a. (85)\\nSimilarly,\\ncos B cos (7 cos yl sin C sin cos b, (86)\\nand cos C cos J. cos sin sin B cos c. (87)\\nThe above proof illustrates a very important application\\nof the theory of polar triangles in Spherical Trigonometry.\\nIf any relation has been found between the elements of a\\nspherical triangle, an analogous relation may be derived\\nfrom it, in which each side or angle is replaced by the\\nopposite angle or side, with suitable modifications in the\\nalgebraic signs.\\n142. To express the sines, cosines, and tangents of the half-\\nangles of a spherical triangle in terms of the sides of the\\ntriangle.", "height": "3703", "width": "2451", "jp2-path": "completetrigonom00well_0130.jp2"}, "131": {"fulltext": "Oblique Spherical Triangles. 121\\nFrom (82), 140,\\nsin h sin c cos cos a cos cos c.\\n-rxTi A cos a cos h cos c\\nWnence, cos A (A)\\nsm b sm c\\nSubtracting both members from 1, we have\\nA cos a cos 6 cos c\\n1 cos A l\\nsm b sm c\\n_ cos b cos c sin b sin c cos a\\nsin 6 sin c\\nWhence, by (28),\\n2 gin^ A\\nsin b sin c\\nBut by (20),\\ncos y cos x 2 sin (a? 2/) sin V)- (B)\\nWhence,\\n2 gjn^ 1 ,1 (P-^)1 s^^ i c)\\nsin sin c\\nsin i (a -1- 5 c) sin i (a 6 c)\\nor sm^ ^A i\\nsm sm c\\nDenoting the sum of the sides, a-\\\\-b c^ by 2 s, we have\\na -c (a 64-c)-2c 2s-2c 2(s-c),\\nand a-6 c (a 6 c)-2 2s-2 2(s- 6).\\nsin is b) sin (s c)\\nWhence, sin-\\nsin b sin c\\nOr, sm|^=J\u00c2\u00abi iii^4^i!lii^\\nsm o sm c\\n(88)\\nT T1 1 T^ /sm (s c) sm (s a) ^-.v\\nIn like manner, sm i J5 V (Q^)\\nsm n sm a\\nsm c sm a\\n/sin (s a) sin (s b)\\nand sm i \u00e2\u0080\u0094^1 (90)\\nsm a sm h", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0131.jp2"}, "132": {"fulltext": "122 Spherical Trigonometry.\\nAgain, adding both, members of (A) to 1, we have\\nA -i cos a cos b cos c\\n1 cos 1 -I\\nsm b sm c\\n_ cos a (cos b cos c sin 6 sin c)\\nsin 6 sin c\\nWhence, by (29),\\nn cos a cos (6 c)\\n2cosn^\\n2\\nsin 6 sin c\\n2 sm 4 f c a) sm 1 f c a) ,_,x\\nr^ -5 by (B).\\nsm sm c v\\nPutting a-\\\\-b-\\\\-c 2s, whence 5 c a 2 (s a),\\nsin s sin (s a)\\nsm sm c\\n/sin s sin (s a)\\nOr, cos i ^--^-^-\\\\-\u00e2\u0080\u0094-l. (91)\\nsin 6 sin c\\nT /sin s sin (s 6)\\nIn like manner, cos B (92)\\nsm c sm a\\n1 /sin s sin (s c)\\nand cos i O (93)\\nsm a sm 6\\nDividing (88) by (91), we have\\nin i /sin (s 5) sin (s c) sin sin c\\nOS -J- J. sin 5 sin c sin s sin (s a)\\nsm\\ncos\\n-r-rr, 1 A /Sin (S b) SlTl (S C)\\nWhence, tan i V r (9*)\\nsin s sin (s a)\\nIn like manner, tan IB= (95)\\n^f smssm(.s-\\nand tani(7=JgHL(lz :^)\u00c2\u00ab;n(^-^) (gg)\\ni( sin .s sm (Si rA\\nsin s sm (s c)", "height": "3681", "width": "2449", "jp2-path": "completetrigonom00well_0132.jp2"}, "133": {"fulltext": "Oblique Spherical Triangles. 123\\n143. To express the sines, cosines, and tangents of the half-\\nsides of a spherical triangle in terms of the angles of the\\ntriangle.\\nFrom (85), 141, sin B sin C cos a cos -f cos 5 cos G.\\nWhence, cos a 22^ii+^2i^oosC.\\nsin B sin C\\nThen, l_eosa l-22iA\u00c2\u00b1^5^BcosO.\\nsin B sin C\\nOv 9 oiT^2 1 _ (cos S cos sin B sin O) cos A\\nyji, L sm 2^ a\\nsm B sm G\\ncos C) COS A\\nsin 5 sin (7\\nThen by (19),\\n2cosi(^+0 ^)cosi(^ C-^)\\ni^/ bill -s- (X/ 7^\\nsm B sm C\\nDenoting the sum of the angles, A-\\\\- B C, by 2 /S, we\\nhave 5 O 2 (aS\\nWhence, sin^ \\\\a ^^^^^^^(.S\\nsin S sin (7\\nc\\\\^ 1 cos aS cos iS A)\\nOr, sm 1 a A^ -d n\\nsm S sm C\\nT Ti 1 r COS S cos S -B) /^_x\\nIn like manner, sm 6 (98)\\nsm 6 sm\\ncos /S cos (aS C) /rto\\\\\\nand smic=:\\\\ -A (99)\\nsm J[ sm 5\\nAgain, adding both members of (A) to 1, we have\\ncos^ 4- cos B cos C\\n1 H- cos a 1 4-\\nsin 5 sin C\\ncos cos B cos sin B sin C\\nsin B sin C", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0133.jp2"}, "134": {"fulltext": "124 Spherical Trigonometry.\\nThen, 2 cos^ i a cos cos (ij C)\\nsin B sin (7\\n2cos|[^ ^-0]cosi[^-(^-C)]\\nsin B sin (7\\nOr, cosHc\u00c2\u00bb ^t(-^-^+^).\\nsm i? sm (7\\nY^vA A+ B C =2{S G), 2.nd. A- B C =2{8 B).\\nWhence, cos^ i a^ c os cos 0)\\nsm sm (7\\nOr, cosi\u00c2\u00ab=xp(^^^) \u00c2\u00b0^(f-^X\\nsm 5 sm (7.\\nIn like manner,\\n(100)\\nsin r/sin /I\\nsin C sin\\nand\\ncos J os( S-^)cos(^-i?) (102)\\nsin sin B\\nDividing (97) by (100), we have\\ntania=J cos S cos (.S\\n2 COS (/S j5) cos (7)\\nIn like manner.\\n1 COS S cos 5) /-.r./iN\\n=V- eos(^-a)cos(^-^)\\nand tanic=JIZ^p^M^^. (105)\\ncos cos (/S\\nNAPIER S ANALOGIES.\\n144. Dividing (94) by (95), we have\\ntan iA_ /sin (s b) sin (s c) sin s sin (s b)\\ntan 5 ^Z sin s sin (s a) sin (s c) sin (s a)\\nsin cos jB _ /sin^ _ sin (s\\ncos 4 sin -1- ^Z sin^ (s a)~ sin (s a)", "height": "3681", "width": "2459", "jp2-path": "completetrigonom00well_0134.jp2"}, "135": {"fulltext": "Oblique Spherical Triangles. 125\\nWhence bj composition and division,\\nsin i A cos ^B cos A sin B _ sin (s b) sin (s a)\\nsin 1 J. cos i 5 cos -L yl sin i sin (s 5) sin (s a)\\nThen by (9), (U), and (21),\\nsin {lA-{-^B) _ tan i [s s a]\\nsin (i^ i jB) tani^[s 6 (s a)]*\\nBut s 6 8 rt 2s a 6 c.\\nsini(^ tanic\\nAMience, f-^-j z rv^ t^ (^06)\\nsm ^(A B) tan i (a 6;\\n145. Multiplying (94) by (95), Tve have\\ntan i tan i B= sin (s-b) sin (.9- c) sin (s-c) sin (s- a)\\nsin s sin (s cO sin s sin (s b)\\n{s sin s sin (s b)\\nsin -L sin B _ /sin- (s c) _ sin (s c)\\nOr _ _\\ncos ^4 cos -J- -B sin- s sin\\nWhence by composition and division,\\ncos A cos Y sin A sin 1? _ sin s sin (s e)\\ncos cos -B sin sin -i- 5 sin s sin (s c)\\nOr, by (21),\\ncos(i^4-i^) _ tani[g-(s-c)]\\ncos (^A ^B)~ tan [s 4- s c]\\nBut s-\\\\-s c 2s c a-\\\\-b.\\nTTT-L cos-i-(^ tanic\\nWhence, a -ni i w (107)\\ncos 5) tan i (a 6)\\n146. Applying formula (106) to triangle A B C, in the\\nfigure of 141, we obtain\\nsin-i(^^ _ tanic\\nsin i (A tan i (a 6", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0135.jp2"}, "136": {"fulltext": "126 spherical Trigonometry.\\nBut,\\ni{A B ^(180\u00c2\u00b0 a 4- 180\u00c2\u00b0 6)= 180\u00c2\u00b0 -i.(a 6);\\n_ 5 1(180\u00c2\u00b0 -a- 180\u00c2\u00b0 -i-b) -^(a-b);\\n1 c i(180\u00c2\u00b0 0)= 90\u00c2\u00b0 1 (7;\\nand i(a 6 J(180\u00c2\u00b0 ~A- 180\u00c2\u00b0 5) _ i (^4 J5).\\nWhence,\\nsin[180\u00c2\u00b0-i(a _ tan (90\u00c2\u00b0 -^C)\\nsin[- i-(a 5)] tan i-(^ _ 5)]\\nTherefore, by 28, 31, and 32,\\nsin i (a 5) _ cot C\\nsin i- (a b) tan -i- J.\\nsin|(a 6) cotiC\\nsin i (a -6) tan 1(^-5)\\nIn like manner, from (107), we obtain\\ncosi(^ 4-5 )_ tanic\\ncos i (A 5 tan i (a 5\\nBut,\\n1 (a 5 ^(180\u00c2\u00b0 180\u00c2\u00b0 5)= 180\u00c2\u00b0 5).\\nWhence,\\ncos [180\u00c2\u00b0 (g 6)] tan (90\u00c2\u00b0 -jC)\\ncos 1 (a 6)] tan [180\u00c2\u00b0 5)]\\nTherefore, by 28, 31, and 32,\\ncos |-(a _ cot ^(7\\ncos^(a 6) tan i- 5)\\ncosi(a 6) _ cot -1(7\\ncos I (a -6) tan K^ 4- 5^", "height": "3679", "width": "2448", "jp2-path": "completetrigonom00well_0136.jp2"}, "137": {"fulltext": "Oblique Spherical Triangles. 127\\n147. The foriniilse exemplified in 144, 145, and 146\\nare known as Napier^ s Analogies. In each case there may\\nbe other forms according as other elements are used.\\nSOLUTION OF OBLIQUE SPHERICAL TRIANGLES.\\n148. In the solution of oblique spherical triangles, we\\nmay distinguish six cases\\n1. Given a side and the adjacent angles.\\n2. Given two sides and their included angle.\\n3. Given the three sides.\\n4. Given the three angles.\\n5. Given two sides and the angle opposite to one of them.\\n6. Given two angles and the side opposite to one of them.\\nBy application of the principles of 119, 5, the solution\\nof an example under Case 2, 4, or 6, may be made to depend\\nupon the solution of an example under Case 1, 3, or 5,\\nrespectively and vice versa.\\nHence, it is not essential to consider more than three cases\\nin the solution of oblique spherical triangles.\\nThe student must carefully bear in mind the remarks\\nmade in 134 and 135.\\n149. Case I. Given a side and the adjacent angles.\\n1. Given A 70\u00c2\u00b0, B 131\u00c2\u00b0 20 c 116\u00c2\u00b0; find a, b, and O.\\nBy Napier s Analogies 144, 145), we have\\nsm^(B-\\\\-A)_ tan|c cos^(_S J.)_ tanic\\nsini(^ tan 1(6\u00e2\u0080\u0094 a) cosl(B\u00e2\u0080\u0094A) tan 1(6 a)\\nWhence, tan (6 a) sin ^(B A) esc (_B J.) tan ic,\\nand tan (6 a) cos \\\\{B A) sec \\\\{B A) tan i c.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0137.jp2"}, "138": {"fulltext": "128 spherical Trigonometry.\\nFrom the data, l(B- A)= 30\u00c2\u00b0 40 l(B+ A)= 100\u00c2\u00b0 40 c 58\u00c2\u00b0.\\nlog sin 1 9.7076 10 log cos i 5 9.9346 10\\nlog CSC 1{B A)= 0.0076 log sec ^B A) 0.7326\\nlog tan 1 c 0. 2042 log tan i c 0. 2042\\nlog tan K 9.9194 10 log tan K 0.8714\\n1(6 -a) =39\u00c2\u00b0 42.8 180\u00c2\u00b0 i (6 a)= 82\u00c2\u00b020.5\\n=97\u00c2\u00b0 39.5\\nThen, i (5 _ i (6 _ 57\u00c2\u00b0 56.7\\nand 5 3^ 1 (6 a) 1 (6 _ a) 137\u00c2\u00b0 22.3\\nTo find C, we have by 146,\\nsin i(6 a)^ ._\\ncot 1 O ^tan UB A)\\nsm 1 (6 a)\\nsini(6 a) esc a) tani(5 A).\\nlog sin 1 (6 9.9961 10\\nlog CSC i (6 a) =0.1946\\nlogtani(5-Yl)= 9.7730- 10\\nlogcoti(7 9.9637 10\\n1(7 47\u00c2\u00b0 23.6 and C= 94\u00c2\u00b0 47.2\\nNote 1. The value of C may also be determined by the formula\\ncoti(7=:-^^f4^^tani(5 (\u00c2\u00a7146).\\ncos 1(6 a)\\nNote 2. The triangle is alv^ays possible for any values of the\\ngiven elements.\\nEXAMPLES.\\nSolve the following spherical triangles\\n2. Given A 87\u00c2\u00b0, B 61\u00c2\u00b0, c 112\u00c2\u00b0.\\n3. Given 5 41\u00c2\u00b0, 0=122\u00c2\u00b0, a 37\u00c2\u00b0.\\n4. Given A 135\u00c2\u00b0, O 51\u00c2\u00b0, b 69\u00c2\u00b0.\\n5. Given 147\u00c2\u00b0 30 B 163\u00c2\u00b0 10 c 76\u00c2\u00b0 20\\n(For additional examples under Case I., see 155.)", "height": "3708", "width": "2458", "jp2-path": "completetrigonom00well_0138.jp2"}, "139": {"fulltext": "Oblique Spherical Triangles. 129\\n150. Case II. Given two sides and their included angle.\\n1. Given 6=: 137\u00c2\u00b0 20 c=116\u00c2\u00b0, ^-70\u00c2\u00b0; find 5, O, and a.\\nBy Napier s Analogies 146), we have\\nsini(6 c)_ cot^^ cosi(6 c)_ coti^\\nsini( -c) tani(S-C) cosi(6-c) tan 1(5+0)\\nWhence, tan \\\\{B C) sin |(6 c) esc ^(6 c) cot 1 A,\\nand tan \\\\{B C)= cos l(b c) sec J c) cot 1 A.\\nFrom the data, 1 (6 c) 10\u00c2\u00b0 40 1 (6 c) 126\u00c2\u00b0 40 ^A 35\u00c2\u00b0.\\nlog sin 1(6 c) 9.2674 10 log cos i( -c) 9.9924 10\\nlog esc 1(6 c) 0.0958 log sec J (6 c) 0.2239\\nlog cot 1^=0. 1548 log coti^ 0.1548\\nlogtan^(J?- (7) =9.5180 -10 logtani(5+ 0=0.3711\\nK-B- 0)= 18\u00c2\u00b0 14.5 180\u00c2\u00b0 -1(5+ 66\u00c2\u00b0 57.1\\n1(5+ C)= 113\u00c2\u00b0 2.9\\nThen, 5 i(^ C K^ 131\u00c2\u00b0 17.4\\nand C=l(B+ C)-U^- C)= 94\u00c2\u00b0 48.4\\nTo find a, we have by 144,\\ntania ^Hlii^^t_^tani( -c).\\nsin 1(5-0)\\nlog sin i(^ C) 9.9639 10\\nlog CSC K^- 0)= 0.5044\\nlog tan i( c) 9.2750 10\\nlogtania 9.7433 10\\nla 28\u00c2\u00b0 58.3 and a =57\u00c2\u00b0 56.6\\nNote. The triangle is possible for any values of the given elements,\\nEXAMPLES.\\nSolve the following spherical triangles\\n2. Given a 64\u00c2\u00b0, 34\u00c2\u00b0, C 48\u00c2\u00b0.\\n3. Given 6 42\u00c2\u00b0, c 96\u00c2\u00b0, 110\u00c2\u00b0.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0139.jp2"}, "140": {"fulltext": "ijo Spherical Trigonometry.\\n4. Given a 146\u00c2\u00b0, c 69\u00c2\u00b0, B 125\u00c2\u00b0.\\n5. Given a 90\u00c2\u00b0 50 6 117\u00c2\u00b0 50 (7=120\u00c2\u00b0.\\n(For additional examples under Case II., see 155.)\\n151. Case III. Given the three sides.\\nThe angles may be calculated by the formnlse of 142.\\nIf all the angles are to be computed, the tangent formulae\\nare the most convenient, since only four different angles\\noccur in the second members.\\nIf but one angle is required, the cosine formula involves\\nthe least work.\\nThe triangle is possible for any values of the data, pro-\\nvided that no side is greater than the sum of the other two,\\nand that the sum of the sides is less than 360\u00c2\u00b0 119, 1\\nand 2).\\nIf all the angles are required, and the tangent formulae\\nare used, it is convenient to modify them as follows.\\nBy (91),\\ntan i J. r^^\\nsin s sin^ (s a)\\n1 sin (s a) sin (s b) sin (s \u00e2\u0080\u0094c)\\nsin (s a) sin s\\nDenoting J ^m sm sin c)\\nSin .Q\\ntan ^A\\nsm\\nsin (s a)\\nk k\\nSimilarly, tan IB and tan \\\\C= 7\\nsm (s 6) sm (s c)\\n1. Given a 57\u00c2\u00b0, b 137\u00c2\u00b0, c 116\u00c2\u00b0 find A, B, and C.\\nHere, 2 s a b c S\\\\0\u00c2\u00b0.\\nWhence, s 155\u00c2\u00b0, s~ a 98\u00c2\u00b0, s-b 18\u00c2\u00b0, s-c 39\u00c2\u00b0.", "height": "3681", "width": "2457", "jp2-path": "completetrigonom00well_0140.jp2"}, "141": {"fulltext": "Oblique Spherical Triangles. 131\\nlog sin(s a) 9.9958 10 log k 9.8294 10\\nlog sin(s 6) 9.4900 10 log sin(s -b)= 9.4900 10\\nlog sin (s c)= 9.7989- 10\\nlog CSC s 0.3741\\nlog\\ntan 1 5\\n0.3394\\ni^\\n66\u00c2\u00b0 24.2\\nB\\n130\u00c2\u00b0 48.4\\n\\\\ogk\\n9.8294\\n10\\nlog sin (s c)\\n9.7989\\n10\\nlog\\ntan 1 C\\n0.0305\\n^c\\n47 0.8\\nC\\n94\u00c2\u00b0 1.6\\n2 )19.6588 20\\nlogA; 9.8294- 10\\nlog sin(s a)= 9.9958 10\\nlog tan 1^ 9.8336 10\\n1^ 34\u00c2\u00b0 17.0\\n68\u00c2\u00b0 34.0\\nEXAMPLES.\\nSolve the following spherical triangles\\n2. Given a 69\u00c2\u00b0, 6 74\u00c2\u00b0, c 63\u00c2\u00b0.\\n3. Given a 103\u00c2\u00b0, 6 53\u00c2\u00b0, c 61\u00c2\u00b0.\\n4. Given a 91\u00c2\u00b0, b li8\u00c2\u00b0, c 132\u00c2\u00b0.\\n5. Given a o8\u00c2\u00b0, 6 138\u00c2\u00b0, c 116\u00c2\u00b0; find A.\\n(For additional examples under Case III., see 155.)\\n152. Case IV. Given the three angles.\\nThe sides may be calculated by the formulse of 143.\\nIf all the sides are to be computed, the tangent formulae\\nare the most convenient, since only four different angles\\noccur in the second members.\\nIf but one angle is required, the sine formula involves the\\nleast work.\\nThe triangle is possible for any values of the data, pro-\\nvided that the sum of the angles is between 180\u00c2\u00b0 and 540\u00c2\u00b0\\n119, 3), and that each of the quantities B C A,\\nC A-B, and A-\\\\-B-C is between 180\u00c2\u00b0 and -180\u00c2\u00b0\\n122).\\nFor such values of the angles, S is between 90\u00c2\u00b0 and 270\u00c2\u00b0,\\nand each of the quantities S \u00e2\u0080\u0094A, S \u00e2\u0080\u0094B, and S C between\\n90\u00c2\u00b0 and 90\u00c2\u00b0.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0141.jp2"}, "142": {"fulltext": "132 Spherical Trigonometry.\\nThen, cos jS is while the cosines of S A, S B, and\\n_ O are 21).\\nHence, the expressions under the radical signs in the\\nformulae are essentially positive, and no attention need be\\npaid to the algebraic signs.\\nIf all the sides are required, and the tangent formulae are\\nused, it is convenient to modify them as follows\\nBy (103),\\nI _ I cos S cos^ (S A)\\n/Qf A\\\\ I COS xS\\nCOS (/iS a/\\nCOS COS S 5) COS 0)\\ncos(S-A)J(S-B)cos(8-C)\\ntan \\\\a /i cos {S A).\\nIn like manner,\\ntan \\\\h Kgos {S B), and tan \\\\c Kcos (S C).\\n1. Given A 150 B lSr, (7=115\u00c2\u00b0; find a, b, and c.\\nHere, 2S A B+ C 396\u00c2\u00b0.\\nWhence, S 198\u00c2\u00b0, S- A 48\u00c2\u00b0, S- B 67\u00c2\u00b0, S-C= 83\u00c2\u00b0.\\nlog cos aS 9.9782 10 log ^=0.7375\\nlog sec (/S 0.1745 lOg cos (S -B)= 9.5919 10\\nlog sec J5) 0.4081 log tan 1 0.3294\\nlog sec (a9-0)= 0.9141 ^6 64\u00c2\u00b0 54.2\\n2)JaU9__ 129\u00c2\u00b0 48.4\\nlog K 0. 7375 loo- K 0. 7375\\nlog cos (/S-^)^ 9.8255 10 log cos{S- C)= 9.0859 10\\nlog tan 1 a 0. 5630 log tan 1 c 9.8234 10\\nla 74\u00c2\u00b0 42.2 ic 33\u00c2\u00b039.6\\na 149\u00c2\u00b0 24.4 c 67\u00c2\u00b0 19.2", "height": "3708", "width": "2460", "jp2-path": "completetrigonom00well_0142.jp2"}, "143": {"fulltext": "Oblique Spherical Triangles. 133\\nNote 1. By 84, cos 198\u00c2\u00b0 sin 108\u00c2\u00b0 cos 18\u00c2\u00b0 whence, with-\\nout regard to algebraic sign, log cos 198\u00c2\u00b0 log cos 18\u00c2\u00b0.\\n2. Given 123\u00c2\u00b0, 5-45\u00c2\u00b0, 0=58\u00c2\u00b0; find a.\\nBy (97) sin 1 a V-\\nsm B sm C\\nHere,\\n2S=A-^B-h C= 226\u00c2\u00b0 whence, S 113\u00c2\u00b0, and ^S 10\u00c2\u00b0.\\nlog cos ^r= 9.5919-10\\nlog cos aS 9. 9934 10\\nlog CSC 0.1505\\nlog CSC C 0.0716\\n2 )19.8074-20\\nlog sin I a 9.9037 10\\n1 a 53\u00c2\u00b0 14.4 and a 106\u00c2\u00b0 28.8\\nNote 2. By 28, cos 10\u00c2\u00b0) cos 10\u00c2\u00b0.\\nEXAMPLES.\\nSolve the following spherical triangles\\n3. Given A 52\u00c2\u00b0, B 59\u00c2\u00b0, C 83\u00c2\u00b0.\\n4. Given A 143\u00c2\u00b0, B 28\u00c2\u00b0, C 32\u00c2\u00b0.\\n5. Given J[ 142\u00c2\u00b0, 5 159\u00c2\u00b0, 0=133\u00c2\u00b0.\\n6. Given A 70\u00c2\u00b0, B 122\u00c2\u00b0, O 95\u00c2\u00b0 find b,\\n(For additional examples under Case IV., see 155.)\\n153. Case V. Given two sides and the angle opposite to\\none of them.\\n1. Given a 58\u00c2\u00b0, h 138\u00c2\u00b0, B 134\u00c2\u00b0 50 find A, C, and c.\\nT. /r.\u00c2\u00abN sin sin a 7 -r\\nBy (79) or sm A sina esc b sni B.\\nsm B sin\\nlog sin a 9.9284 10\\nlog CSC 0.1745\\nlog sin B 9.8507 10\\nlogsin^ 9.9536 10\\nA 63\u00c2\u00b0 58.6 or 116\u00c2\u00b0 1.4 135).", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0143.jp2"}, "144": {"fulltext": "134 Spherical Trigonometry.\\nTo find C and c, we have by 144 and 146,\\ncot J O sin ^(6 a) esc J(6 a) tan \\\\{B A),\\nand tan J c sin ^(B ^4) esc ^(B A) tan J(6 a).\\nUsing the first value of A^\\n^{B -{-A)= Q9\u00c2\u00b0 24.3 and ^(B -A) 35\u00c2\u00b0 25.7^\\nAlso, ^{b -j-a)= 98\u00c2\u00b0, and ^(b -a)= 40\u00c2\u00b0.\\nlog sin iib a)= 9.9958 10 log sin ^(B A)= 9.9941 10\\nlog esc K a) 0.1919 log esc (J5 0. 2368\\nlog tan ^(B A)= 9.8521 10 log tan a) 9.9238 10\\nlog cot O 0.0398 log tan c 0. 1547\\nJ (7= 42\u00c2\u00b0 22. 7 ^c 54\u00c2\u00b0 59.6\\n(7 =84\u00c2\u00b0 45.4 c 109\u00c2\u00b0 59.2\\nUsing the second value of A,\\n^{B A)= 125\u00c2\u00b0 25.7 and ^{B 9\u00c2\u00b0 24.3\\nlog sin J(5 a) =9.9958- 10 log sin ^(5 ^)r= 9.9111 10\\nlog CSC ^(b -a)= 0.1919 log esc ^{B A)= 0.7867\\nlog tan ^{B A)= 9.2192 10 log tan ^{b a) 9.9238 10\\nlog cot J O 9.4069 10 logtanjc 0.6216\\nJ a 75\u00c2\u00b0 40.9 Jc 76\u00c2\u00b033.6\\n0=151\u00c2\u00b0 21. 8 c 153\u00c2\u00b0 7.2\\nThus the two solutions are\\n1. 63\u00c2\u00b0 58.6 C 84\u00c2\u00b0 45.4 c 109\u00c2\u00b0 59.2\\n2. J. 116\u00c2\u00b0 1.4 C= 151\u00c2\u00b0 21. 8 c 153\u00c2\u00b0 7.2\\nAs in the corresponding case in the solution of plane\\noblique triangles (compare 108 and 109), there may\\nsometimes be two solutions, sometimes only one, and some-\\ntimes none, in an example under Case Y.\\nAfter the two values of A have been obtained, the num-\\nber of solutions may be determined by inspection for, by\\n119, 6, if a is 6, must be 5 and if a is 6, must\\nbe 5.", "height": "3681", "width": "2453", "jp2-path": "completetrigonom00well_0144.jp2"}, "145": {"fulltext": "Oblique Spherical Triangles. 135\\nHence, only those values of A can he retained which are\\ngreater or less than B according ccs a is greater or less than b.\\nThus, in Ex. 1, a is given b and since both values of\\nA are B, we have two solutions.\\nAgain, if the data are such as to make log sin positive,\\nthere will be no solution corresponding.\\n2. Given a 58\u00c2\u00b0, c 116\u00c2\u00b0, 0=94\u00c2\u00b0 50 find A\\nIn this case, or sin sin a esc c sin C.\\nsin C sin c\\nlog sin a =9.9284 10\\nlog CSC c =0.0463\\nlogsin 0=9.9985- 10\\nlogsin^ 9.9732 -10\\n70\u00c2\u00b0 5.0 or 109\u00c2\u00b0 55.0\\nSince a is given c, only values of A which are C can be re-\\ntained then the only solution is J. 70\u00c2\u00b0 5.0\\n3. Given b 126\u00c2\u00b0, c 70\u00c2\u00b0, 5 57\u00c2\u00b0 find C.\\nIn this case, 5i5_^ ^^5L5, or sin C sin c esc b sin B.\\nsin B sin b\\nlog sine =9.9730- 10\\nlog CSC =0.0920\\nlogsin5 9.9236- 10\\nlog sin (7 9.9886 10\\n(7= 76\u00c2\u00b0 56.7 or 103\u00c2\u00b0 3.3\\nSince both values of C are B, while c is given there is no\\nsolution.\\nEXAMPLES.\\nSolve the following spherical triangles\\n4. Given a 29\u00c2\u00b0, 6 14\u00c2\u00b0, 49\u00c2\u00b0.\\n5. Given a 98\u00c2\u00b0, c 36\u00c2\u00b0, C 163\u00c2\u00b0.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0145.jp2"}, "146": {"fulltext": "136 Spherical Trigonometry.\\n6. Given 6 132\u00c2\u00b0, c=:56\u00c2\u00b0, 116\u00c2\u00b0 18\\n7. Given a 104\u00c2\u00b0 50 c 153\u00c2\u00b020 70\u00c2\u00b0.\\n8. Given a 111\u00c2\u00b0 20 b 41\u00c2\u00b0 40 B 25\u00c2\u00b0.\\n(For additional examples under Case V., see 155.)\\n154. Case VI. Given tivo angles and the side opposite to\\none of them.\\n1. Given 110\u00c2\u00b0, 5 131\u00c2\u00b0 20 5 137\u00c2\u00b0 20 find a, c,\\nand G.\\nIn this case, ^^5^, or sin a sin esc B sin h.\\nsin h sin B\\nlogsinvl 9.9730- 10\\nlogcscJ5 0.1244\\nlog sin =9.8311 -10\\nlog sin a =9.9285- 10\\na 58\u00c2\u00b0 1.2 or 121\u00c2\u00b0 58.8\\nTo find c and C, we have by 144 and 146,\\ntanic sin 1(5+ A) esc |(5 tan^ (6 a),\\nand cot i C sin \\\\(h a) esc \\\\{h a) tan \\\\{B A).\\nUsing the first value of a,\\n97\u00c2\u00b0 40.6 and \\\\(h-a)= 39\u00c2\u00b0 39.4\\nAlso, \\\\{B A)= 120\u00c2\u00b0 40 and \\\\{B-A)= 10\u00c2\u00b0 40\\nlog sin 1{B A)= 9.9346 10 log sin 1 (6 a) 9.9961 10\\nlog CSC 1{B-A)^0. 7326 log esc 1 a) 0. 1951\\nlog tan 1 (6 a) 9.9185 10 log tan \\\\{B A)= 9.2750 10\\nlogtanic 0.5857 logcoti 0= 9.4662 10\\nic 75\u00c2\u00b026.9 1(7= 73\u00c2\u00b0 41.5\\nc 150\u00c2\u00b0 53.8 0=147\u00c2\u00b0 23.0\\nUsing the second value of a,\\nl(h a) 129\u00c2\u00b0 39.4 and \\\\{h a) 7\u00c2\u00b0 40.6", "height": "3707", "width": "2462", "jp2-path": "completetrigonom00well_0146.jp2"}, "147": {"fulltext": "Oblique Spherical Triangles. 137\\nlog sin ^{B A)= 9.9346 10 log sin 1 (6 9.8865 10\\nlog CSC i 0. 7326 log esc (5 a) 0. 8742\\nlog tan 1 (5 a) 9. 1297 10 log tan i(B-A)= 9.2750 10\\nlog tan 1 c 9.7969 10 log cot (7 0.0357\\nic 32=3.9 1(7 42\u00c2\u00b0 38.8\\nc 64\u00c2\u00b07.8 85\u00c2\u00b0 17.6\\nThus the two solutions are\\n1. a 58\u00c2\u00b0 1.2 c 150\u00c2\u00b0 53.8 O 147\u00c2\u00b0 23.0\\n2. a 121\u00c2\u00b0 58.8 c 64\u00c2\u00b0 7.8 C 85\u00c2\u00b0 17.6\\nIn examples in Case VI., as in Case V., there may some-\\ntimes be two solutions, sometimes only one, and sometimes\\nnone.\\nAs in Case Y., only tJiose values of a can be retained which\\nare greater or less than b according as A is greater or less\\nthan B.\\nAlso, if log sin a is positive, the triangle is impossible.\\nEXAMPLES.\\nSolve the following spherical triangles\\n2. Given A 84\u00c2\u00b0, B 19\u00c2\u00b0, a 28\u00c2\u00b0.\\n3. Given B 159\u00c2\u00b0, C 36\u00c2\u00b0, b 9\u00c2\u00b0.\\n4. Given A 25\u00c2\u00b0 20 C 153\u00c2\u00b0 27 a 73\u00c2\u00b0 10\\n5. Given A 142\u00c2\u00b0 40 C 71\u00c2\u00b0 10 c 39\u00c2\u00b0 30\\n6. Given A 110\u00c2\u00b0, B 123\u00c2\u00b0 20 b 127\u00c2\u00b0.\\n(For additional examples under Case VI., see 155.)\\nMISCELLANEOUS EXAMPLES.\\n155. Solve the following spherical triangles\\n1. Given a 38\u00c2\u00b0, 6 51\u00c2\u00b0, c 42\u00c2\u00b0.\\n2. Given B 116\u00c2\u00b0, C 80\u00c2\u00b0, c 83\u00c2\u00b0.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0147.jp2"}, "148": {"fulltext": "ijS Spherical Trigonometry.\\n3. Given A 78\u00c2\u00b0, B 41\u00c2\u00b0, c 108\u00c2\u00b0,\\n4. Given 5 99\u00c2\u00b0 40 c 64\u00c2\u00b020 96\u00c2\u00b0 10\\n5. Given A 76\u00c2\u00b0, B 81\u00c2\u00b0, C 61\u00c2\u00b0.\\n6. Given A 62% C 102\u00c2\u00b0, a 64\u00c2\u00b0 30\\n7. Given a 72\u00c2\u00b0, 6 47\u00c2\u00b0, (7=33\u00c2\u00b0.\\n8. Given 133\u00c2\u00b0 50 ^=66\u00c2\u00b0 30 a 81\u00c2\u00b0 10\\n9. Given a 101\u00c2\u00b0, 5 49\u00c2\u00b0, c 60\u00c2\u00b0.\\n10. Given 5 135\u00c2\u00b0, C=50\u00c2\u00b0, a 70\u00c2\u00b0 20\\n11. Given a 162\u00c2\u00b0 20 6 15\u00c2\u00b0 40 5 125\u00c2\u00b0.\\n12. Given ^=138\u00c2\u00b0 20 5 31\u00c2\u00b0 10 (7=35\u00c2\u00b0 50\\n13. Given a 109\u00c2\u00b0 20 c 82\u00c2\u00b0, 107\u00c2\u00b0 40\\n14. Given A 132\u00c2\u00b0, B 140\u00c2\u00b0, 6 127^\\n15. Given ct 60\u00c2\u00b0, c 98\u00c2\u00b0, 5=110\u00c2\u00b0.\\n16. Given a 55\u00c2\u00b0, c 138\u00c2\u00b010 42\u00c2\u00b0 30\\n17. Given 61\u00c2\u00b0 40 (7 =140\u00c2\u00b0 20 c 150\u00c2\u00b020\\n18. Given a 61\u00c2\u00b0, 5 39\u00c2\u00b0, c 92\u00c2\u00b0.\\n19. Given a 40\u00c2\u00b0, 5 118\u00c2\u00b0 20 29\u00c2\u00b0 20\\n20. Given A 110\u00c2\u00b0, B 131\u00c2\u00b0, C 147\u00c2\u00b0.\\n21. Given a 115\u00c2\u00b0 20 c 146\u00c2\u00b020 (7 =141\u00c2\u00b0 10\\n22. Given B 73\u00c2\u00b0, C 81\u00c2\u00b0 20 b 122\u00c2\u00b0 40\\n23. Given ^=31\u00c2\u00b0 40 O=122\u00c2\u00b020 5 40\u00c2\u00b040\\n24. Given 5 108\u00c2\u00b0 30 c 40\u00c2\u00b050 (7 =39\u00c2\u00b0 50\\n25. Given 6 120\u00c2\u00b0 20 c 70\u00c2\u00b040 50\u00c2\u00b0.\\n26. Given B 22\u00c2\u00b0 20 C 146\u00c2\u00b0 40 c 138\u00c2\u00b0 20", "height": "3681", "width": "2445", "jp2-path": "completetrigonom00well_0148.jp2"}, "149": {"fulltext": "Applicarions.\\n^39\\nXII. APPLICATIONS.\\n156. Shortest Distance between Two Points on the Sur-\\nface of the Earth.\\nIn problems concerning navigation, the earth may be\\nregarded as a sphere.\\nThe shortest distance between any two points on the sur-\\nface is the arc of a great circle which joins them; the angles\\nbetween this arc and the meridians of the points determine\\nthe hearings of the points from each other.\\nThus, if Q and R are the points, and PQ and PR their\\nmeridians, Z. PQR determines the bearing of R from Q, and\\nZ PRQ the bearing of Q from R,\\nIf the latitudes and longitudes of Q and R are known,\\nthe arc QR and angles PQR and PRQ may be determined\\nby the solution of a spherical triangle.\\nFor if EE is the equator, and PG the meridian of\\nGreenwich,\\nZ QPR Z RPG Z QPG longitude R longitude Q.\\nAlso, PQ PE -QE 90\u00c2\u00b0 latitude Q,\\nand PR PE RE 90\u00c2\u00b0 latitude R.\\nThus, in spherical triangle PQR, two sides and the in-\\ncluded angle are known, and the remaining elements may\\nbe computed.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0149.jp2"}, "150": {"fulltext": "140 Spherical Trigonometry.\\nWhen QB has been found in degrees, its length in miles\\nmay be calculated by finding its ratio to 360\u00c2\u00b0, and multiply-\\ning the result by the length of the circumference of a great\\ncircle in the following problems, the radius of the earth is\\ntaken as 3956 miles.\\nEXAMPLES.\\n157. In each of the following examples, find the shortest\\ndistance in miles between the places named, and the bearing\\nof each from the other\\n1. Havana (lat. 23\u00c2\u00b0 9 N., Ion. 82\u00c2\u00b0 23 W.), and Gibraltar\\n(lat. 36\u00c2\u00b0 9 N., Ion. 5\u00c2\u00b0 21 W.).\\n2. Batavia (lat. 6\u00c2\u00b0 8 S., Ion. 106\u00c2\u00b0 bT E.), and San Fran-\\ncisco (lat. 37\u00c2\u00b0 48 N., Ion. 122\u00c2\u00b0 24 W.).\\n3. Vera Cruz (lat. 19\u00c2\u00b0 12 N., Ion. 96\u00c2\u00b0 9 W.), and Cape\\nof Good Hope (lat. 34\u00c2\u00b0 22 S., Ion. 18\u00c2\u00b0 29 E.).\\n4. Auckland (lat. 36\u00c2\u00b0 51 S., Ion. 174\u00c2\u00b0 50 E.), and Callao\\n(lat. 12\u00c2\u00b0 4 S., Ion. 77\u00c2\u00b0 13 W.).\\n5. Boston lies in lat. 42\u00c2\u00b0 20 N., Ion. 71\u00c2\u00b0 4 W., and Glas-\\ngow in lat. 55\u00c2\u00b0 52 N., Ion. 4\u00c2\u00b0 16 W. In what latitude does\\na great circle course from Boston to Glasgow cross the meri-\\ndian of 40\u00c2\u00b0 W.\\n6. Yokohama lies in lat. 35\u00c2\u00b0 27 N., Ion. 139\u00c2\u00b0 41 E., and\\nCape Horn in lat. 59 S., Ion. 67\u00c2\u00b0 16 W. In what lati-\\ntude does a great circle course from Yokohama to Cape\\nHorn cross the meridian of 160\u00c2\u00b0 W.\\n158. The Astronomical Triangle.\\nLet be the position of an observer on the surface of\\nthe earth P the celestial north-pole Z the zenith.\\nThe great circle EE having P for its pole, is called the\\ncelestial equator; and the great circle HH having Z for its\\npole, is called the horizon.", "height": "3701", "width": "2447", "jp2-path": "completetrigonom00well_0150.jp2"}, "151": {"fulltext": "Applications.\\n141\\nLet S be the position of a star PSM a meridian through\\nS ZSJSf a quadrant of a great circle through Z and S.\\nThe arc SM is called the declination of the star it is\\ncalled declination north or south, according as the star is\\nnorth or south of the celestial equator.\\nThe angle SPZ is called the hour-angle of the star the\\narc SN, its altitude; the angle PZS, its hearing or azimuth.\\nThe arc EZ is the latitude of the observer.\\nThe spherical triangle SPZ is called the Astronomical\\nTriangle.\\nIts sides have the following values\\nSP=PM- jSM= 90\u00c2\u00b0 the declination of the star\\nSZ ZN SN 90\u00c2\u00b0 the altitude of the star\\nPZ=EP-EZ 90\u00c2\u00b0 the latitude of the observer.\\nIts angle SPZ is the hour-angle of the star, and its angle\\nSZP the azimuth.\\nIf any three of these five elements are known, the solu-\\ntion of a spherical triangle serves to determine the other\\ntwo.\\n159. Determination of Longitude and Time.\\nIf the altitude and declination of the sun are known, and\\nthe latitude of the observer, the three sides of triangle SPZ\\nare known, and the hour-angle SPZ may be computed.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0151.jp2"}, "152": {"fulltext": "142 spherical Trigonometry.\\nIf 24 hours be multiplied by the ratio of this angle to\\n360\u00c2\u00b0, we obtain the time required for the sun to move from\\nS to the meridian EP.\\nIf this time be subtracted from 12 o clock, if the observa-\\ntion is made in the morning, or added, if made in the after-\\nnoon, we obtain the Jiour of the day at the time and place of\\nobservation.\\nIf the Greenwich time of the observation be noted on a\\nchronometer, the difference between this and the local time\\nas calculated above serves to determine the longitude of the\\nplace of observation.\\nIn reducing time to longitude, it should be remembered\\nthat 24 hours of time correspond to 360\u00c2\u00b0 of longitude that\\nis, one hour of time corresponds to 15\u00c2\u00b0 of longitude, one\\nminute to 15 and one second to 15\\nEXAMPLES.\\n160. 1. At a certain place in latitude 40\u00c2\u00b0 N., the altitude\\nof the sun was found to be 41\u00c2\u00b0. If its declination at the\\ntime of the observation was 20\u00c2\u00b0 N., and the observation was\\nmade in the morning, how long did it take the sun to reach\\nthe meridian\\n2. A mariner observes the altitude of the sun to be 60\u00c2\u00b0,\\nits declination at the hour of observation being 6\u00c2\u00b0 IST. If\\nthe latitude of the vessel is 12\u00c2\u00b0 S., and the observation is\\nmade in the morning, find the hour of the day. If the\\nobservation is taken at 11.40 a.m., Greenwich time, find the\\nlongitude of the vessel.\\n(In this case, the side PZ of the astronomical triangle is 90\u00c2\u00b0 plus\\n12\u00c2\u00b0.)\\n3. At what hour will the sun set in Montreal (lat. 45\u00c2\u00b0\\n30 N.), if its declination at sunset is 18\u00c2\u00b0 N.\\n(At sunset, the sun s altitude is 0\u00c2\u00b0, so that the side 8Z of the\\nastronomical triangle becomes 90\u00c2\u00b0.", "height": "3681", "width": "2562", "jp2-path": "completetrigonom00well_0152.jp2"}, "153": {"fulltext": "Applications. 143\\n4. A mariner observes the altitude of the sun to be 35\u00c2\u00b0\\n23 its declination being 10\u00c2\u00b0 48 S. If the latitude of the\\nvessel is 26\u00c2\u00b0 13 N., and the observation is made in the\\nafternoon, find the hour of the day. If the observation is\\ntaken at 7.13 p.m., Greenwich time, find the longitude of\\nthe vessel.\\n5. At what hour will the sun rise in Panama (lat. 8\u00c2\u00b0 57\\nN.), if its declination at sunrise is 23\u00c2\u00b0 2 S.\\n6. What will be the bearing of the sun at 4 p.m. in Kio\\nJaneiro (lat. 22\u00c2\u00b0 54 S.), if its declination at that time is\\n3\u00c2\u00b0 S.\\n7. What will be the bearing of the sun at sunrise in\\nBoston (lat. 42\u00c2\u00b021 N.), if its declination at that time is\\n13\u00c2\u00b024 K?\\n8. What will be the altitude of the sun at 9 a.m. in\\nMexico (lat. 19\u00c2\u00b0 25 N.), if its declination at that time is\\n8\u00c2\u00b0 23 N.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0153.jp2"}, "154": {"fulltext": "144 Plane Trigonometry.\\nFORMULAE.\\nPLANE TRIGONOMETRY.\\n\u00c2\u00a728. sin(\u00e2\u0080\u0094 ^1) sin A cos(\u00e2\u0080\u0094 cos J.\\ntan ^4) tan J.. cot ^4) cot A\\nsec(\u00e2\u0080\u0094 J_)= sec A esc J.) esc A\\n29.\\nsin(90\u00c2\u00b0-h^)= cos A cos (90\u00c2\u00b0 sin A\\ntan(90\u00c2\u00b0 cot A. cot (90\u00c2\u00b0 yl) tan A.\\nsec (90\u00c2\u00b0 -CSC A esc (90\u00c2\u00b0 -4)= sec A\\n\u00c2\u00a735.\\n1 1 1\\nsec a;\\nsin X\\n1\\nCSC a?\\ntan X\\ncot a;\\ncosa^\\n1\\nsecic\\ncot \u00c2\u00a3C 7\\ntana;\\n\u00c2\u00a736.\\nsin X\\ncosa;\\n\u00c2\u00a737.\\ncos a;\\nsma;\\n\u00c2\u00a738.\\nsin^a; cos2a^ 1.\\n\u00c2\u00a740.\\nsec^ a; 4 tan^\\ncsc^ a^ 4 cot^\\nCSC a;\\ncos a;\\n4\\nsin a;\\n(1)\\n(2)\\n(3)\\n(4)\\n(5)\\n(6)\\n(8)\\n\u00c2\u00a741. sin (a; sin a; cos cos a^ sin y. (9)\\ncos (a; y) cos x cos y sin x sin y. (10)\\n43. sin (x y) sin x cos y cos qi sin y. (11)\\ncos (x y) cos x cos y sin x sin 2/. (12)\\nX tan X tan v /_ _x\\n\u00c2\u00a744 tan(x (13)\\ntan X tan\\n4 tan a; tan y", "height": "3707", "width": "2453", "jp2-path": "completetrigonom00well_0154.jp2"}, "155": {"fulltext": "Formulae. 145\\ncot X cot y r-,\\ncot {x^y) (15)\\ncotiy cotx\\ncoti\u00c2\u00abcot?/ l\\ncot (x y) (16)\\ncot?/ cot cc\\n\u00c2\u00a745. sin ;K 4- sin 2sini(a; ?/)cos^(a; (17)\\nsin X sin 2/ 2 cos {x /y) sin (x y). (18)\\ncos X -f cos 2 cos (;c cos ^{x (19J\\ncos X cos y 2 sin i (a.- sin \\\\(x y). (20)\\ng sin a; sin y tan|(x y) /g^x\\nsin a. sin 1/ tani(a: y)\\n\u00c2\u00a747.\\nsin 2 a? 2 sin x cos ic. (22) cos 2 x cos^ ic sin^ x. (23)\\ncos 2 X 1 2 sin^ x.\\n(24)\\ncos2a. 2cos^a7 1.\\n(25)\\ntan2a.=\\nl-tan^x\\n\u00c2\u00a748.\\n(26)\\n2 cot a;\\n(27)\\n2siTi^^x=l coscc.\\n(28)\\n2cos2-ia. =l cosa.\\\\\\n(29)\\n1 1 cos X\\ntania^=:\\nsmx\\n\u00c2\u00a797.\\n(30)\\ncotU-=^+\\nsm X\\n(31)\\n4 ^=c- sin 2 A\\n(32)\\n4.K=G sm2B,\\n(33)\\n2/ir=a2cotA\\n(34)\\n2 K=b cot B.\\n(35)\\n2A^=aHan5.\\n(36)\\n2K=l) td.nA.\\n(37)\\n2 K= a^\\\\c a){c- a). (38) 2 K= bV(c b)(c-b). (39)\\n2 j5r= ab, (40)\\n99. a 6 sin sin B. (41)\\nc sin 5 sin C. (42)\\nc a sin C sin A. (43)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0155.jp2"}, "156": {"fulltext": "146 Plane Trigonometry.\\na-b tanK^-5)\\n6 _ tan i (5 0)\\n6-c~tani(^-C)*\\n5c\\n(45)\\nc-a tani((7-^)\\n101. a^ 52 c^ 2 6c cos A. (47)\\n52 c^ a^ 2 ca cos J5. (48)\\nc2 a^ 52 _ 2 ct^ cos a (49)\\n\u00c2\u00a7102. cos^^^^y^. (50)\\ncos jB (51)\\n2 ca\\ncos G (52)\\n2a6\\n\u00c2\u00a7103. si\u00e2\u0080\u009e|^=^(i^4^. (53)\\n1 T. /(s c) (s a)\\nsmiB=yl^ (54)\\nsm i O (55)\\ncosiA=\\\\ (56)\\ncosi5=V^-\\n-^*^=V^-", "height": "3700", "width": "2448", "jp2-path": "completetrigonom00well_0156.jp2"}, "157": {"fulltext": "Formulae. 147\\nj(s b)(s c)\\ntani^ (59)\\n1 -r^ (s c)(s a)\\nt^-ic=^l^\\n)(s-b)\\ns (s c)\\n\u00c2\u00a7124.\\nta\\n\u00c2\u00a7125.\\nsin-B _ sin 5\\nsin C sin c\\n(61)\\n^04.\\n-r^ a- sin B sin (7\\n2^=:6csinA (62) 2K= (65)\\nsm^\\no D /^ox o T^ 6^ sin C sin J.\\n2A^casin5. (63) 2/f= (66)\\nsin5\\n_. ^__ -f /_ r-k TT- c sm sm x\\n2K=ab^mC. (64) 2^= (67)\\nsm (7\\n/iT Vs (s a) (s 5) (s c). (68)\\nSPHERICAL TRIGONOMETRY.\\n123. cos c cos a cos 5. (69)\\nA sin a /_-,x -D sin\\nsm (70) sm B (72)\\nsm e sm c\\ntan 6 ,__. tana .__^\\nC0SJ. (71) COS 5 (73)\\ntan c tan c\\ntan^ *^5^. (74) tan 5 (75)\\nsm h sin a\\ncos 5 x__s cos^\\nsm^ (76) sm5 (77)\\ncos 6 cos a\\n126. cos c cot ^4 cot B. (78)\\n\u00c2\u00a7139. sjn^^sina,\\nsin sm 5\\n(80)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0157.jp2"}, "158": {"fulltext": "148 Spherical Trigonometry.\\nsin C sin c\\nsm c sm a\\ncos 4-B\\nsm c sm a\\n(81)\\nsm^tL sin a\\n140. cos a cos h cos c -f- sin h sin c cos (82)\\ncos h cos c cos a sin c sin a cos (83)\\ncos c cos a cos 6 sin a sin 6 cos 0. (84)\\n\u00c2\u00a7141. cos cos 5 cos sin 5 sin (7 cos a. (85)\\ncos B cos C cos ^1 sin C sin cos h. (86)\\ncos C cos A cos 5 sin A sin 5 cos c. (87)\\n\u00c2\u00a7142. sml^^A H^^T^^ (88)\\nsin sin a\\n1 -D sin (s c) sin a) /__.\\nsm 1 B (89)\\n^1 sm sin nr,\\n1 /sm (s a) sm (s b) .^_x\\nimi(7 (90)\\nsm a sm o\\n1 ismssm(s a) ,\u00e2\u0080\u009es\\ncos i J. (91)\\nsm sm p.\\nkin s sin (s-b)^ (92)\\nsin f^ sin a\\nT /sm s sm (s c)\\ncos i (7 (93)\\nsm a sm\\nsm s sm (s a)\\ntan 1 g^ sm(\u00c2\u00ab-c)sm(s-\u00c2\u00ab) ^gg^\\nsin s sin (s b)\\ntan^C Jgi^ (96)\\nsm s sm (s c", "height": "3709", "width": "2461", "jp2-path": "completetrigonom00well_0158.jp2"}, "159": {"fulltext": "Formulae.\\n\u00c2\u00a7143.\\n\u00c2\u00a7144.\\n\u00c2\u00a7145.\\n\u00c2\u00a7146.\\nsm\\nsm\\n1 _ co s S cos (S A)\\nsin^sinO\\nJ. _ cos S cos (S B)\\nsin (7 sin\\nsm^c\\ncos cos 1^-0)\\nsin A sin B\\ncos\\nXa= cos B) cos S C)\\nsin 5 sin (7\\n/c_oK^-01cosl,S^^.\\nsin C sin\\nsin ^1 sin\\ntani\\ncos aS cos (S A)\\ngos(S-B)gos(S- C)\\ni7^_ I COS ^S cos {S\\n^^2 cos(^-(7)cos(^-^)\\ntan^c\\ncos /iS cos {S 0)\\ncos( S-^)cos( S- JB)\\nsin \\\\{A-]- B) _ tan c\\nsin i 5) tan (a b)\\ncos -i-\\nB)\\ntan i c\\nCOS (A\\n-B)\\ntan 1 (a 6)\\nsin (a\\n^)_\\ncot i C\\nsin i (a\\n-5)\\ntan i\\nCOS (a\\n)_\\ncot 10\\ncos ^(a\u00e2\u0080\u0094 b) tan i 5)\\n149\\n(97)\\n(98)\\n(99)\\n(100)\\n(101)\\n(102)\\n(103)\\n(104)\\n(105)\\n(106)\\n(107)\\n(108)\\n(109)", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0159.jp2"}, "160": {"fulltext": "", "height": "3681", "width": "2446", "jp2-path": "completetrigonom00well_0160.jp2"}, "161": {"fulltext": "ANSWERS.\\n54 pag-e 40.\\n13. 85\u00c2\u00b0 56 37.32 15. 95\u00c2\u00b0 29 34.8\\n14. 14\u00c2\u00b0 19 26.22 16. 22\u00c2\u00b0 55 5.952\\n72; pag-e 53.\\n2. 1.5441. 6. 2.1673. 10. 2.4592. 14. 3.3434.\\n3. 1.4771. 7. 2.3522. 11. 2.8363. 15. 3.8963.\\n4. 1.9912. 8. 2.2431. 12. 2.7023. 16. 3.7656.\\n5. 1.9242. 9. 2.6232. 13. 2.5741. 17. 4.1494.\\n74; page 54.\\n2. .1549. 5. 1.5229. 8. .5192. 11. 1.3734.\\n3. .2431. 6. .2273. 9. .6478. 12. .8942.\\n4. 1.6532. 7. 2.0212. 10. 2.7202. 13. 1.9842.\\n77 page 55.\\n3. 2.4080. 8. 2.2415. 13. .2510. 19. .9132.\\n4. .6036. 9. .0954. 14. .4095. 20. .1643.\\n5. 1.0485. 10. .1409. 16. .0409. 21. .3726.\\n6. 8.1160. 11. .0777. 17. .7264. 22. .1118.\\n7. .4704. 12. .3618. 18. .1511. 23. .8618.\\n81; page 57.\\n2. 0.3801. 5. 8.2831 10. 8. 8.3892 10. 11. 2.3043.\\n3. 1.2252. 6. 7.1303-10. 9. 6.6865-10. 12. 0.1459.\\n4. 9.9084 10. 7. 3.7693. 10. 9.0124 10. 13. 1.6505", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0161.jp2"}, "162": {"fulltext": "Answers.\\n4. 2.8878.\\n6. 3.0237.\\n6. 8.5177 10.\\n7. 9.7164 -10.\\n8. 1.3028.\\n9. 4.9659.\\n82 pag-e 58.\\n10. 7.6055- 10. 18. 1.646.\\n11. 6.8560 10. 19. 8886.\\n12. 0.7144. 20. 545.9.\\n13. 3.0155. 21. .01461.\\n14. 8.9379 10. 22. .003318.\\n15. 5.0610 10. 23. 102.2.\\n30. .00005029.\\n24.\\n9.493.\\n25.\\n.2079.\\n26.\\n44.48.\\n27.\\n.0001109.\\n28.\\n63330.\\n29.\\n.01301.\\n87; pages\\n61 1\\nbo 63.\\n1.\\n2.151.\\n16.\\n.002555.\\n31.\\n.3702.\\n48.\\n.2985.\\n2.\\n19.38.\\n17.\\n3692.\\n34.\\n13.83.\\n49.\\n.04477.\\n3.\\n3135.\\n18.\\n.2777.\\n35.\\n2.487.\\n50.\\n.7945.\\n4.\\n.09778.\\n19.\\n15890.\\n36.\\n1.056.\\n51.\\n1.805.\\n5.\\n.009213.\\n20.\\n.03162.\\n37.\\n.00002143.\\n52.\\n179.5.\\n6.\\n.1088.\\n21.\\n244.1.\\n38.\\n.007105.\\n53.\\n1.883.\\n7.\\n6.359.\\n22.\\n.002791.\\n39.\\n.6955.\\n54.\\n8894.\\n8.\\n.03017.\\n23.\\n.0000002373.\\n40.\\n.5428.\\n55.\\n1.344.\\n9.\\n3.119.\\n24.\\n2.236.\\n41.\\n36.03.\\n56.\\n.01335.\\n10.\\n1327.\\n25.\\n1.149.\\n42.\\n11.11.\\n57.\\n37.82.\\n11.\\n847.8.\\n26.\\n1.220.\\n43.\\n.9432.\\n58.\\n.00001146,\\n12.\\n.005421.\\n27.\\n1.778.\\n44.\\n2.627.\\n59.\\n.1782.\\n13.\\n1.205.\\n28.\\n.6683.\\n45.\\n2.534.\\n60.\\n4.698.\\n14.\\n.2357.\\n29.\\n.6458.\\n46.\\n1.795.\\n61.\\n.03402.\\n15.\\n11.54.\\n30.\\n.1378.\\n47.\\n1.032.\\n88; page 64.\\n9.8556 10.\\n9.9458 10.\\n0.6518.\\n9.9501 10.\\n0.8550.\\n9.7070 10.\\n9.7547 10.\\n8. 9.9535-\\n9. 0.7654.\\n10. 0.0420.\\n11. 83\u00c2\u00b0 5.2\\n12. 33\u00c2\u00b0 17.8\\n13. 46\u00c2\u00b0 40.9\\n14. 73\u00c2\u00b0 33.4\\n10.\\n15. 80\u00c2\u00b0 26.3\\n16. 31\u00c2\u00b0 20.4\\n17.\\n18.\\n19.\\n20.\\n21.\\n8\u00c2\u00b0 53.5\\n5\u00c2\u00b0 17.6\\n40\u00c2\u00b0 20.7\\n66\u00c2\u00b0 43.3\\n.2960.\\n22.\\n23.\\n24.\\n25.\\n26.\\n27.\\n.2482.\\n.7033.\\n.3886.\\n47\u00c2\u00b0 36.3\\n21\u00c2\u00b0 52.7\\n49\u00c2\u00b0 57.0\\n28\u00c2\u00b0 46.7", "height": "3681", "width": "2453", "jp2-path": "completetrigonom00well_0162.jp2"}, "163": {"fulltext": "Answers.\\n94; pag-es 67 to\\n1. a\\nh\\n3. a\\n4. A\\n12. a\\n13. A\\n14.\\n15.\\n16.\\n36.\\n1.812, 6.761.\\n12.38, c 13.35.\\n16.78, c 26.11.\\n34\u00c2\u00b0 22.2 b .5118.\\n32\u00c2\u00b0 44.4 c 49.92.\\n10.35, c 13.14.\\n.005916, b .01269.\\n39\u00c2\u00b0 49.1 a 488.7.\\n148.4, c 948.6.\\n49\u00c2\u00b0 55.0 c 4.457.\\n77.38, c 91.08.\\n3814, b 3651.\\n24\u00c2\u00b0 23.3 a .02126.\\n156.6, c 856.4.\\n.003607, b .008830\\n24840, c 36090.\\n949.8. 34. 34\u00c2\u00b0 36.7\\n1.491. 37. a .03446.\\n17.\\n18.\\n19.\\n20.\\n21.\\n22.\\n23.\\n24.\\n25.\\n26.\\n27.\\n30.\\n31.\\n32.\\nA\\nb\\n.4\\na\\na\\n6\\nA\\na\\nb\\na\\nA\\na\\na\\nA\\na\\n69.\\n55\u00c2\u00b0 44.8 c 4116.\\n.6441, c .6503.\\n76\u00c2\u00b0 34.0 a 2423.\\n.2072, b .4212.\\n5091, c 5268.\\n.8478, c 1.234.\\n39\u00c2\u00b0 22.0 121.2.\\n8.243, c 9.275.\\n.000005736, c .00002118.\\n.0006772, b .0003899.\\n43\u00c2\u00b0 45.7 6 66650.\\n30.51, b 18.59.\\n24540, c 30010.\\n60\u00c2\u00b0 14.1 c=. 007745.\\n25.40.\\n.2923.\\n35. c 4.488.\\n14.\\n53\u00c2\u00b0 31.8\\n15.\\n135.2 ft.\\n16.\\n5.036.\\n17.\\n53\u00c2\u00b0 8.1\\n96 pages 69 to 72.\\n2. 416.1 ft. 6. 23.26. 10. 15.27.\\n3. 651.8. 7. 285.1 ft. 11. 70.91 ft.\\n4. 34.07. 8. 1.235. 12. 121\u00c2\u00b0 0.8\\n5. 6\u00c2\u00b0 2.3 9. 52\u00c2\u00b0 4.2 13. 39\u00c2\u00b0 12.0\\n18. Perimeter, 3.1908; diameter circumscribed circle, 1.2278.\\n19. Eadius inscribed circle, 28.58 circumscribed, 30.94.\\n20. 740.2. 21. Height of cliff, 144.4 ft.; of lighthouse, 153.6 ft.\\n22. 1131.3 ft. 25. 17\u00c2\u00b0 1.6 28. 14.9 ft. 31. 3\u00c2\u00b0 43.9\\n23. 62.9 ft. 26. 18.68. 29. 1575 mi. 32. 195.9 ft.\\n24. 12\u00c2\u00b028.9 27. 229.02. 30. 109.0 mi. 33. 60.14 ft.\\n34. Bearing, S. 42\u00c2\u00b0 28.8 W.; distance, 17.77 mi.\\n3.564. 4. 13440.\\n.1098. 5. 12.64.\\n10. .000001323.\\n98 page 74.\\n6. 46.0. 8. .02036\\n7. .0004838. 9. 795.\\n11. 2840.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0163.jp2"}, "164": {"fulltext": "4 Answers.\\n105; pages 82, 83.\\n2. 6 282.9, c 268.5. 5. a 31.49, c 49.88.\\n3. a 3.384, c 9.828. 6. .5042, 6 =.3618.\\n4. a .02893, 6 .01825. 7. 5499, c 2959.\\n106; pag-e 84.\\n2. 118\u00c2\u00b0 18.0 6 44.73. 5. 76\u00c2\u00b0 12.9 c 6.362.\\n3. A 29\u00c2\u00b0 59.5 c 1419. 6. C 96\u00c2\u00b0 3.3 6 5141.\\n4. (7 88\u00c2\u00b0 34.8, a .4038. 7. 5= 146\u00c2\u00b0 26.3 a .01044.\\n107 page 86.\\n3. 44\u00c2\u00b0 25.0 5 78\u00c2\u00b0 28.0 0=57\u00c2\u00b0 7.4\\n4. 71\u00c2\u00b0 47.4 B 58\u00c2\u00b0 45.6 C 49\u00c2\u00b0 27.6\\n5. 29\u00c2\u00b0 55.2 5 22\u00c2\u00b0 31.4 0=127\u00c2\u00b0 34.4\\n6. 71\u00c2\u00b0 33.4 7. 31\u00c2\u00b0 6.6 8. 134\u00c2\u00b0 29.4\\n111 pag-e 90.\\n1. 5 48\u00c2\u00b0 32.7 c 8.522. 6. (7 90\u00c2\u00b0, 6 5939.\\n2. 29\u00c2\u00b0 21.4 a 102.2. 7. 14\u00c2\u00b0 5.4 6 .1435.\\n3. Impossible. 8. 25\u00c2\u00b0 31.9 c 278.4.\\n4. C 44\u00c2\u00b0 56.2 a 66.68; 9. (7 =46\u00c2\u00b0 21.7 6 .8728\\nor, (7= 135\u00c2\u00b0 3.8 a 12.89. or, C= 133\u00c2\u00b0 38.3 6 .2351.\\n5. Impossible.\\n112 pages 90, 91.\\n1. 61\u00c2\u00b0 25.7 c= 1018. 2. a 15.52, 6 10.29.\\n3. 44\u00c2\u00b0 37.8 5 =101\u00c2\u00b0 50.0 (7 33\u00c2\u00b0 33.4\\n4. 34\u00c2\u00b025.0 6 .7135. 6. (7 =14\u00c2\u00b0 57. 7 6 5074.\\n5. 46\u00c2\u00b0 2.1 a .05676. 7. a .0004395, c .0002092.\\n8. 51\u00c2\u00b0 52.6 5 42\u00c2\u00b0 59.0 (7 85\u00c2\u00b0 8.8\\n9. 5 34\u00c2\u00b0 7.8 c 1223; 11. (7 48\u00c2\u00b0 37.3 a 7.597.\\nor, B 145\u00c2\u00b0 52.2 c 269.3. 12. 6 55610, c 79270.\\n10. (7= 46\u00c2\u00b0 37.8 a =7577. 13. Impossible.\\n14. 54\u00c2\u00b0 5.4 6 .01073.\\n15. ^=72\u00c2\u00b0 43.8 5 =47\u00c2\u00b0 40.6 C= 59\u00c2\u00b0 36.6\\n16. 90\u00c2\u00b0, c .1379. 20. 27\u00c2\u00b034.2 c .01875.\\n17. Impossible. 21. 0= 134\u00c2\u00b0 36.9 6 273.3.\\n18. 6 .4392, c .4723. 22. B 63\u00c2\u00b0 58.6 a 25660\\n19. 95\u00c2\u00b0 22.2 c 38.25. or, i5 n6\u00c2\u00b01.4 a 7573.", "height": "3681", "width": "2545", "jp2-path": "completetrigonom00well_0164.jp2"}, "165": {"fulltext": "Answers^ 5\\n113; pages 91, 92.\\n2. 582. 7. 24530. 12. 10.28. 17. 7255.\\n3. 53.8. 8. .000003186. 13. 883.2. 18. 19.19.\\n4. 20.98. 9. .1682. 14. 34840. 19. 47210.\\n5. .0780. 10. .0000002941. 15. .03519. 20. .00003759.\\n6. 271.3. 11. 28.77. 16. .003042. 21. .000675.\\n114; pages 92 to 95.\\n1. 608.4 ft. 2. 420.0 sq.rd. 3. 525.8 ft. 4. 34\u00c2\u00b0 22.2 or 145\u00c2\u00b0 37.8\\n5. Height of tower, 212.8 ft.; distances, 236.4 ft., 436.4 ft.\\n6. Distance, 21.20 mi.; bearing of first from second, N. 74\u00c2\u00b0 1.7 E.\\n7. Opposite angles, 76\u00c2\u00b0 9.2 57\u00c2\u00b0 42.2 remaining side, .6313.\\n8. _B 51\u00c2\u00b030.5 c= 53.51, 46.80. 9. 1.658.\\n10. 7.087, 11.30. 11. 3995 sq. ft. 12. 61.51, 58.48.\\n13. 14.922 miles an hour. 14. Height, 69.71 ft.; distance, 86.08 ft.\\n15. 82.70 ft. 16. 173.2 ft. 17. 19.92, 16.62.\\n18. One angle, 118\u00c2\u00b0 6.6 diagonal, 91.02. 19. 9.012 mi.\\n20. ^i 9.282, CZ 10.65. 21. 1538 ft.\\n22. ^i) 74.98, 68\u00c2\u00b0 58.3\\n23. One angle, 59\u00c2\u00b0 44.8 sides, 66.99, 37.77. 24. 84.28 ft.\\n25. 272.4 ft.\\n26. Bluff, 438.7 ft.; lighthouse, 280.1 ft.\\n136; pages 113 to 115.\\n5.\\n36\u00c2\u00b0 29.4\\n5 69\u00c2\u00b0 42.1\\n54\u00c2\u00b0 18.9\\n6.\\na 21\u00c2\u00b0 18.0\\n5 49\u00c2\u00b0 54.7\\nc 53\u00c2\u00b08.2\\n7.\\n20\u00c2\u00b0 33.8\\nJ5 70\u00c2\u00b059.5\\nc 23\u00c2\u00b0 18.3\\n8.\\n68\u00c2\u00b0 51.6\\na 31\u00c2\u00b0 26.4\\n13\u00c2\u00b0 40.2\\n9.\\n5 58\u00c2\u00b0 27.5\\n6 35\u00c2\u00b0 32.4\\nc 42\u00c2\u00b0 59.3\\n5 121\u00c2\u00b0 32.5\\n6 144\u00c2\u00b0 27. 6\\nc 137\u00c2\u00b0 0.7\\n10.\\n83\u00c2\u00b0 38.8\\n127\u00c2\u00b0 36.2\\nc 94\u00c2\u00b0 52.3\\n11.\\n97\u00c2\u00b0 36.4\\na 113\u00c2\u00b0 22.4\\n^=19\u00c2\u00b0 29.4\\n12.\\na 164\u00c2\u00b0 20\\n31\u00c2\u00b0 16.9\\n9\u00c2\u00b019.1\\n13.\\na 165\u00c2\u00b0 18.8\\nB 104\u00c2\u00b0 13.4\\nc 46\u00c2\u00b0 50.4\\n14.\\n48\u00c2\u00b0 10.9\\na 44\u00c2\u00b0 29.2\\nc 109\u00c2\u00b0 52.5\\n131\u00c2\u00b0 49.1\\na 135\u00c2\u00b0 30.8\\nc 70\u00c2\u00b07.5", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0165.jp2"}, "166": {"fulltext": "Answers.\\n15.\\nA\\n49\u00c2\u00b0 35.8\\nB\\n97\u00c2\u00b0 36.0\\nc\\n96\u00c2\u00b0 31.2\\n16.\\na\\n172\u00c2\u00b0 28.1\\nB\\n84\u00c2\u00b0 45.4\\nc\\n124\u00c2\u00b0 39.3\\n17.\\na\\n26\u00c2\u00b0 42.8\\nB\\n99\u00c2\u00b0 47.4\\nb\\n110\u00c2\u00b0 59.7\\n18.\\na\\n129\u00c2\u00b0 66.7\\nb\\n161\u00c2\u00b0 32.5\\nc\\n52\u00c2\u00b0 28.2\\n19.\\nA\\n157\u00c2\u00b0 46.8\\nB\\n74\u00c2\u00b0 12.0\\nb\\n43\u00c2\u00b0 56.9\\n20.\\nA\\n165\u00c2\u00b0 0.6\\na\\n168\u00c2\u00b0 29.2\\nb\\n130\u00c2\u00b0 28.2\\n21.\\nA\\n114\u00c2\u00b0 49.3\\na\\n121\u00c2\u00b0 22.9\\nB\\n126\u00c2\u00b0 14.4\\n22.\\nA\\n100\u00c2\u00b0 38.0\\nB\\n163\u00c2\u00b0 8.0\\nc\\n51\u00c2\u00b0 44.4\\n23.\\na\\n8\u00c2\u00b0 30. 5\\nB\\n66\u00c2\u00b0 55.5\\nc\\n20\u00c2\u00b0 53.7\\n24.\\nB\\n30\u00c2\u00b0 53.3\\nb\\n30\u00c2\u00b0 12.9\\nc\\n78\u00c2\u00b0 35.0\\nB\\n149\u00c2\u00b0 6.7\\nh\\n149\u00c2\u00b0 47.1\\nc\\n101\u00c2\u00b0 25.0\\n25.\\nA\\n138\u00c2\u00b0 15.5\\nb\\n130\u00c2\u00b0 2.3\\nc\\n57\u00c2\u00b0 55.4\\n26.\\nA\\n59\u00c2\u00b0 17.1\\na\\n49\u00c2\u00b0 26.0\\nB\\n51\u00c2\u00b0 46.0\\n27.\\na\\n78\u00c2\u00b0 32.1\\nb\\n132\u00c2\u00b0 25.0\\nc\\n97\u00c2\u00b0 42.6\\n28.\\nA\\n137\u00c2\u00b0 17.7\\nB\\n119\u00c2\u00b0 29.5\\nb\\n136\u00c2\u00b0 31.7\\n29.\\na\\n144\u00c2\u00b0 0.6\\nB\\n110\u00c2\u00b0 57.9\\nb\\n123\u00c2\u00b0 6.1\\n30.\\nA\\n68\u00c2\u00b0 10.6\\nb\\n34\u00c2\u00b0 3.0\\nc\\n61\u00c2\u00b0 11.3\\n31.\\nA=:\\n71\u00c2\u00b0 45.5\\na\\n53\u00c2\u00b0 44.7\\nB\\n148\u00c2\u00b0 2.5\\n32.\\na\\n16\u00c2\u00b0 48.5\\nB\\n124\u00c2\u00b0 31.6\\nc\\n151\u00c2\u00b0 56.7\\n33.\\nA\\n25\u00c2\u00b0 4.8\\na\\n4\u00c2\u00b0 53.9\\nb\\n169\u00c2\u00b0 27.2\\n34.\\nA\\n76\u00c2\u00b0 16.7\\nB\\n144\u00c2\u00b0 1.1\\nb\\n146\u00c2\u00b0 26.2\\n35.\\na\\n6\u00c2\u00b0 50.4\\nB\\n59\u00c2\u00b0 54.0\\nb\\n11\u00c2\u00b0 36.6\\n36.\\nA\\n152\u00c2\u00b0 7.1\\nb\\n40\u00c2\u00b0 48.8\\nc\\n135\u00c2\u00b0 39.6\\n37.\\nA\\n142\u00c2\u00b0 5.8\\nB\\n82\u00c2\u00b0 43.4\\nc\\n99\u00c2\u00b0 26.4\\n38.\\na\\n144\u00c2\u00b0 24.4\\nb\\n32\u00c2\u00b0 28.8\\nc\\n133\u00c2\u00b0 19.3\\n39.\\nA\\n102\u00c2\u00b0 48.8\\na\\n141\u00c2\u00b0 46.9\\nb\\n10\u00c2\u00b0 19.1\\n40.\\nA\\n10\u00c2\u00b0 22.5\\na\\n6\u00c2\u00b0 16.1\\nc\\n142\u00c2\u00b0 41.2\\nA\\n169\u00c2\u00b0 37.5\\na\\n137;\\n173\u00c2\u00b0 43.9\\npage 116.\\nc\\n37\u00c2\u00b0 18.8\\n2.\\na\\n152\u00c2\u00b0 52.8\\nb\\n104\u00c2\u00b0 46.7\\nB\\n124\u00c2\u00b0 1.1\\n3.\\nA\\n138\u00c2\u00b0 42.7\\nB\\n153\u00c2\u00b0 25.0\\nC\\n132\u00c2\u00b0 14.3\\n4.\\na\\n44\u00c2\u00b0 8.1\\nb\\n101\u00c2\u00b0 3.9\\nC^\\n78\u00c2\u00b0 22.3\\n5.\\na\\n82\u00c2\u00b0 14.4\\nA\\n63\u00c2\u00b0 34.8\\nB\\n164\u00c2\u00b0 4.8\\n6.\\n6\\n20\u00c2\u00b0 21.2\\nB\\n10\u00c2\u00b0 33.7\\n148\u00c2\u00b0 11.9\\n6\\n159\u00c2\u00b0 38.8\\nB\\n169\u00c2\u00b0 26.3\\nc\\n31\u00c2\u00b0 48.1", "height": "3681", "width": "2549", "jp2-path": "completetrigonom00well_0166.jp2"}, "167": {"fulltext": "Answers.\\n7. a 108\u00c2\u00b0 29.5 5 =141\u00c2\u00b0 31.9 C= 111\u00c2\u00b0 45.7\\n8. 19\u00c2\u00b0 56.3 b 138\u00c2\u00b0 36.4 B 163\u00c2\u00b0 15.7\\n9. 46\u00c2\u00b0 49.3 A 33\u00c2\u00b0 43.3 b 124\u00c2\u00b0 37.8\\n10. 11\u00c2\u00b0 12.1 6 =76\u00c2\u00b0 13.8 (7 =140\u00c2\u00b0 14.8\\n11. a 50\u00c2\u00b0 56.0 23\u00c2\u00b0 47.2 (7=31\u00c2\u00b0 17.6\\n31, a 129\u00c2\u00b0 4.0 156\u00c2\u00b0 12.8 C 148\u00c2\u00b0 42.4\\n138; pag-e 117.\\n2. C= 159\u00c2\u00b0 59.8 a 69\u00c2\u00b0 44.6 4. (7 145\u00c2\u00b0 45.0 141\u00c2\u00b0 17.0\\n3. 120\u00c2\u00b0 46.6 c 30\u00c2\u00b0 26.6 5. O 148\u00c2\u00b0 37.6 c 80\u00c2\u00b0 36.8\\n149 page 128.\\n2. a 98\u00c2\u00b0 19.9 5 60\u00c2\u00b0 3.9 C= 110\u00c2\u00b0 38.6\\n3. c 72\u00c2\u00b0 14.1 6 47\u00c2\u00b0 27.3 ^=32\u00c2\u00b0 24.2\\n4. a 120\u00c2\u00b0 34.6 c 71\u00c2\u00b0 7.6 jB 50\u00c2\u00b03.6\\n5. 6 153\u00c2\u00b0 48.4 a 125\u00c2\u00b0 0.8 C 140\u00c2\u00b0 24.6\\n150; pages 129, 130.\\n2. A 110\u00c2\u00b0 46.6 B 35\u00c2\u00b0 34.2 c 45\u00c2\u00b0 36.0\\n3. (7 =78\u00c2\u00b0 65.5 i? 41\u00c2\u00b019.7 a 107\u00c2\u00b0 47.6\\n4. 145\u00c2\u00b0 11.8 C 107\u00c2\u00b0 39.8 b 126\u00c2\u00b0 37.4\\n5. jB 121\u00c2\u00b0 43.2 J. 105\u00c2\u00b0 52.8 c= 115\u00c2\u00b0 48.6\\n151; pag-e 131.\\n2. 74\u00c2\u00b0 12.4 J5 82\u00c2\u00b012.0\\n3. 78\u00c2\u00b0 32.0 5 87\u00c2\u00b0 7.0\\n4. 120\u00c2\u00b0 21.4 5 130\u00c2\u00b0 21.8 (7\\n5. 70\u00c2\u00b0 8.8\\n152 pag-e 133.\\n3. a 37\u00c2\u00b0 7.2 6 41\u00c2\u00b0 1.6\\n4. a 101\u00c2\u00b0 34.4 b 49\u00c2\u00b0 50.4\\n5. a 125\u00c2\u00b0 16.2 b 151\u00c2\u00b0 37.4\\n6. 126\u00c2\u00b0 43.4\\n153 pag-es 135, 136.\\n4. 5 22\u00c2\u00b0 7,4 (7= 112\u00c2\u00b0 17.0 c 36\u00c2\u00b0 28.0\\n5. Impossible. 6. C=90\u00c2\u00b0, 138\u00c2\u00b0 31.6 a 146\u00c2\u00b0 41.9\\n7. (7= 154\u00c2\u00b0 7.7 J5 60\u00c2\u00b045.8 5 =63\u00c2\u00b0 52.2\\n8. 36\u00c2\u00b0 18.2 (7= 160\u00c2\u00b0 52.8 c 148\u00c2\u00b0 58.4\\nor, A 143\u00c2\u00b0 41.8 C 38\u00c2\u00b0 23.0 c 77\u00c2\u00b0 39.2\\nc\\nm\\n41.4\\nc\\n93\u00c2\u00b0\\n29.6\\nc\\n14C\\n\u00c2\u00b07.0\\nc\\n49\u00c2\u00b0\\n28.2\\nc\\n59\u00c2\u00b0\\n37.2\\nc\\n75\u00c2\u00b0\\n55.4", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0167.jp2"}, "168": {"fulltext": "8 Answers.\\n154; page 137.\\n2. 5 8\u00c2\u00b0 50.4 c 27\u00c2\u00b0 37.0 O 79\u00c2\u00b0 9.2\\n3. Impossible. 4. c 90\u00c2\u00b0, B 8\u00c2\u00b0 14.0 b 18\u00c2\u00b0 41.2\\n5. a 155\u00c2\u00b0 56.8 b 188\u00c2\u00b0 33.6 B 100\u00c2\u00b0 0\\n6. a 63\u00c2\u00b0 55. 7 c 156\u00c2\u00b05.8 C 154\u00c2\u00b0 55.0\\nor, a 116\u00c2\u00b0 4.3 c=72\u00c2\u00b043.8 87\u00c2\u00b0 24.0\\n155 pag-es 137, 138.\\nor,\\nor,\\nor,\\n1.\\nA =51\u00c2\u00b0 58.8\\nB^\\n8.3\u00c2\u00b0 55.2\\n58\u00c2\u00b0 53.8\\n2.\\nb 115\u00c2\u00b0 3.3\\na\\n85\u00c2\u00b0 16.0\\n81\u00c2\u00b0 24.0\\n3.\\na 95\u00c2\u00b0 37. 8\\n41\u00c2\u00b0 52.2\\n110\u00c2\u00b0 48.2\\n4.\\nC 65\u00c2\u00b0 23.3\\nA\\n96\u00c2\u00b0 8.0\\na 99\u00c2\u00b0 40.0\\n5.\\na 68\u00c2\u00b0 25.2\\nb^\\n71\u00c2\u00b0 10.8\\nc 56\u00c2\u00b0 56.8\\n6.\\nc 90\u00c2\u00b0,\\nB^\\n63\u00c2\u00b0 43.4\\n66\u00c2\u00b0 26.8\\n7.\\n121\u00c2\u00b0 32.8\\nB\\n40\u00c2\u00b0 56.8\\nc 37\u00c2\u00b0 25.8\\n8.\\nImpossible.\\n9.\\nA 142\u00c2\u00b0 32.8\\nB\\n27\u00c2\u00b0 52.6\\n0=82\u00c2\u00b0 27.2\\n10.\\nb 120\u00c2\u00b0 16.6\\nc\\n69\u00c2\u00b0 19.6\\n50\u00c2\u00b0 26.2\\n11.\\nImpossible.\\n12.\\na 100\u00c2\u00b0 8.4\\n6\\n50\u00c2\u00b0 1.8\\nc 60\u00c2\u00b06.0\\n13.\\nC 90\u00c2\u00b0,\\nB\\n113\u00c2\u00b0 36.5\\n6 114\u00c2\u00b0 51.9\\n14.\\na 67\u00c2\u00b0 24.0\\n164\u00c2\u00b0 6.4\\nc 160\u00c2\u00b0 6.4/\\na 112\u00c2\u00b0 36.0\\n128\u00c2\u00b0 20.6\\nc 103\u00c2\u00b0 2.4\\n15.\\nC= 86\u00c2\u00b0 59.7\\n60\u00c2\u00b0 50.9\\n6 111\u00c2\u00b0 17.0\\n16.\\n0=146\u00c2\u00b0 37.9\\njB\\n55\u00c2\u00b0 1.2\\n5 96\u00c2\u00b0 34.4\\n17.\\na 43\u00c2\u00b0 3.1\\n129\u00c2\u00b0 8.4\\ni 89\u00c2\u00b023.8\\na 136\u00c2\u00b0 56.9\\nb--\\n20\u00c2\u00b0 35.8\\n5 26\u00c2\u00b0 58.6\\n18.\\n35\u00c2\u00b0 31.0\\nBz\\n24\u00c2\u00b0 42.6\\n0=138\u00c2\u00b0 24.8\\n19.\\n5 42\u00c2\u00b0 7.9\\n160\u00c2\u00b0 12.8\\nc 153\u00c2\u00b0 37.8\\n5 137\u00c2\u00b0 52.1\\nC--\\n49\u00c2\u00b0 38.8\\nc 89\u00c2\u00b0 51.2\\n20.\\na 59\u00c2\u00b0 34.4\\nb--\\n136\u00c2\u00b0 10.6\\nc 150\u00c2\u00b0 1.6\\n21.\\nImpossible.\\n22\\nImpossible.\\n23.\\nc 64\u00c2\u00b0 19.4\\na\\n34\u00c2\u00b0 3.0\\n5 37\u00c2\u00b0 39.6\\n24.\\n5 68\u00c2\u00b0 18.0\\nA--\\n132\u00c2\u00b0 33.8\\na 131\u00c2\u00b0 15.8 i\\nB= 111\u00c2\u00b0 42.0\\nA--\\n77\u00c2\u00b0 4.6\\na 95\u00c2\u00b0 50.0\\n25.\\n5 134\u00c2\u00b0 57.3\\nC--\\n50\u00c2\u00b041.1\\na 69\u00c2\u00b0 8.8\\n26.\\n6 27\u00c2\u00b0 22.1\\na\\n117\u00c2\u00b0 9.2\\n47\u00c2\u00b0 21.2", "height": "3700", "width": "2446", "jp2-path": "completetrigonom00well_0168.jp2"}, "169": {"fulltext": "Answers. 9\\n157 page 140.\\n1. Bearing of Havana from Gibraltar, N. 77\u00c2\u00b0 40.3 W. of Gibraltar\\nfrom Havana, N. 59 5.1 E. distance, 4593 mi.\\n2. Bearing of Batavia from San Francisco, N. 67\u00c2\u00b0 25.5 W. of San\\nFrancisco from Batavia, N. 47\u00c2\u00b0 12.3 E. distance, 8650 mi.\\n3. Bearing of Cape of Good Hope from Vera Cruz, S. 60\u00c2\u00b0 45.6 E.\\nof Vera Cruz from Cape of Good Hope, N. 86\u00c2\u00b0 42.4 W. distance,\\n8330 mi.\\n4. Bearing of Callao from Auckland, S. 69\u00c2\u00b0 29.9 E. of Auckland\\nfrom Callao, S. 50\u00c2\u00b0 2.5 W. distance, 6671 mi.\\n5. 54\u00c2\u00b0 35.9 N. 6. 51\u00c2\u00b0 48.0 S.\\n160; pages 142, 143,\\n1. 3 h. 30 m. 40 s.\\n2. Hour of the day, 10 h. 23 m. 33.6 s., a.m. longitude, 19\u00c2\u00b0 6.6 W.\\n3. 7 h. 17 m. 14.4 s., p.m.\\n4. Hour of the day, 2 h. 45 m. 4 s., p.m. longitude, 66\u00c2\u00b0 59 W.\\n5,. 6 h. 15 m. 21.6 s., a.m. 6. N. 80\u00c2\u00b0 24.6 W. 7. N. 71\u00c2\u00b0 43.7 E.\\n8. 45\u00c2\u00b0 5", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0169.jp2"}, "170": {"fulltext": "", "height": "3681", "width": "2542", "jp2-path": "completetrigonom00well_0170.jp2"}, "171": {"fulltext": "FOUR PLACE\\nLOGARITHMIC TABLES\\nTOGETHER WITH A\\nTABLE OE NATURAL SINES, COSINES,\\nTANGENTS, AND COTANGENTS\\nPEEPARED BY\\nWEBSTER WELLS, S.B.\\nPROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS\\nINSTITUTE OF TECHNOLOGY\\nBOSTON, U.SA.\\nD. C. HEATH CO., PUBLISHERS\\n1900", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0171.jp2"}, "172": {"fulltext": "Copyright, 1887 and 1900,\\nBy WEBSTER WELLS.\\nTTPOGKAPHY BY J. S. CUSHING CO., NOEWOOD, MASS.", "height": "3681", "width": "2528", "jp2-path": "completetrigonom00well_0172.jp2"}, "173": {"fulltext": "Use of the Tables. i\\nUSE OF THE TABLES.\\nI. USE OF THE TABLE OF LOGARITHMS OF NUMBERS.\\nThis table (pages 12 and 13) gives the mantissse of the\\nlogarithms of all integers from 100 to 1000, calculated to\\nfour places of decimals.\\nTo FIND THE Logarithm of a Number of Three\\nFigures.\\nLook in the column headed No. for the first two signifi-\\ncant figures of the given number.\\nThen the required mantissa will be found in the corre-\\nsponding horizontal line, in the vertical column headed by\\nthe third figure of the number.\\nFinally, prefix the characteristic in accordance with the\\nrules of 66 or 67.\\nFor example, log 168 2.2253;\\nlog .0344 8.5366 10 etc.\\nFor a number consisting of one or two significant figures,\\nthe column headed may be used.\\nThus, let it be required to find log 83 and log 9.\\nBy 80, log 83 has the same mantissa as log 830, and log\\n9 the same mantissa as log 900.\\nHence, log 83 1.9191, and log 9 0.9542.^\\nTo FIND THE Logarithm of a Number of more than\\nThree Figures.\\nEx. 1. Required the logarithm of 327.6.\\nFrom the table, log 327 2.5145,\\nand log 328 2.5159.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0173.jp2"}, "174": {"fulltext": "2 Use of the Tables.\\nThat is, an increase of one unit in the number produces\\nan increase of .0014 in the logarithm.\\nTherefore, an increase of .6 of a unit in the number will\\nproduce an increase of .6 x .0014 in the logarithm, or .0008\\nto the nearest fourth decimal place.\\nHence, log 327.6 2.5145 .0008 2.5153.\\nNote I. The above method is based on the assumption that the\\ndifferences of logarithms are proportional to the differences of their\\ncorresponding numbers which, though not strictly accurate, is suffi-\\nciently exact for practical purposes.\\nNote II. The difference between any mantissa in the table and the\\nmantissa of the next higher number of three figures, is called the tabu-\\nlar difference. The subtraction may be performed mentally.\\nThe following rule is derived from the above\\nFind from the table the mantissa of the first three significant\\nfigures^ and the tabular difference.\\nMultiply the latter by the remaining figures of the number\\nwith a decimal point before them. (See Note III.)\\nAdd the residt to the mantissa of the first three significant\\nfigures^ and prefix the proper characteristic.\\nNote III. In finding the correction to the nearest unit s figure, the\\ndecimal portion should be omitted provided that, if it is .5 or more\\nthan 5, the unit s figure is increased by 1. Thus, 13.26 would be taken\\nas 13 30.5 as 31 22.803 as 23.\\nEx. 2. Find the logarithm of .021508.\\nMantissa 215 3324 Tabular difference 21\\n2 _^\\n3326 Correction 1.68 2, nearly.\\nResult, 8.3326 10.\\nTo FIND THE Number corresponding to a Logarithm.\\nEx. 1. Required the number whose logarithm is 1.6571.\\nFind in the table the mantissa 6571.", "height": "3700", "width": "2446", "jp2-path": "completetrigonom00well_0174.jp2"}, "175": {"fulltext": "Use of the Tables. 3\\nIn the corresponding line, in the column headed No.,\\nwe find 45, the first two figures of the required number, and\\nat the head of the column we find 4, the third figure.\\nSince the characteristic is 1, there must be two places to\\nthe left of the decimal point 66).\\nHence, the number corresponding to 1.6571 is 45.4\\nEx. 2. Required the number whose logarithm is 2.3934.\\nWe find in the table the mantissse 3927 and 3945, whose\\ncorresponding numbers are 247 and 248, respectively.\\nThat is, an increase of 18 in the mantissa produces an\\nincrease of one unit in the number corresponding.\\nTherefore, an increase of 7 in the mantissa will produce an\\nincrease of y^g of a unit in the number, or A, nearly.\\nHence, the number corresponding is 247 .4, or 247.4.\\nThe following rule is derived from the above\\nFind from the table the next less mantissa, the three figures\\ncorresponding, and the tabular difference.\\nSubtract the next less from the given mantissa, and divide\\nthe remainder by the tabular difference. (See Note V.)\\nAnnex the quotient to the first three figures of the number,\\nand point off the result. (See Note IV.)\\nNote IV. The rules for pointing off are the reverse of the rules\\nfor characteristic given in ^^QQ and 67.\\n1. If \u00e2\u0080\u00941^ is not written after the mantissa, add 1 to the character-\\nistic, giving the number of places to the left of the decimal point.\\n2. If \u00e2\u0080\u009410 is written after the mantissa^ subtract the positive part of\\nthe characteristic from 9, giving the number of ciphers to be placed\\nbetiveen the decimal point and first significant figure.\\nEx. 3. Find the number whose logarithm is 8.5265 10.\\n5265\\nNext less mantissa 5263 three figures corresponding, 336.\\nTabular difference 13)2.00(.15 .2, nearly.\\n13\\n70", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0175.jp2"}, "176": {"fulltext": "4 Use of the Tables.\\nBy the rule of Note IV. there will be one cipher between the decimal\\npoint and first significant figure.\\nHence, the number corresponding .03362.\\nNote V. The correction can usually be depended upon to one\\ndecimal place the division should be carried out to two decimal places\\nin order to determine the last figure accurately. (See Note III.)\\nII. USE OF THE TABLE OF LOGARITHMIC SINE\\nCOSINES, ETC.\\nThis table (pages 14 to 19) gives the logarithms of the\\nsines, cosines, tangents, and cotangents of all angles at inter-\\nvals of 10 minutes from 0\u00c2\u00b0 to 90\u00c2\u00b0.\\nFor angles between 0\u00c2\u00b0 and 45\u00c2\u00b0, the degrees and minutes\\nwill be found in the left-hand column, and the functions in\\nthe columns designated by the names at the top; that is,\\nsines in the first column, cosines in the second, tangents in\\nthe third, and cotangents in the fourth.\\nFor angles between 45\u00c2\u00b0 and 90\u00c2\u00b0, the degrees and minutes\\nwill be found in the right-hand column, and the functions in\\nthe columns designated by the names at the foot; that is,\\ncosines in the first column, sines in the second, cotangents\\nin the third, and tangents in the fourth.\\nIf only the mantissa of the logarithm is found, the charac-\\nteristic may be determined from the nearest logarithm in the\\nsame column in which the characteristic is given.\\nSince the sines and cosines of all acute angles, the tan-\\ngents of angles between 0\u00c2\u00b0 and 45\u00c2\u00b0, and the cotangents of\\nangles between 45\u00c2\u00b0 and 90\u00c2\u00b0, are less than unity, the charac-\\nteristics of their logarithms have been increased by 10, and\\n10 must be written after the mantissa; in all other cases,\\nthe true value of the characteristic is given in the table.\\nThus, log sin 38\u00c2\u00b0 30 9.7941 10\\nlog tan i55\u00c2\u00b0 20 0.3380\\nlog cot 79\u00c2\u00b0 10 -9.2819 -10;\\nlog cos 89\u00c2\u00b0 40 7.7648 10 etc.", "height": "3705", "width": "2370", "jp2-path": "completetrigonom00well_0176.jp2"}, "177": {"fulltext": "Use of the Tables. 5\\nTo FIND THE Logarithmic Sine, Cosine, Tangent, or\\nCotangent, of any Acute Angle expressed\\nIN Degrees and Minutes.\\nFind from the table the logarithmic sine, cosine, tangent, or\\ncotangent of the next less m^dtiple of ten minutes, and the\\ndifference for V corresponding. (See Note VI.)\\nMultiply this difference by the number of minutes remaining.\\nIf sine or tanqent, add\\n\\\\this correction.\\nIf cosine or cotangent, subtract)\\nNote VI. The columns immediately to the right of those headed\\nSin., Cos., and Tan., contain the respective differences for V\\nthe right-hand column of differences is also to be used with the column\\nheaded Cot.\\nIt will be observed that the differences do not stand in the same\\nhorizontal line with the logarithms, but opposite the intervals between\\nconsecutive logarithms. For angles below 45\u00c2\u00b0, the difference next he-\\nlow should be taken for angles above 45\u00c2\u00b0, the difference next above.\\nNote VII. The rule assumes that the differences of the logarithmic\\nfunctions are proportional to the differences of their corresponding\\nangles, which, unless the angle is near to 0\u00c2\u00b0 or 90\u00c2\u00b0, is in general suffi-\\nciently exact for practical purposes. (See page 9.)\\nNote VIII. If the angle is expressed in degrees, minutes, and\\nseconds, the seconds should be reduced to the decimal part of a minute\\nbefore applying the rule.\\nEx. 1. Find Ior tan 17\u00c2\u00b0 14\\nlogtan 17\u00c2\u00b0 10 =9.4898-10\\nD. V 4.5\\n18\\n4\\nResult, 9.4916 10\\nCorr. 18.0\\n:x. 2. Find log cos 58\u00c2\u00b0 33.5\\nlog cos 58\u00c2\u00b0 30 9.7181 10\\nD. V 2.1\\n7\\n3.5\\nResult, 9.7174 10\\n105\\n63\\n7.35\\n7, nearly", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0177.jp2"}, "178": {"fulltext": "6 Use of the Tables.\\nTo FIND THE Acute Angle corresponding to a given\\nLogarithmic Sine, Cosine, Tangent, or Cotangent.\\nTake from the table, if sine or tangent the next less, if cosine\\nor cotangent the next greater, logarithmic function, the angle\\ncorresponding, and the difference for V. (See Note IX.)\\nFind the difference between the given logarithm and that\\ntaken from the table, and divide it by the difference for V,\\ngiving the correction in minutes.\\nAdd the result to the angle corresponding to the next less, or\\nnext greater, function.\\nNote IX. Ill searching for the next less (or greater) logarithm,\\nattention must be paid to the fact that the functions are found in\\ndifferent columns according as the angle is below or above 45\u00c2\u00b0.\\nIf, for example, the next less logarithmic sine is found in the column\\nwith Sin. at the top^ the angle corresponding must be taken from\\nthe left-hand column but if it is found in the column with Sin. at\\nthe foot^ the angle corresponding must be taken from the right-hand\\ncolumn. Similar considerations hold with respect to the other three\\nfunctions.\\nEx. 1. Eind the angle whose log sin 9.9594 10.\\n9.9594 10\\nNext less log sin 9.9590 10 angle corresponding 65\u00c2\u00b0 30\\nD. 1 .6)4.0(6.66 6.7, nearly.\\nAdding the correction, the result is 65\u00c2\u00b0 36.7\\nEx. 2. Eind the angle whose log cot 0.1696,\\nNext greater log cot 0.1710 angle corresponding 34\u00c2\u00b0 0\\n0.1696\\n1\\n2.7)14.0(5.18::\\n13 5\\n50\\n27\\n5.2\\nnearly.\\n230\\nKesult, 34\u00c2\u00b0\\n5.2", "height": "3702", "width": "2438", "jp2-path": "completetrigonom00well_0178.jp2"}, "179": {"fulltext": "Use of the Tables. 7\\nTo FIND THE Logarithmic Secant or Cosecant of\\nANY Acute Angle.\\nSince sec x and csc x we have by 85,\\ncos X sm X\\nlog sec X colog cos x, and log csc x colog sin x.\\nHence, to find the logarithmic secant, subtract the logarith-\\nmic cosine from 10 10; and to find the logarithmic cosecant,\\nsubtract the logarithmic sine from 10 10.\\nEx. Find log sec 22\u00c2\u00b0 38\\nFrom the table, log cos 22\u00c2\u00b0 38 9.9652 10\\nSubtracting from 10 10, we have\\nlog sec 22\u00c2\u00b0 38 0.0348.\\nNote X. The logarithmic cotangent of an angle may be obtained\\nby subtracting the logarithmic tangent from 10 10.\\nTo FIND THE Logarithmic Functions op an Angle not\\nLYING BETWEEN THE LiMITS 0\u00c2\u00b0 AND 90\u00c2\u00b0.\\nBy 34, any function of any angle may be expressed as a\\nfunction of a certain acute angle and hence the table of the\\nfunctions of acute angles serves to determine the functions\\nof angles of any magnitude whatever, positive or negative.\\nEx. Find log sin 152\u00c2\u00b0 16\\nBy 34, sin 152\u00c2\u00b0 16 cos 62\u00c2\u00b0 16\\nThen, log sin 152\u00c2\u00b0 16 log cos 62^ 16 9.6678 10.\\nAnother method would be to find the logarithmic sine of 27\u00c2\u00b0 44\\nthe supplement of 152\u00c2\u00b0 16 32).\\nNote XI. If the natural function is negative, as for example in\\nthe case of the cosine of an angle between 90\u00c2\u00b0 and 180\u00c2\u00b0, there is no\\nlogarithmic function, strictly speaking. (See Note before 87.)\\nIn solving examples involving such functions, we proceed as if the\\nfunctions were positive, and determine the algebraic sign of the result\\nirrespective of the logarithmic work. Illustrations of this will be found\\nin Chapters X. and XI.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0179.jp2"}, "180": {"fulltext": "8 Use of the Tables.\\nin. USE OF THE TABLE OF NATURAL SINES AND\\nCOSINES.\\nThis table (pages 20 and 21) gives the natural vahies of\\nthe sines and cosines of all angles at intervals of 10 minutes\\nfrom 0\u00c2\u00b0 to 90\u00c2\u00b0, calculated to four places of decimals.\\nIts use is similar to that of the table of logarithmic func-\\ntions, except that the tabular differences for 1 are not given,\\nbut are to be calculated from the table when required.\\nEx. 1. Find sin 48\u00c2\u00b0 52\\nThe difference between sin 48\u00c2\u00b0 50 and sin 49\u00c2\u00b0 0 is .0019, one-tentli\\nof whicli is .00019.\\nsin 48\u00c2\u00b0 50 .7528 D. 1 1.9\\n4 2\\n.7532, A71S. Corr. 3,8 4, nearly.\\nEx. 2. Find the angle whose cos .5506.\\nThe difference between the next greater and next less functions,\\n.5519 and .5495, is .0024 one-tenth of which is .00024.\\nNext greater cos .5519 angle corresponding 56\u00c2\u00b0 30\\n.5506\\n1\\n2.4)13.0(5.41 z\\n12\\n100\\n96\\n5.4,\\nnearly.\\n40\\nResult,\\n56\u00c2\u00b0\\n35.4\\nIV. USE OF THE TABLE OF NATURAL TANGENTS\\nAND COTANGENTS.\\nThis table (pages 22 and 23) gives the tangents and\\ncotangents of all angles at intervals of 10 minutes from 0\u00c2\u00b0\\nto 90\u00c2\u00b0 its use is similar to that of the table of natural sines\\nand cosines.", "height": "3681", "width": "2435", "jp2-path": "completetrigonom00well_0180.jp2"}, "181": {"fulltext": "Use of the Tables. 9\\nV. MORE ACCURATE METHOD FOR FINDING THE\\nLOGARITHMIC FUNCTIONS OF ANGLES NEAR\\nTO 0\u00c2\u00b0 OR 90\u00c2\u00b0.\\nIt was stated in Note YII., page 5, that in general the\\ndifferences of the logarithmic functions are approximately\\nproportional to the differences of their corresponding angles.\\nIt will be seen from the table that this is not the case with\\nthe logarithmic sines, tangents, and cotangents of angles\\nnear to 0\u00c2\u00b0, nor with the logarithmic cosines, tangents, and\\ncotangents of angles near to 90\u00c2\u00b0.\\nThus, the difference for 1 in the case of the logarithmic\\nsine or tangent of an angle between 40 and 50 is 96.9,\\nwhile for an angle between 50 and 1\u00c2\u00b0 it is 79.2.\\nA very accurate method for finding the logarithmic sine or\\ntangent of an angle near to 0\u00c2\u00b0, or the logarithmic cosine or\\ncotangent of an angle near to 90\u00c2\u00b0, is to first calculate the\\nnatural function by aid of the table of natural sines and\\ncosines, or of natural tangents and cotangents, and then\\nfind the logarithm of the result.\\nTo find the angle corresponding in similar cases, find the\\nnumber corresponding to the logarithmic function, and then,\\nby aid of the tables of natural functions, calculate the angle\\ncorresponding to the result.\\nEx. 1. Eind log sin 0\u00c2\u00b0 m\\\\\\nFrom the table of natural sines and cosines, we obtain\\nnatural sin 0\u00c2\u00b0 56 .016289.\\nWhence, log sin 0\u00c2\u00b0 56 8.2119 10.\\nThis result is correct to the last place of decimals by the ordinary\\nmethod we should have obtained 8.2102 10.\\nEx. 2. Eind the angle whose log tan 8.0302 10.\\nThe number corresponding to this logarithm is .01072.\\nFrom the table of natural tangents and cotangents, the angle whose\\nnatural tangent is .01072 is 36.85", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0181.jp2"}, "182": {"fulltext": "lo Use of the Tables.\\nThis is correct to the last place of decimals the ordinary method\\nwould have given 37.15\\nNote XII. To find with accuracy the log cotangent of an angle\\nnear to 0\u00c2\u00b0, find the log tangent of the angle by the above method, and\\nthen take the cologaritlim of the result. (See Note X., page 7.)\\nTo find the angle corresponding to a log cotangent in a similar case,\\nfind the log tangent of the angle (Note X.) and then find the angle\\ncorresponding as above.\\nNote XIII. To find the log tangent of an angle near to 90\u00c2\u00b0, find\\nthe log tangent of its complement^ and take the cologarithm of the\\nresult. (See Note XIL)\\nTo find the angle corresponding in a similar case, find the angle\\ncorresponding to its cologarithm, and take the complement of the\\nresult.\\nNote XIV. The more accurate method should be employed in\\nfinding the log sines, tangents, or cotangents of angles between 0\u00c2\u00b0 and\\n5\u00c2\u00b0, or the log cosines, tangents, or cotangents of angles between 85\u00c2\u00b0\\nand 90\u00c2\u00b0, and the angles corresponding in similar cases. For angles\\nbetween 5\u00c2\u00b0 and 85\u00c2\u00b0 the ordinary method is sufficiently exact.", "height": "3681", "width": "2555", "jp2-path": "completetrigonom00well_0182.jp2"}, "183": {"fulltext": "FOUR PLACE LOGARITHMIC TABLES", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0183.jp2"}, "184": {"fulltext": "12\\nLOGARITHMS OF NUMBERS.\\nNo.\\n1\\n2\\n3\\n4\\n5\\n6\\n7\\n8\\n9\\nlO\\n0043\\n0086\\n0128\\n0170\\n0212\\n0253\\n0294\\n0334\\n0374\\nII\\n0414\\n0453\\n0492\\n0531\\n0569\\n0607\\n0645\\n0682\\n0719\\n0755\\n12\\n0792\\n0828\\n0864\\n0899\\n0934\\n0969\\n1004\\n1038\\n1072\\n1 106\\n13\\n39\\n1173\\n1206\\n1239\\n1271\\n1303\\n1335\\n1367\\n1399\\n1430\\n14\\n1 461\\n1492\\n1523\\n1553\\n1584\\n1614\\n1644\\n1673\\n1703\\n1732\\n15\\n1761\\n1790\\n1818\\n1847\\n1875\\n1903\\n1931\\n1959\\n1987\\n2014\\n16\\n2041\\n2068\\n2095\\n2122\\n2148\\n2175\\n2201\\n2227\\n2253\\n2279\\n17\\n2304\\n2330\\n2355\\n2380\\n2405\\n2430\\n2455\\n2480\\n2504\\n2529\\n18\\n2553\\n2577\\n2601\\n2625\\n2648\\n2672\\n2695\\n2718\\n2742\\n2765\\n19\\n2788\\n2810\\n2833\\n2856\\n2878\\n2900\\n2923\\n2945\\n2967\\n2989\\n20\\n3010\\n3032\\n3054\\n3075\\n3096\\n3118\\n3139\\n3160\\n3181\\n3201\\n21\\n3222\\n3243\\n3263\\n3284\\n3304\\n3324\\n3345\\n3365\\n3385\\n3404\\n22\\n3424\\n3444\\n3464\\n3483\\n3502\\n3522\\n3541\\n3560\\n3579\\n3598\\n23\\n3617\\n3636\\n3655\\n3674\\n3692\\n3711\\n3729\\n3747\\n3766\\n3784\\n24\\n3802\\n3820\\n3838\\n3856\\n3874\\n3892\\n3909\\n3927\\n3945\\n3962\\n25\\n3979\\n3997\\n4014\\n4031\\n4048\\n4065\\n4082\\n4099\\n41 16\\n4133\\n26\\n4150\\n4166\\n4183\\n4200\\n4216\\n4232\\n4249\\n4265\\n4281\\n4298\\n27\\n4314\\n4330\\n4346\\n4362\\n4378\\n4393\\n4409\\n4425\\n4440\\n4456\\n28\\n4472\\n4487\\n4502\\n4518\\n4533\\n4548\\n4564\\n4579\\n4594\\n4609\\n29\\n4624\\n4639\\n4654\\n4669\\n4683\\n4698\\n4713\\n4728\\n4742\\n4757\\n30\\n4771\\n4786\\n4800\\n4814\\n4829\\n4843\\n4857\\n4871\\n4886\\n4900\\n31\\n4914\\n4928\\n4942\\n4955\\n4969\\n4983\\n4997\\n501 1\\n5024\\n5038\\n32\\n5051\\n5065\\n5079\\n5092\\n5105\\n5119\\n5132\\n5145\\n5159\\n5172\\n33\\n5185\\n5198\\n5211\\n5224\\n5237\\n5250\\n5263\\n5276\\n5302\\n34\\n5315\\n5328\\n5340\\n5353\\n5366\\n5378\\n5391\\n5403\\n5416\\n5428\\n35\\n5441\\n5453\\n5465\\n5478\\n5490\\n5502\\n5514\\n5527\\n5539\\n5551\\n36\\n5563\\n5575\\n5587\\n5599\\n5611\\n5623\\n5635\\n5647\\n5658\\n5670\\n37\\n5682\\n5694\\n5705\\n5717\\n5729\\n5740\\n5752\\n5763\\n5775\\n5786\\n38\\n5798\\n5809\\n5821\\n5832\\n5843\\n5855\\n5866\\n5877\\n5888\\n5899\\n39\\n5911\\n5922\\n5933\\n5944\\n5955\\n5966\\n5977\\n5988\\n5999\\n6010\\n40\\n6021\\n6031\\n6042\\n6053\\n6064\\n6075\\n6085\\n6096\\n6107\\n6117\\n41\\n6128\\n6138\\n6149\\n6160\\n6170\\n6180\\n6191\\n6201\\n6212\\n6222\\n42\\n6232\\n6243\\n6253\\n6263\\n6274\\n6284\\n6294\\n6304\\n6314\\n6325\\n43\\n6335\\n6345\\n6355\\n6365\\n6375\\n6385\\n6395\\n6405\\n6415\\n6425\\n44\\n6435\\n6444\\n6454\\n6464\\n6474\\n6484\\n6493\\n6503\\n6513\\n6522\\n45\\n6532\\n6542\\n6551\\n6561\\n6571\\n6580\\n6590\\n6599\\n6609\\n6618\\n46\\n6628\\n6637\\n6646\\n6656\\n6665\\n6675\\n6684\\n6693\\n6702\\n6712\\n47\\n6721\\n6730\\n6739\\n6749\\n6758\\n6767\\n6776\\n6785\\n6794\\n6803\\n48\\n6812\\n6821\\n6830\\n6839\\n6848\\n6857\\n6866\\n6875\\n6884\\n6893\\n49\\n6902\\n691 1\\n6920\\n6928\\n6937\\n6946\\n6955\\n6964\\n6972\\n6981\\n50\\n6990\\n6998\\n7007\\n7016\\n7024\\n7033\\n7042\\n7050\\n7059\\n7067\\n51\\n7076\\n7084\\n7093\\n7101\\n7110\\n7118\\n7126\\n7135\\n7143\\n7152\\n52\\n7160\\n7168\\n7177\\n7185\\n7193\\n7202\\n7210\\n7218\\n7226\\n7235\\n53\\n7243\\n7251\\n7259\\n7267\\n7275\\n7284\\n7292\\n7300\\n7308\\n7316\\n54 7324\\n7332\\n7340\\n7348\\n7356\\n7364\\n7372\\n7380\\n7388\\n7396\\nNo.\\n1\\n2\\n3\\n4\\n5\\n6\\n7\\n8\\n9", "height": "3681", "width": "2435", "jp2-path": "completetrigonom00well_0184.jp2"}, "185": {"fulltext": "LOGARITHIMS OF 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V.\\nTan.\\nAngle.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0187.jp2"}, "188": {"fulltext": "16\\nLOGARITHMIC SINES, COSINES,\\nAngle.\\nSin.\\nD.l\\nCos.\\nD. 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1\\n69\u00c2\u00b0 40\\n20\u00c2\u00b0 30\\n\u00e2\u0080\u00a25443\\n.9716\\n\u00e2\u0080\u00a25727\\n\u00e2\u0080\u00a24273\\n69\u00c2\u00b0 30\\n20\u00c2\u00b0 40\\n\u00e2\u0080\u00a25477\\n.9711\\n.5766\\n.4234\\n69\u00c2\u00b0 20\\n20\u00c2\u00b0 50\\n21\u00c2\u00b0\\n21\u00c2\u00b0 10\\n\u00e2\u0080\u00a25510\\n.9706\\n\u00e2\u0080\u00a25804\\n9.5842\\n.4196\\n69\u00c2\u00b0 10\\n69\u00c2\u00b0 0\\n68\u00c2\u00b0 50\\n9^5543\\n\u00e2\u0080\u00a25576\\n9.9702\\n0.4158\\n.9697\\n\u00e2\u0080\u00a25879\\n4121\\n21\u00c2\u00b0 20\\n.5609\\n3-3\\n3-2\\n3^2\\n3^1\\n3^2\\n3^1\\n3-1\\n.9692\\n\u00e2\u0080\u00a25917\\n.4083\\n68\u00c2\u00b0 40\\n21\u00c2\u00b0 30\\n\u00e2\u0080\u00a2S641\\n.9687\\n\u00e2\u0080\u00a25954\\n.4046\\n68\u00c2\u00b0 30\\n21\u00c2\u00b0 40\\n\u00e2\u0080\u00a25673\\n.9682\\n\u00e2\u0080\u00a25991\\n.4009\\n68\u00c2\u00b0 20\\n21\u00c2\u00b0 50\\n22\u00c2\u00b0 0\\n\u00e2\u0080\u00a25704\\n.9677\\n9.9672\\n.6028\\n\u00e2\u0080\u00a23972\\n68\u00c2\u00b0 10\\n68\u00c2\u00b0 0\\n9^5736\\n9.6064\\n0.3936\\n22 10\\n\u00e2\u0080\u00a2S767\\n.9667\\n.6100\\n.3900\\n67\u00c2\u00b0 50\\n22\u00c2\u00b0 20\\n\u00e2\u0080\u00a25798\\n.9661\\n\u00e2\u0080\u00a25\\n.6136\\n.3864\\n67\u00c2\u00b0 40\\n22\u00c2\u00b0 30\\n.5828\\n3-0\\n.9656\\n.6172\\n.3828\\n67\u00c2\u00b0 30\\nCos.\\nD.l\\nSin.\\nD. 1\\nCot.\\nT .V.\\nTan.\\nAngle.", "height": "3705", "width": "2370", "jp2-path": "completetrigonom00well_0188.jp2"}, "189": {"fulltext": "TANGENTS AND COTANGENTS.\\n17\\nAngle.\\nSi*\u00c2\u00bb.\\nD.l\\nCos. D\\n1\\nTan.\\nD. \\\\t.\\nCot.\\n22\u00c2\u00b0 30\\n9.5828\\n3-1\\n9.9656\\n5\\nI\\n5\\n9.6172\\n3-6\\n3-5\\n3-6\\n3-5\\n3-4\\n3-5\\n3-4\\n3-5\\n3-4\\n3-4\\nZA\\nZ-Z\\n3-4\\nl-Z\\nZ-Z\\n3-2\\n3^2\\nZ-Z\\nZ-2\\n3^2\\n3^2\\n3-1\\n3-2\\n3-1\\n3^2\\n3^1\\n3^1\\n3^1\\n3^1\\n3-0\\n3-1\\n3-0\\n3-0\\n0.3828\\n67\u00c2\u00b0 30\\n22\u00c2\u00b0 40\\n\u00e2\u0080\u00a25859\\n\u00e2\u0080\u00a29651\\n.6208\\n\u00e2\u0080\u00a23792\\n67\u00c2\u00b0 20\\n22\u00c2\u00b0 50\\n23\u00c2\u00b0 0\\n23\u00c2\u00b0 10\\n.5889\\n3-0\\n3-0\\n2.9\\n3-0\\n.9646\\n.6243\\n\u00e2\u0080\u00a23757\\n67\u00c2\u00b0 10^\\n67\u00c2\u00b0 0\\n66\u00c2\u00b0 50\\n9-5919\\n9.9640\\n\u00e2\u0080\u00a29635\\n9.6279\\n0.3721\\n\u00e2\u0080\u00a25948\\n.6314\\n.3686\\n23\u00c2\u00b0 20\\n.5978\\n.9629\\n.6348\\n.3652\\n66\u00c2\u00b0 40\\n23\u00c2\u00b0 30\\n.6007\\n2.9\\n2.9\\n.9624\\n\u00e2\u0080\u00a26383\\n-3617\\n66\u00c2\u00b0 30\\n23\u00c2\u00b0 40\\n.6036\\n.9618\\n5\\n6\\n.6417\\n\u00e2\u0080\u00a23583\\n66\u00c2\u00b0 20\\n23\u00c2\u00b0 50\\n.6065\\n2.9\\n2.8\\n.9613\\n.6452\\n-3548\\n66\u00c2\u00b0 10\\n24\u00c2\u00b0 0\\n9.6093\\n2.8\\n9.9607\\n5\\n6\\n9.6486\\n0.3514\\n66\u00c2\u00b0 0\\n24\u00c2\u00b0 10\\n.6121\\n2.8\\n2 8\\n.9602\\n,6520\\n.3480\\n65\u00c2\u00b0 50\\n24\u00c2\u00b0 20\\n.6149\\n.9596\\n6\\n\u00e2\u0080\u00a26553\\n\u00e2\u0080\u00a23447\\n65\u00c2\u00b0 40\\n24\u00c2\u00b0 30\\n.6177\\n2.8\\n2.7\\n2.7\\n2.7\\n2.7\\n2.7\\n9 6\\n\u00e2\u0080\u00a29590\\n6\\n5\\n6\\n.6587\\n\u00e2\u0080\u00a23413\\n65\u00c2\u00b0 30\\n24\u00c2\u00b0 40\\n.6205\\n\u00e2\u0080\u00a29584\\n.6620\\n\u00e2\u0080\u00a23380\\n65\u00c2\u00b0 20\\n24\u00c2\u00b0 50\\n.6232\\n\u00e2\u0080\u00a29579\\n.6654\\n\u00e2\u0080\u00a23346\\n65\u00c2\u00b0 10\\n25\u00c2\u00b0 0\\n9.6259\\n9-9573\\n6\\n9.6687\\n0^3313\\n65\u00c2\u00b0 0\\n25\u00c2\u00b0 10\\n.6286\\n-9567\\n6\\n.6720\\n.3280\\n64\u00c2\u00b0 50\\n25\u00c2\u00b0 20;\\n\u00e2\u0080\u00a26313\\n.9561\\n6\\n.6752\\n.3248\\n64\u00c2\u00b0 40\\n25\u00c2\u00b0 30\\n.6340\\n-9555\\n(=1\\n\u00e2\u0080\u00a2678s\\n\u00e2\u0080\u00a23215\\n64\u00c2\u00b0 30\\n25\u00c2\u00b0 40\\n.6366\\n2.6\\n26\\n-9549\\n6\\n.6817\\n\u00e2\u0080\u00a23183\\n64\u00c2\u00b0 20\\n25\u00c2\u00b0 50\\n.6392\\n\u00e2\u0080\u00a29543\\n6\\n.6850\\n\u00e2\u0080\u00a23150\\n64\u00c2\u00b0 10\\n26\u00c2\u00b0 0\\n9.6418\\n2.6\\n9-9537\\n7\\n9.6882\\n0.31 18\\n64\u00c2\u00b0 0\\n26\u00c2\u00b0 10\\n.6444\\n6\\n\u00e2\u0080\u00a29530\\n.6914\\n.3086\\n63\u00c2\u00b0 50\\n26\u00c2\u00b0 20\\n.6470\\n2.5\\n2 6\\n\u00e2\u0080\u00a29524\\n6\\n6\\n.6946\\n\u00e2\u0080\u00a23054\\n63\u00c2\u00b0 40\\n26\u00c2\u00b0 30\\n\u00e2\u0080\u00a26495\\n.9518\\n.6977\\n\u00e2\u0080\u00a23023\\n63\u00c2\u00b0 30\\n26\u00c2\u00b0 40\\n.6521\\n2-5\\n2.4\\n2.5\\n2-5\\n\u00e2\u0080\u00a29512\\n7\\n6\\n.7009\\n.2991\\n63\u00c2\u00b0 20\\n26\u00c2\u00b0 50\\n.6546\\n\u00e2\u0080\u00a29505\\n.7040\\n.2960\\n63\u00c2\u00b0 10\\n27\u00c2\u00b0 0\\n9.6570\\n9-9499\\n7\\n6\\n9.7072\\n0.2928\\n63\u00c2\u00b0 0\\n27\u00c2\u00b0 icy\\n\u00e2\u0080\u00a26595\\n.9492\\n-7103\\n\u00e2\u0080\u00a22897\\n62\u00c2\u00b0 50\\n27\u00c2\u00b0 20\\n.6620\\n.9486\\n7\\n\u00e2\u0080\u00a27134\\n.2866\\n62\u00c2\u00b0 40\\n27\u00c2\u00b0 30\\n.6644\\n2.4\\n2.4\\n\u00e2\u0080\u00a29479\\n-7165\\n\u00e2\u0080\u00a22835\\n62\u00c2\u00b0 30\\n27\u00c2\u00b0 40\\n.6668\\n\u00e2\u0080\u00a29473\\n7\\n7\\n6\\n.7196\\n.2804\\n62\u00c2\u00b0 20\\n27\u00c2\u00b0 50\\n28\u00c2\u00b0 0\\n.6692\\n9.6716\\n2,4\\n2.4\\n2.4\\n2-3\\n.9466\\n.7226\\n.2774\\n62\u00c2\u00b0 10\\n62\u00c2\u00b0 0\\n9^9459\\n9.7257\\n0.2743\\n28\u00c2\u00b0 10\\n.6740\\n\u00e2\u0080\u00a29453\\n7\\n7\\n7\\n.7287\\n\u00e2\u0080\u00a22713\\n61\u00c2\u00b0 50\\n28\u00c2\u00b0 20\\n.6763\\n.9446\\n\u00e2\u0080\u00a27317\\n.2683\\n61\u00c2\u00b0 40\\n28\u00c2\u00b0 30\\n.6787\\n2.4\\n2-3\\n.9439\\n\u00e2\u0080\u00a27348\\nZ-^\\n3-0\\n.2652\\n61\u00c2\u00b0 30\\n28\u00c2\u00b0 40\\n.6810\\n\u00e2\u0080\u00a29432\\n\u00e2\u0080\u00a27378\\n.2622\\n61\u00c2\u00b0 20\\n28\u00c2\u00b0 50\\n.6833\\n2-3\\n2.2\\n-9425\\n7\\n7\\n7\\n7\\n7\\n7\\n7\\n8\\n.7408\\n3-0\\n3-0\\n2.9\\n3-0\\n2.9\\n3-0\\n2.9\\n2.9\\n\u00e2\u0080\u00a22592\\n61\u00c2\u00b0 10\\n29\u00c2\u00b0 0\\n9.6856\\n9.9418\\n9^7438\\n0.2562\\n61 \u00c2\u00b00\\n29\u00c2\u00b0 10\\n.6878\\n.9411\\n.7467\\n\u00e2\u0080\u00a22533\\n60\u00c2\u00b0 50\\n29\u00c2\u00b0 2.0\\n.6901\\n.9404\\n\u00e2\u0080\u00a27497\\n\u00e2\u0080\u00a22503\\n60\u00c2\u00b0 40\\n29\u00c2\u00b0 30\\n.6923\\n^3\\n\u00e2\u0080\u00a29397\\n.7526\\n\u00e2\u0080\u00a22474\\n60\u00c2\u00b0 30\\n29\u00c2\u00b0 40\\n.6946\\n\u00e2\u0080\u00a29390\\n\u00e2\u0080\u00a27S56\\n.2444\\n60\u00c2\u00b0 20\\n29\u00c2\u00b0 50\\n30\u00c2\u00b0 0\\n.6968\\n2.2\\n\u00e2\u0080\u00a29383\\n\u00e2\u0080\u00a27585\\n.2415\\n60\u00c2\u00b0 10/\\n60\u00c2\u00b0 0\\n9.6990\\n9-9375\\n9.7614\\n0.2386\\nCos.\\nD.l\\nSin. D.\\n1\\nCot.\\nD.l\\nTan.\\nAngle.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0189.jp2"}, "190": {"fulltext": "18\\nLOGARITHMIC SINES, COSINES,\\nAngle.\\nSin.\\nD.l\\nCos.\\nB. 1\\nTan.\\nD. 1\\nCot.\\n30\u00c2\u00b0 0\\n9.6990\\n2 2\\n9-9375\\n\u00e2\u0080\u00a27\\n-7\\n.8\\n9.7614\\n3-0\\n2.9\\n2.8\\n0.2386\\n60\u00c2\u00b0 0\\n30\u00c2\u00b0 10\\n.7012\\n.9368\\n.7644\\n.2356\\n59\u00c2\u00b0 50\\n30\u00c2\u00b0 20\\n.7033\\n.9361\\n\u00e2\u0080\u00a27673\\n\u00e2\u0080\u00a22327\\n59\u00c2\u00b0 40\\n30\u00c2\u00b0 30;\\n\u00e2\u0080\u00a27055\\n.9353\\ni\\n.7701\\n2.9\\n2-9\\n2.9\\n2.8\\n2.9\\n2.8\\n2.9\\n28\\n.2299\\n59\u00c2\u00b0 30\\n30\u00c2\u00b0 40\\n7076\\n2 I\\n\u00e2\u0080\u00a29346\\n\u00e2\u0080\u00a27730\\n.2270\\n59\u00c2\u00b0 20\\n30\u00c2\u00b0 50\\n.7097\\n2.1\\n-9338\\n\u00e2\u0080\u00a27\\n.8\\n8\\n-7759\\n.2241\\n59\u00c2\u00b0 10\\n31\u00c2\u00b0 0\\n31\u00c2\u00b0 10\\n9.7118\\n7139\\n2.1\\n2 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40\\n37\u00c2\u00b0 3o\\n.7844\\n\u00e2\u0080\u00a28995\\n.8850\\n.1150\\n52\u00c2\u00b0 30\\nCos.\\nD.l\\nSin.\\nD.l\\nCot.\\nD. 1\\nTan.\\nAngle.", "height": "3705", "width": "2370", "jp2-path": "completetrigonom00well_0190.jp2"}, "191": {"fulltext": "TANGENTS AND COTANGENTS..\\n19\\nAngle.\\nSin.\\nD.l\\nCos.\\nD.l\\nTan. 1\\nD.l\\nCot.\\n370 30\\n9.7844\\nA\\n9.8995\\n9.8850\\n2.6\\n2.6\\n2.6\\n2.6\\n2 6\\n0.II50\\n52\u00c2\u00b0 30\\n37\u00c2\u00b0 40\\n.7861\\n.8985\\n.8876\\n.1124\\n52\u00c2\u00b0 20\\n37\u00c2\u00b0 50\\n38\u00c2\u00b0 0\\n38\u00c2\u00b0 10\\n.7877\\n1.6\\n1^7\\nI 6\\n.8975\\n9.8965\\nI.O\\nI.O\\n.8902\\n.1098\\n52\u00c2\u00b0 10\\n52\u00c2\u00b0 0\\n51\u00c2\u00b0 50^\\n9-7893\\n9.8928\\n0.1072\\n.7910\\n-8955\\n.8954\\n.1046\\n38\u00c2\u00b0 id\\n.7926\\n1^5\\nI 6\\n.8945\\n.8980\\n2 6\\n.1020\\n51\u00c2\u00b0 40\\n38\u00c2\u00b0 30\\n.7941\\n.8935\\n.9006\\n2 6\\n.0994\\n51\u00c2\u00b0 30\\n38\u00c2\u00b0 40\\n\u00e2\u0080\u00a27957\\nI 6\\n.8925\\n.9032\\n26\\n.0968\\n51\u00c2\u00b0 20\\n38\u00c2\u00b0 50\\n39\u00c2\u00b0 0\\n\u00e2\u0080\u00a27973\\n1.6\\n1.6\\n1-5\\n1-5\\nT 6\\n.8915\\n9.8905\\n1.0\\nI\\n.9058\\n2.6\\n26\\n.0942\\n51\u00c2\u00b0 10\\n51\u00c2\u00b0 0\\n9.7989\\n9.9084\\n0.0916\\n39\u00c2\u00b0 10\\n.8004\\n.8895\\n.9110\\n2 6\\n.0890\\n50\u00c2\u00b0 50\\n39\u00c2\u00b0 20\\n.8020\\n.8884\\n\u00e2\u0080\u00a29135\\n.0865\\n50\u00c2\u00b0 40\\n39\u00c2\u00b0 30\\n\u00e2\u0080\u00a28035\\n-8874\\n.9161\\n.0839\\n50\u00c2\u00b0 30\\n39\u00c2\u00b0 40\\n.8050\\n.8864\\n.9187\\nri\\n.0813\\n50\u00c2\u00b0 20\\n39\u00c2\u00b0 50\\n.8066\\n1-5\\n1-5\\n1-5\\n1^4\\n15\\n1^5\\n1.4\\n1-5\\n1.4\\n1-5\\n1.4\\n1.4\\n1.4\\n1.4\\n1.4\\n1.4\\n1.4\\n1-3\\n1.4\\n1-3\\n1-4\\n1-3\\n1-3\\n1-4\\n1-3\\n1-3\\n1-3\\n1-3\\n\u00e2\u0080\u00a28853\\nI.O\\n.9212\\n.0788\\n50\u00c2\u00b0 10\\n40\u00c2\u00b0 0\\n9.8081\\n9-8843\\n9.9238\\n26\\n0.0762\\n50\u00c2\u00b0 0^\\n40\u00c2\u00b0 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"192": {"fulltext": "20\\nNATURAL SINES AND COSINES.\\nA.\\nSin.\\nCos.\\nA.\\nSin.\\nCos.\\nA.\\nSin.\\nCos.\\n0\u00c2\u00b0\\nlo\\n20\\n30;\\n40\\n50\\n1\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n2\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n3\u00c2\u00b0\\n10\\n20\\n30\\n40;\\n50\\n40\\n10\\n20\\n30\\n40\\n50\\n5\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n6\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n7\u00c2\u00b0\\n10\\n20\\n30\\n.000000 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1 609\\n.11898\\n9942\\n9939\\n9936\\n9932\\n9929\\n.3611\\n.3638\\n\u00e2\u0080\u00a23665\\n.3692\\n\u00e2\u0080\u00a23719\\n\u00e2\u0080\u00a29325\\n\u00e2\u0080\u00a29315\\n\u00e2\u0080\u00a29304\\n\u00e2\u0080\u00a29293\\n.9283\\n.2419\\n\u00e2\u0080\u00a29703\\n.2447\\n.2476\\n\u00e2\u0080\u00a22504\\n\u00e2\u0080\u00a22532\\n.2560\\n.9696\\n.9689\\n.9681\\n\u00e2\u0080\u00a29674\\n.9667\\n.12187\\n9925\\n\u00e2\u0080\u00a23746\\n.9272\\n.12476\\n.12764\\n\u00e2\u0080\u00a213053\\n9922\\n9918\\n.9914\\n\u00e2\u0080\u00a23773\\n.3800\\n.3827\\n.9261\\n\u00e2\u0080\u00a29250\\n\u00e2\u0080\u00a29239\\n.2588\\n\u00e2\u0080\u00a29659\\nCos.\\nSin.\\nA.\\nCos.\\nSin.\\nA.\\nCos.\\nSin.", "height": "3700", "width": "2370", "jp2-path": "completetrigonom00well_0192.jp2"}, "193": {"fulltext": "NATURAL SINES AND COSINES.\\n21\\nA.\\nSin.\\n40\\n50\\n23\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n24\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n25\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n26\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n27\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n28\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n29\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n30\u00c2\u00b0\\n3827\\n3854\\n3881\\n3907\\n3934\\n3961\\n3987\\n4014\\n4041\\n4067\\n4094\\n4120\\n4147\\n4173\\n4200\\n4226\\n4253\\n4279\\n4305\\n4331\\n4358\\n4384\\n4410\\n4436\\n4462\\n4514\\n4540\\n4566\\n4592\\n4617\\n4643\\n4669\\n4695\\n4720\\n4746\\n4772\\n4797\\n4823\\n4874\\n4899\\n4924\\n4950\\n4975\\n5000\\nCos.\\n9239\\n9228\\n9216\\n9135\\n9205\\n9194\\n9182\\n9171\\n9159\\n9147\\n9124\\n9112\\n9100\\n9075\\n9063\\n9051\\n9038\\n9026\\n9013\\n9001\\n8975\\n8962\\n8949\\n8936\\n8923\\n8910\\n8897\\n8857\\n8843\\n8816\\n8802\\n8774\\n8760\\n8746\\n8732\\n8718\\n8704\\n8675\\n8660\\nCos.\\nSin.\\n30\\n20\\n10\\n67\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n66\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n65\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n64\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n63\u00c2\u00b0\\nscy\\n40\\n30\\n20\\n10\\n62\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n61\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n60^\\nA.\\n30\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n31\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n32\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n33\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n34\u00c2\u00b0\\n30\\n40\\n50/\\n35\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n36\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n37\u00c2\u00b0\\n10\\n20\\n30\\nSin. Cos.\\n5000\\n5025\\n5050\\n5075\\n5100\\n5125\\n5150\\n5175\\n5200\\n5225\\n5250\\n5275\\n5299\\n5324\\n5348\\n5373\\n5398\\n5422\\n5446\\n5471\\n5495\\n5519\\n5544\\n5568\\n5592\\n5616\\n5640\\n5664\\n5688\\n5712\\n5736\\n5760\\n5783\\n5807\\n5831\\n5854\\n5878\\n5901\\n5925\\n5948\\n5972\\n5995\\n6018\\n6041\\n6065\\n8660\\n8646\\n8631\\n8616\\n8601\\n8587\\n8572\\n8557\\n8542\\n8526\\n8511\\n8496\\n8480\\n8465\\n8450\\n8434\\n8418\\n8403\\n8387\\n8371\\n8355\\n8339\\n8323\\n8307\\n8290\\n8274\\n8258\\n8241\\n8225\\n8208\\nI192\\n8175\\n8158\\n8141\\n8124\\n8107\\n8090\\n8073\\n8056\\n8021\\n8004\\n7986\\n7969\\n7951\\n7934\\nCos. Sin.\\n60\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n59\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n58\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n57\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n56\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n55\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n54\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n53\u00c2\u00b0\\n50;\\n40\\n30\\nA.\\nA.\\n3c/\\n40\\n50\\n38\u00c2\u00b0\\n10\\n2(y\\n30\\n40\\n50\\n39\u00c2\u00b0\\n20\\n30\\n40\\n50\\n40\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n41\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n42\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n43\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n44\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n45\u00c2\u00b0\\nSin. Cos.\\n6088\\n61II\\n6134\\n6157\\n6180\\n6202\\n6225\\n6248\\n6271\\n6293\\n6316\\n6338\\n6361\\n6383\\n6406\\n6428\\n6450\\n6472\\n,6494\\n6517\\n6539\\n6561\\n,6583\\n,6604\\n,6626\\n,6648\\n6670\\n669]\\n6713\\n6734\\n,6756\\n6777\\n6799\\n6820\\n.6841\\n,6862\\n,6884\\n,6905\\n.6926\\n6947\\n6967\\n,7009\\n,7030\\n,7050\\n7071\\n7934\\n7916\\n7898\\n7862\\n7844\\n7826\\n7808\\n7790\\n7771\\n7753\\n7735\\n7716\\n7698\\n7679\\n7660\\n7642\\n7623\\n7604\\n7585\\n7566\\n7547\\n7528\\n7509\\n7490\\n7470\\n7451\\n7431\\n7412\\n7392\\n7373\\n7353\\n7333\\n7314\\n7294\\n7274\\n7254\\n7234\\n7214\\n7193\\n7173\\n7153\\n7133\\n7112\\n7092\\n7071\\n30\\n20\\n10\\n52\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n51\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n50\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n49\u00c2\u00b0\\n50\\n40\\n48\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n47\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n46^\\n50\\n40\\n30\\n20\\n10\\n45^\\nCos. Sin.", "height": "3624", "width": "2312", "jp2-path": "completetrigonom00well_0193.jp2"}, "194": {"fulltext": "22\\n^ATURAL TANGENTS AND COTANGENTS.\\nA.\\nTan.\\nCot.\\nA.\\nTan.\\nCot.\\nA.\\nTan.\\nCot.\\n0\u00c2\u00b0\\nlO\\n20\\n30\\n50\\n1\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n2\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n3\u00c2\u00b0\\n10\\n20\\n30\\n40;\\n50\\n4\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n5\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n6\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n7\u00c2\u00b0\\n10\\n20\\n30\\n.000000\\n00\\n90\u00c2\u00b0\\n50;\\n40\\n30^\\n20\\n10\\n89\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n88\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n87\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n86\u00c2\u00b0\\n10\\n85\u00c2\u00b0\\n50;\\n40\\n30;\\n20\\n10\\n84\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n83\u00c2\u00b0\\n50;\\n40\\n30\\n30\\n40\\n50\\n8\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n90\\n10\\n20\\n30;\\n40^\\n50\\n10\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n11\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n12\u00c2\u00b0\\n10\\n20\\n30\\n40;\\n50\\n13\u00c2\u00b0\\n10;\\n20\\n30;\\n40\\n50\\n14\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n15\u00c2\u00b0\\n\u00e2\u0080\u00a2I317\\n.1346\\n-1376\\n7^5958\\n7.4287\\n7.2687\\n30;\\n20\\n10\\n82\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n81\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n80\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n79\u00c2\u00b0\\n50;\\n40\\n30;\\n20\\n10\\n78\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n77\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n76\u00c2\u00b0\\n50\\n40\\n30;\\n20\\n10\\n75\u00c2\u00b0\\n15\u00c2\u00b0\\n10\\n20^\\n30\\n40\\n50\\n16\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n17\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n18\u00c2\u00b0\\n10\\n20\\n30\\n40\\n50\\n19\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n20\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n21\u00c2\u00b0\\n10\\n20\\n30;\\n40\\n50\\n22\u00c2\u00b0\\n10\\n20\\n30\\n.2679\\n3^732i\\n75\u00c2\u00b0\\n50\\n40\\n30\\n20\\n10\\n74\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n73\u00c2\u00b0\\n50;\\n40\\n30\\n20\\n10\\n72\u00c2\u00b0\\n50;\\n40\\n.002909\\n.005818\\n.008727\\n.011636\\n.014545\\n343-7737\\n171.8854\\n114.5887\\n85.9398\\n68.7501\\n.2711\\n.2742\\n-2773\\n.2805\\n.2836\\n3.6891\\n3.6470\\n3-6059\\n3-5656\\n3.5261\\n.1405\\n7-II54\\n\u00e2\u0080\u00a21435\\n.1465\\n\u00e2\u0096\u00a01495\\n.1524\\n\u00e2\u0080\u00a21554\\n6.9682\\n6.8269\\n6.6912\\n6.5606\\n6.4348\\n\u00e2\u0096\u00a0017455\\n57.2900\\n.2867\\n3-4874\\n.02036\\n.02328\\n.02619\\n.02910\\n.03201\\n49.1039\\n42.9641\\n38.1885\\n34-3678\\n31.2416\\n.2899\\n.2931\\n.2962\\n-2994\\n.3026\\n3-4495\\n3.4124\\n3-3759\\n3-3402\\n3-3052\\n.1584\\n6.3138\\n.1614\\n.1644\\n\u00e2\u0096\u00a01673\\n\u00e2\u0096\u00a01703\\n\u00e2\u0080\u00a21733\\n6.1970\\n6.0844\\n5-9758\\n5.8708\\n5-7694\\n.03492\\n28.6363\\n-3057\\n3.2709\\n\u00e2\u0080\u00a203783\\n.04075\\n.04366\\n.04658\\n.04949\\n26.4316\\n24.5418\\n22.9038\\n21.4704\\n20.2056\\n.3089\\n.3121\\n\u00e2\u0080\u00a231^3\\n-3185\\n\u00e2\u0080\u00a23217\\n3-2371\\n3.2041\\n3.1716\\n3-1397\\n3.1084\\n.1763\\n5-6713\\n\u00e2\u0096\u00a01793\\n.1823\\n.1914\\n5-5764\\n5-4845\\n5-3955\\n5-3093\\n5-2257\\n.05241\\n19.0811\\n\u00e2\u0080\u00a23249\\n3-0777\\n\u00e2\u0080\u00a205533\\n.05824\\n.06116\\n.06408\\n.06700\\n18.0750\\n17.1693\\n16.3499\\n15.6048\\n14.9244\\n.3281\\n\u00e2\u0080\u00a23314\\n\u00e2\u0080\u00a23346\\n\u00e2\u0080\u00a23378\\n\u00e2\u0080\u00a2341 1\\n3-0475\\n3.0178\\n2.9887\\n2.9600\\n2.9319\\n.1944\\n5.1446\\n30; 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