QA5'5 Digitized by the Internet Archive in 2010 with funding from Boston Library Consortium IVIember Libraries http://www.archive.org/details/essentialsofplanOOwell WELLS' MATHEMATICAL SERIES. Academic Arithmetic. Academic Algebra. Higher Algebra. University Algebra. College Algebra. Plane Geometry. Solid Geometry. Plane and Solid Geometry. Plane and Spherical Trigonometry. Plane Trigonometry. Essentials of Trigonometry. Logarithms (flexible covers). Elementary Treatise on Logarithms. Special Catalogue and Terms on application. ^- THE ESSENTIALS OF PLAXE AXD SniERICAL TRIGOXOMETllY. BY WEBSTER WELLS, S.B., Associate Professor of Mathematics in the Massachusetts Institute of Technology. coU.»o»,-^' •iSr^^^ ^'""^ irUkTrt . oerT' LEACH, SHEWELL, AND SANBOEN. BOSTON. NEW YORK. CHICAGO. Copyright, 1887, By WEBSTER \YELLS, BOSTON COLLEGE LIBRARY CHESTNUT HILL, MASS. Typography by Presswork by J. S. Gushing & Co. Berwick & Smith. PEEFACE. This volume contains only those portions of Plane and Spherical Trigonometry which are essential in the practical applications of the subject to problems in surveying, geod- esy, and navigation. The attention of teachers and others is invited to the fol- lowing, which may be regarded as constituting the salient features of the work : 1 . The proofs of the functions of 0°, 90°, 180°, and 270° ; Arts. 38 to 41. 2. The figures, and the tabular arrangement of the work, in the discussions of Arts. 42 and 44. 3. The solution of problems in Art. 56 by the construc- tion of a diagram, in a manner analogous to that of Art. 15. 4. The proofs of the fundamental formuhe for any angle ; Arts. 57 to 59. 5. The discussion of the line values of the trigonometric functions, and their application in tracing the changes in the functions as the angle increases from 0° to 360° ; Arts. 62 to 64. 6. The proofs of the formulje for the sines and cosines of X -\- y and x — y for an}' values of x and y ; Arts. 68 to 70. 7. The discussion of the ambiguous case in the solution of plane oblique triangles ; Arts. 124 to 126. iy PREFACE. 8. The geometrical proofs (Art. 138) of the propositions that in any spherical riglit triangle : I. If the sides including the right angle are in the same quadrant, the hypotenuse is < 90° ; if they are in different quadrants, the hypotenuse is > 90°. II. An angle is in the same quadrant as its opposite side. 9. The discussion of the properties of spherical right triangles before those of spherical oblique triangles ; see Chapters XI. and XII. 10. The reduction of the number of cases in the complete demonstration of the fundamental formulae for spherical right triangles, to three, by application of the results proved geometrically in Art. 138 ; see Art. 143. 11. The discussions of the ambiguous cases in the solution of spherical oblique triangles (Arts, 171 and 172) ; espe- cially the rules given on pages 130 and 132 for determining the number of solutions. At the end of the book will be found a collection of for- mulae in form for convenient reference. Teachers who desire a briefer course are recommended to omit Chapter IV., which may be done without interrupting the logical completeness of the rest of the work. Chapter VI. may also be omitted by those who have taken the sub- ject of Logarithms in their course in Algebra. The course might be still further abridged, if desired, by the omission of the exercises at the end of Chapter V. WEBSTER WELLS. Boston, 1887. CONTENTS. PART I. PLANE TRIGONOMETRY. Page T. Definitions ; Measurement of Angles .... 1 Circular Measure of an Angle 1 IT, The Trigonometric Functions 4 Functions of Acute Angles 4 Fundamental Theorems 10 III. Application of Algebraic Signs 12 Trigonometric Functions of Angles in General . . 12 Rectangular Co-ordinates 14 General IVfinitions of the Functions 15 Functions of 0° 90° 180° 270°, and 360° .... 19 Functions of (—yl) in Terms of those of ^ ... 21 Functions of (90° + .4) in Terms of those of ^1 . . 24 Proofs of the Fundamental Formuhu for any Angle 34 IV. Miscellaneous Theorems 36 To express Each of the Six Principal Functions in Terms of the other Five 36 T • •• • -IT 1 (• sin X . tan x oir Limiting Values of and 37 XX Line Values of the Trigonometric Functions ... 38 V. General Formula 42 Functions of 2 a: 48 Functions of ^ a; 49 Inverse Trigonometric Functions 50 VI. Logarithms 54 Properties of Logarithms 56 Applications 61 Arithmetical Complement 63 VII. Solution of Right Triangles 67 Formulae for the Area of a Right Triangle .... 73 yi CONTENTS. Page VIIL General Properties of Triangles 75 Formulae for the Area of an Oblique Triangle . . 81 IX. Solution of Oblique Triangles 83 Area of an Oblique Triangle 92 PART 11. SPHERICAL TRIGONOMETRY. X. Geometrical Definitions and Principles ... 95 XI. Spherical Right Triangles 99 Napier's Rules of Circular Parts 104 Solution of Spherical Right Triangles 106 Solution of Quadrantal Triangles 112 XII. Spherical Oblique Triangles . 114 General Properties of Spherical Triangles ... 114 Napier's Analogies ' . 121 Solution of Spherical Oblique Triangles .... 123 Applications 133 Formula . 137 Answers to the Examples 144 Paet L PLAIs^E TRIGOXOMETRY. JXKc I. DEFINITIONS; MEASUREMENT OF ANGLES. 1. Trigonometi^ is that branch of mathematics in which algebraic processes are used to treat of the properties and measurement of geometrical figures. In Plane Trigonometry we consider j^Zajie figures only. 2. An angle is measured by finding its ratio to anothei angle adopted ar])itrarily as the unit of measurement. 3. The usual unit of measurement for angles is the degree, or an angle equal to the ninetieth part of a right angle. To express fractional parts of the unit, the degree is di- vided into sixty equal parts, called minutes, and the minute into sixty equal parts, called seconds. Degrees, minutes, and seconds are denoted by the sym- bols °, ', ", respectively ; thus, 43° 22' 37" denotes an angle of 43 degrees, 22 minutes, and 37 seconds. CIRCULAR MEASURE OF AN ANGLE. 4. Another method of measuring angles, and one of great importance, is known as tlie circular method, 'in which the unit of measurement is the angle subtended at the centre of a circle by an arc ivhose length is equal to the radius of the circle. ^ PLANE TRIGONOMETRY. 5. Let AOB be any angle, and AOO the unit of circular measure. C B By Geometry, we have angle ^O-g _ arc AB angle ^0(7 slycAG arc AB Whence, circular measure ^05 = — ;-— — (Art. 4). OA That is, the circular measure of an angle is equal to the ratio of its subtending arc to the radius of the circle. For example, the circular measure of a right angle is equal to the ratio of one-fourth the circumference to the radius. But the circumference of a circle is equal to the radius multiplied by 27r, where tt = 3.14159265 ... Hence, if R denotes the radius, we have cu'cular measure of 90° = i0f27ri? TT B 6. Since the circular measure of 90° is -, the circular measure of 180° is tt ; of 60°, - ; of 45°, - ; of 30°, ^ ; etc. 3 4 6 That is, an angle expressed in degrees may be reduced to circular measure by finding its ratio to 180°, and multiplying the result by tt. 23 Thus, sinfte 115° is — of 180°, the circular measure of 36 115° is ?^. 36 DEFIXITI0:N^S ; MEASUREMENT OF AXGLES. 3 7. Conversely, any angle expressed in circular measure may he reduced to degrees by multiplying by 180° and dividing by IT ; or, more briefly, by substituting 180° /or tt. Thus, l!I = -I of 180° = 84°. 15 15 8. In the circular method such expressions may occur as " the angle -," " the angle 1," etc. 3 These refer to the unit of circular measure ; that is, the 2 angle - means an angle whose subtending arc is two-thirda 3 of the radius. The angle 1, that is, the angle whose subtending arc is equal to the radius, or the unit of circular measure, if reduced to degrees l)y the rule of Art. 7, gives 180 180 _„ ^^^_^,^o TT 3.14150265... The rule of Art. 7 may then be modified as follows : Any angle expressed in circular measure may be reduced tc degrees by multiplying by 57.2957795°... Thus, the angle ? = - x 57.2957795°... ^33 = 38.1971863°... = 38° 11' 49.87068"... EXAMPLES. 9. Express the following in circular measure : 1. 135°. 3. 11° 15'. 5. 29° 15'. 7. 128° 34f. 2. 198°. 4. 37° 30'. 6. 174° 22' 30". 8. 92° 48' 45". Express the following in degree measure : 9. 1 11. 11^. 13. ?. 15. ^^-^ 2 30 4 3 10. — . 12. — . 14. 2. 16. ^~^. 5 4 4 PLANE TRIGONOMETRY. II. THE TRIGONOMETRIC FUNCTIONS. FUNCTIONS OF ACUTE ANGLES. 10. Let BAG be any acute angle. From any point in either side, as B, draw BO perpendic- ular to the other side, forming a right triangle ABC. We then have the following definitions, applicable to either of the acute angles A and B : In any right triangle^ The SINE of either acute angle is the ratio of the op- posite side to the hypotenuse. The COSINE is the ratio of the adjacent side to the hypot- enuse. The TANGENT is the ratio of the opposite side to the adja- cent side. The COTANGENT is the ratio of the adjacent side to the oppo- site side. The SECANT is the ratio of the hypotenuse to the adjacent side. The COSECANT is the ratio of the hypotenuse to the opposite side. That is, denoting the sides BC, CA, and AB by a, b, and c, and employing the usual abbreviations, . a sm A = -t c cos A = -'i c a tan A = -') cot A=-') a sec ^ = -5 b csc^ = -• a (1) THE TRIGONOMETRIC FUNCTIONS. And in like manner, sin B = 1 c COS B = -i c tan B = -1 a cot B = —I ^ seeB = --> I a V -n ^ CSC B= -' b (2) 11. The following definitions are also used : The versed sine of an angle is equal to unity minus the cosine of the angle. TJie coversed sine is equal to unity 7ninus the sine. That is, vers A= 1 — ? covers A= 1 » c c vers B=l — a covers B= 1 12. The eight ratios defined in Arts. 10 and 11 are called Trigonometric Functions, or Trigonometric Jiatios, of the angle. It is important to observe that their values depend solely on the magnitude of the angle, and are entirely independent of the lengths of the sides of the right triangle which con- tains it. f yD c c "^T—E Thus, let B and B' be any two points in the side AD of the angle DAE, and di-aw BC and B'C perpendicular to AJE. Then by Art. 10, we have . . BC , . , B'C sm A = — — , and sm A = ^ ^, ♦ AB AB' g PLANE TRIGONOMETRY. But the right triangles ABO and AB'C^ are similar, since they have the angle A common ; and therefore, by Geometry, ' BC ^B'C AB AB'' Thus the two values obtained for sin A are seen to be equal. 13. We obtain from (i) and (2), Art. 10, a = c sin ^, 6 = c sin jB, b = c cos A, a = c cos B. That is, in any right triangle, either side about the right angle is equal to the hypotenuse multiplied by the sine of the opposite angle, or by the cosine of the adjacent angle. 14. We have from Arts. 10 and 11, a h sin ^ = - = cos B, smB = - = cos A, c c a b tan A = T = cot J5, tan B = - = cot A, b a c c seGA = T = CSC B, secB=- = esc A, b a b a vers A=l — = covers B, vers B= 1 = covers A. c c Since the angles A and B are complements of each other, the above results may be expressed as follows : The sine, tangent, secant, and versed sine of an acute angle are respectively the cosine, cotangent, cosecant, and coversed sine of the complement of the angle. It is from this circumstance that the names co-sine, co- tangent, etc., were derived. 15. The Pythagorean Theorem affords a simple method for finding the values of the remaining seven functions of an acute angle, when the value of any one is given. The trigoxometric functions. 1. Given cot ^= 2 ; requii*ed the values of the remaining functions of A. B The equation may be written cot A = — Then since the cotangent is the ratio of the adjacent side to the opposite side, we may regard our vahie as having been taken from a right triangle ABC\ having its side AC adjacent to the angle A equal to 2 units, and its side BC opposite to A equal to 1 unit. But by Geometry, we have AB = V^.lC'-f BC = V^4 H- 1 = V^S. Whence bv definition, sin A = — 1 _Vo cos J. sec A = ^^—^ 2 tan A = -1 2 esc vers A=s/5, A=l-'^. 5 A 1 V^ covers A = l ~ j— . 5 2. Given covers A = -; required the values of the remain- ins functions of A. PLANE TKIGONOMETRY. 2 3 Since covers A=l — sin A^ we have sin A= 1 = -• 5 5 We then take the opposite side BC equal to 3 units, and the hypotenuse AB equal to 5 units. Then, AC==^AB' - BO'^ =^25 -9 = 4. Whence by definition, 4 cos A = -•) 5 tan A = --) 4 cot A — —> 3 sec A = -i 4 CSC A = -f 3 . 1 vers A = -- 5 EXAM PLES. In each case find the values of the remaininor functions : 3. tan^ = -. 3 6. vers A = — 4 X 9. cotA = x. 4. covers ^ = — 7. sin J. =-• 10. cos^ y 17 5. CSC ^ = 4. 8. sec^ = — 11. sec^=-'^ — 5 b 16. To find the values of the sine^ cosine, tangent, cotan^^ gent, secant, and cosecant of 45°. Let ABC be an isosceles right triangle, having each of the sides AC and BC equal to 1. Then, AB = s/aC' + BC' = \/l + 1 = ^2. Also, the angles A and B are equal ; and since their sum is a right angle, each is equal to 45°. THE TRIGOXOMETRIC FUNCTIONS. Then by definition, sin 45° = -4 = ^. tan 45° = 1 . sec 45° = sJ'2. \/2 2 \/2 2 cot 45° = 1 CSC 45° = \/2. The second line might have been derived from the first by aid of Art. 14, since 45° is complement of itself. 17. To Jind the values of the sine, cosine, tangent, cotan- gent, secant, and cosecant of 30° and 60°« Let ABD be an equilateral triangle, having each side equal lo 2. Draw AC perpendicular to BD ; then by Geometr3', BC=IBD=\, and Z BAC =l-Z BAD = 30°. Again, AC = ^ Alf - BC"" = ^4 - 1 = y/3. Then from the triangle xiBC, by definition. sin 30° = = cos 60°. tan30° = i=^ = cot 60°. V3 o o_ 2 _2V^3_ sec 30° = -4= = v/^ = CSC 60 . cos 30° = ^ = sin 60°. 2 cot30° = v/3=tan60°. CSC 30° = 2 = sec 60°. Or the functions of 60° may be derived from those-'of 30^^ by aid of Art. 14, since 60° is the complement of 30°, 10 PLANE TRIGONOMETRY. FUNDAMENTAL THEOREMS. 18. We obtain from the definitions of Art. 10, Sin A CSC ^ = - X - = 1 . c a A A b C 1 COS A sec A = - x - = 1 . c b tan A cotA = -x -=1. These results may be written sin A = cos A = 1 CSC A 1 sec A tan^ = cot A = cot ^' 1 tan^ sec A — cos A ^ }■ (3^ CSC A = sin A That is, tJie sine of an acute angle is the reciprocal of the cosecant^ the tangent is the reciprocal of the cotangent^ and the secant is the reciprocal of the cosine. 19. To j)rove the formula sin- A + cos- ^4=1. Note. Sin^^ signifies (sin A)'^; that is, the square of the sine of A. By Geometry, cr + 6^ = c^. Dividing by c^, gY + ('^Y=l. THE TRIGONOMETRIC FUNCTIONS. H Whence by definition, {sin Ay -\-{qos Ay- = 1; or, sin^ A + cos- A= 1 . (4) The result may be written in the forms sin ^1 = V 1 — cos'- A, acd cos A = y/l — sin- A. 20. To prove the formulcB , . sin A 1 . A cos A tan A = , and cot A = cos A sin A a Bv definition, tan .1 = - = ^ = ^HL4, (5) c -, 4. < ^ '" cos ^ , X and cot ^-1 = - = - = (g) a t< sin J. c 21. 7'o prove /7ie formulce sec- .l = 1 4- tan- A, and csc^ ^= 1 + cot" A. By G eom etry , c- = cr -f b- . Dividing by 5-, ^ = 1 + ^. That is, sec- .4 = 14- tan-' A. (7) Dividing by a% - = 1 + -.• a- a- That is, csc^^l = 1 + cot- A. (s) 12 PLANE TKIGONOMETRY. III. APPLICATION OF ALGEBRAIC SIGNS. TRIGONOMETRIC FUNCTIONS OF ANGLES IN GENERAL. 22. In Geometry we are, as a rule, concerned with angles less than two right angles ; but in Trigonometry it is con- venient to regard them as unrestricted in magnitude. A"' In the circle AA", let AA" and A' A'" be a pair of perpen- dicular diameters. Suppose a radius OB to start from the position OA, and revolve about the point as a pivot in the direction of OA'. When it coincides with OA', it has described an angular magnitude of 90° ; when it coincides with OA", of 180° ; with OA'", of 270° ; with OA, its starting-point, of 360° ; with OA' again, of 450° ; etc. We thus see that a signifi- cance may be attached to any positive angular magnitude. 23. The interpretation of an angle as the measure of the amount of rotation of a moving radius, enables us to distin- guish between positive and negative angles. Thus, if a x)ositive angle is taken to indicate revolution from the position OA in the direction of OA', a negative angle would naturally be taken as signifying revolution from the position OA' in the opposite direction, or towards OA'". APPLICATIOX OF ALGEBRAIC SIGNS. 13 That is, if the angles AOB and AOB' are each equal to 30° in absolute value, we should sa}- that AOB = +30°, and AOB' = -S(f. We may thus conceive of negative angles of any magni- tude whatever. It is immaterial which du'ection we consider the positive direction of rotation ; but having adopted a certain direction as positive, our subsequent operations must be. in accordance. 24. The fixed line OA, from which the rotation is sup- posed to commence, is called the initial line, and the final position of the rotating radius is called the terminal line. Either side of an angle may be taken as the initial line, the other being then the terminal line ; thus, in the angle AOB, we may consider OA the initial line and OB the terminal line, in which case the angle is positive ; or we may consider OB the initial line and OA the terminal line, in which case the anoxic is negative. 25. In designating an angle, we shall always write first the letter at the extremity of the initial line ; thus, in desig- nating the angle AOB, if we regard OA as the initial line, we should call it AOB, and if we regard OB as the initial line, we should call it BOA. 26. There are always two angles in absolute value less than 3G0°, one positive and the other negative, formed by a given initial and terminal line. Thus, there are formed by OA and OB' the positive angle AOB' greater than 270°, and the negative angle AOB' less than 90°. We shall distinguish between such angles by referring to them as "the positive angle AOB','' and "the negative angle AOB'," respectively. 27. It is evident that the terminal lines of two angles which differ by a multiple of 360° are coincident ; thus, the angles 30°, 390°, —330°, etc., have the same terminal line. 14 PLANE TRIGONOMETRY. RECTANGULAR CO-ORDINATES. Y a h N a -P3 i-bra) P,{h,a) a M a -X PiihrO) 28. Let XX' and YY' be a pair of straight lines at right angles to each other ; let Fi be any point in their plane, and draw PiiHf perpendicular to XX'. The distances OM and PiM are called the rectangular co- ordinates of Pi with reference to the lines XX' and YY'. OJf is called the abscissa, and FiMthe ordinate; the lines XX' and YY' are called the axes of X and Y, respectively, and their intersection is called the origin. 29. If in the above figure, OM— 0N= &, and the per- pendiculars PiM= P2N— PzN= P^M= a, the points Pi, Pg? P3, and P4 will have the same co-ordinates. To avoid this ambiguity, the following conventions have been adopted : Abscissas measured to the right of are considered posi- tive., and to the left., negative. Ordinates measured above the line XX' are considered positive,^ and below, negative. Then the co-ordinates of the four points will be : Point. Abscissa. Ordinate A b a P2 -h a A -h — a A b — a APPLICATION OF ALGEBRAIC SIGNS. 15 Note. In the figures of this chapter, the small letters denote the lengths of the lines, without regard to their algebraic sign. 30. It is customary to denote the abscissa and ordinate of a point by the letters x and ?/, respectively. Thus the fact that the abscissa of a point is equal to h and its ordinate to a, is expressed by the saying that for the point in question x=h and y^=a. The same fact ma}- be stated more concisely by referring to the point as " the point (6, a)", where the first quantity in the parenthesis is understood to be the abscissa, and the second the ordinate. 31. If a point lies upon the axis of X, its ordinate is zero ; and the same is true of the abscissa of a point upon the axis of Y. GENERAL DEFINITIONS OF THE FUNCTIONS. 32. We will now give general delinitions for the trigono- metric functions, applicable to any angle whatever. Take the initial line as the positive direction of the axis of X, the vertex being the origin. Take any point in the terminal line, and construct its rec- tangular co-ordinates by dropping a perpendicular to the initial line, produced if necessary. Then, designating the distance of the assumed point from the origin as the "• distance " of the point. The SINE is the ratio of the ordinate to the distance. The cosine is the ratio of the abscissa to the distance. The tangent is the ratio of the ordinate to the abscissa. The COTANGENT is the ratio of the abscissa to the ordinate. Tlie secant is the ratio of the distance to the abscissa. The cosecant is the ratio of the distance to the ordinate. Note. These definitions include those of Art. 10. The definitions of the versed sine and coversed sine, given in Art. 11, are sufficiently general to apply to any angle whatever. 16 PLANE TRIGONOMETRY. 33. We will now apply the definitions of Ai't. 32 to the angles XOPj, XOP^, XOP^, and XOP^ in the following- figures : Let Pi, Po, P3, and P4 be any points on the terminal lines OPi, OP21 OP^^ and OP4, and let their co-ordinates be (6, a), (—6, ci), (— ?>, — «), and (6, —a), respectively. Let OPi = OPo = OPs = OP4 = c. Then by the definitions, sin XOPi = -• c tanXOPi = -. cosXOPi = cot XOPi = -• sec XOP. = -. ^ 5 esc XOPi = -. a sin XOP2 = -. c cosXOP2 = -b b c c tan XOPo— ^ — ^, cot XOP2 = -b b — 5 5 a a -b b CSCXOP2 __ c a ArPLlCATIOX OF ALGEBRAIC SIGNS. 17 X- PA-lj, CSC A cot A cos A cos A = ^ cot A = 1 esc A = -• sec A tan A sm A APPLICATIOX OF ALGEBRAIC SIGNS. 35 58. To prove the formulae,, bin-^1 + cos^.l = 1 , sec- .4 = 1 + tan-^, csc^^ = 1 + cot- .4, for any value of A. Since the distance of the point of reference is the hypote- nuse of a riglit triangle wliose sides are equal to the absolute values of the abscissa and ordinate, we have by Geometry, (ordinate)- + (abscissa)- = (distance)-. This ma}' be written in the forms, /ordinateV / absciss aY _ ^ \distancey \distance/ /distance Y_ -, /ordinaceV \abscissay \abscissa/ and /'distanceY-_ , /abscissa^^ ordinate/ \ordinatey Hence for any value of A, we have sin- .4 + cos-yl = 1 , sec- .4 = 1 + tan-yl, csc-yl = 1 + cotM. 59. To prove the formulae 4 sin A J .J cos A tan A = ana cot A = -i cos A sin A for any value of A. By definition, ordinate . ordinate distance sin A tan A = = -; — : = -> abscissa abscissa cos J. distance and abscissa , . abscissa distance cos^ cot A = = — -. — — = -• ordinate ordinate sin A distance 36 PLANE TRIGONOMETRY. IV. MISCELLANEOUS THEOREMS. TO EXPRESS EACH OF THE SIX PRINCIPAL FUNCTIONS IN TERMS OF THE OTHER FIVE. 60. The following table expresses the value of each of the six principal functions in terms of the other five : sin COS tan cot sec CSC tan 1 y/l + cot- cot 1 CSC y/sec2 - sec 1 sec 1 \/l - cos2 y/l — cos- cos cos y/l + tan- 1 y/csc- — 1 CSC 1 y/l - sin^ sin \/l — sin--2 sjl + tan^ 1 tan y/l + cot2 1 cot y/l + cot2 cot y/sec"-2 - 1 1 y/csc'- — 1 y/l — sin- sin 1 y/csc- — 1 CSC y/l - cos2 1 COS 1 y/l — cos- y/sec2 - sec a y/l + tan"-^ y/l — sin^ 1 sin y/csc'- — 1 y^l + tan- tan y/l + cot- y/sec^ - ^ The reciprocal forms were proved in Art. 57 ; the others may be derived by aid of Arts. 57, 58, and 59, and are left as exercises for the student. As an illustration, we will give a proof of the formula . Vcsc^ A-1 cos A = ' CSC A By Art. 58, cos A=\/l- s'm^A = i 1 1 ^ y/csc^ A-l csc^ A CSC A MISCELLAXEOUS THEOREMS. 3T LIMITING VALUES OF ^^^ AND ^^5^- 61. To find the limiting values of the fractions tan X X sin X X arid as X approaches the limit 0. Note. AVe suppose x to be expressed in circular measure (Art. 4). Let OPXP' be a circular sector; draw PT and P'T tan- gent to the arc at P and P', and join OT and PP'. By Geometry, PT=P'T. Then OT is perpendicuhir to PP' at its middle point M, and bisects the arc PP' at X. Let Z XOP=Z XOP' = X. By Geometry, arc PP' > chord PP', and < PTP. Whence, arc PX > PJ/, and < PT. Therefore. arc PX ^ PM . PT OP OP OP or, by Art. 5, ciic. meas. x > sin.i-, and < tanoj. (a) Representing the circular measure of x by x simply, and dividing through by sin x, we have 1 •'^' ^ 1 1 ^ tan X > 1, and < or That is. sin X sin X X sm X cos X < 1 , and > cos x. 38 PLANE TRIGONOMETRY. But as X approaches the limit 0, cos x approaches the hmit 1 (Art. 38). Hence, approaches the limit 1 as i» approaches 0. X tana; sin a; since 1 X Affain, ° XX cos X X cos X But each factor approaches the limit 1 as a; approaches 0. Hence, -^ — approaches the limit 1 as a; approaches 0. X LINE VALUES OF THE TRIGONOMETRIC FUNCTIONS. 62. Let AOB be any angle. With as a centre, and a radius equal to 1, describe the circle AB; draw BD and AE perpendicular to XX', and CF perpendicular to YY'. Then by Art. 32, the functions of AOB are : MISCELLANEOUS THEOREMS. 39 ; Si'n. 1 Cos. 1 1 Tan. Cot. Sec. Csc. Fig. 1 Fig. 2 Fig. 3 Fig. 4 BD OB BD OB BD OB BD OB OD OB OD OB OD OB OD OB BD OD BD OD BD OD BD OD 1 bobobobo bbbbbbbb OB \ OB OD BD OB j OB OD ' BD OB OB OD BD OB OB OD BD P>ut since the right triangles OBD^ OEA, and OCF are similar, and OA = OC = 1 , we have, BD^AE OD ~ OA AE, OB^OE OD ~ OA OE, Qll=.9l=CF BD OC BD OC Whence, since 0B= 1, the fnnctions of AOB are Sin. Cos. Tan. Cot. Sec. Csc. Fig. 1 BD OD AE CF OE OF Fig. 2 BD -OD -AE -GF -OE OF Fig. 3 -BD -OD AE CF - OE -OF Fig. 4 -BD OD -AE -CF OE - OF That is, if the radius of the circle is unity, The sine is the perpendicular drawn to XX' from the intersection of the circle with the terminal line. The cosine is the line drawn from the centre to the foot of the sine. The tangent is that portion of the geometrical tangent to the circle at its intersection with OX included between OX and the terminal line, produced if necessary. 40 PLANE THIGOXOMETRY. The cotangent is that portion of the geometrical tangent to the circle at its intersection with Y included between Y and the terminal line, produced if necessary. The secant is that portion of the terminal line, or terminal line produced, included between the centre and tlie tangent. The cosecant is that portion of the terminal line, or terminal line produced, included between the centre and the cotangent. And with regard to algebraic signs, Sines and tangents measured above XX' are positive^ and beloiv, negative ; cosines and cotangents measured to the right of YY' are positice^ and to the left^ negative; secants and cosecants measm'ed on the terminal line itself are posi- tive^ and on the terminal line produced, negative. 63. The above are called the line vcdues of the trigono- metric functions. They simply represent the values of the functions when the radius is unity ; that is, the numerical value of the sine of an angle is the same as the number which expresses the length of the perpendicular drawn to XX' from the intersection of the circle and terminal line. 64. To trace the changes in the six principcd trigonometric functions of an angle as the angle increases from 0° to 360°. Let the terminal line start from the position OA, and revolve about as a pivot in the direction of 1", occupying successively the positions OB^. OB... OC, OB^, OB^, etc. MISCELLANEOUS THEOREMS. 41 Then since the sine of the angle commences with the vakie 0, and assumes in succession the values ByD^^ BoDo, OC, B^D^, B^D^^ etc. (Art. GO), it is evident that as the angle increases from O'' to 90°, the sine increases fron to 1 ; from 90° to 180°, it decreases from 1 to : from 180° to 270°, it decreases (algebraically) from to —1; and from 270° to 360°, it increases from —1 to 0. Since the cosine commences with the value OA^ and as- sumes in succession the values OZ>i, OD.j-, 0, —01);^^, —OD^, etc., from 0° to 90°, it decreases from 1 to ; from 90° to 180°, it decreases from to -1 ; from 180° to 270°, it in- creases from —1 to ; and from 270° to 360°, it increases from to 1. Since the tangent commences with the value 0, and as- sumes in succession the values ^-LE'i, AE.2-, ^c, — vLEg, — AE^, etc., from 0° to 90°, it increases from to x ; from 90° to 180°, it increases from — x to 0; from 180° to 270°, it increases from to oo ; and from 270° to 360°, it in- creases from — X to 0. Since the cotangent commences at x, and assumes in suc- cession the values CFy, CR^, 0, -OT;, —CF^, etc., from 0° to 90°, it decreases from x to ; from 90° to 180°, it de- creases from to — x; from 180° to 270°, it decreases from X to ; and from 270° to 360°, it decreases from to — x. Since the secant commences at OA, and assumes in suc- cession the values OE^^ OE.,, x, —OE., —OE^, etc., from 0° to 90°, it increases from 1 to x ; from 90° to 180°, it increases from —X to — 1 ; from 180° to 270°, it decreases from —1 to — X ; and from 270° to 360°, it decreases from x to 1. Since the cosecant commences at x, and assumes in suc- cession the values OF,, OF., OC, OF^, OF^, etc., from 0° to 90°, it decreases from x to 1 ; from 90° to 180°, it increases from 1 to X ; from 180° to 270°, it increases from — x to — 1 ; and from 270° to 360°, it decreases from —1 to — x. Note. Wherever the symbol oo occurs in the foregoing discussion, it must be interpreted in accordance with the Note to Art. 41. 42 PLANE TRIGONOMETKY. V. GENERAL FORMULAE. 65. To fiyid the values of sin {x + y) and cos {x + y) in terms of the sines and cosines of x and y. C/ A k y / / B M^ -^^ /<» A . D Let AOB and BOC denote the angles x and ?/, respec- tively ; then, AOC=x-{-y. From any point C in OC draw QA and CB perpendicular to OA and OB ; and draw BD and BE perpendicular to OA and AC. Since EC and BC are perpendicular to 0^ and OB, the angles jBOjEJ and AOB are equal ; that is, BCE = x. We then have . , , . AC BD-^ CE BD , CE But and BD BD ^ OB = X = sm X cos y. OC OB OC ^ CE CE BC — - = — - X — - = COS a? sinw. OC BC OC Whence, sin (x -\-y) = sin x cos y + cos xsiny. . . , , . OA OD-BE OD BE Again, cos(. + 2/)=^=^^- = ^-^- (11) OD^OD OB OC OB OC cosx COS 2/, But and Whence, cos (ic -|- ?/) = cos x cos y — sin x sin 2/. 5^ BE BC . - — = X = sni X sin y. OC BC OC ^ (12) GENERAL FORMULA. 43 66. To find the values of sin {x — y) and cos {x — y) in terms of the sines and cosines of x ojid y. Let AOB and BOC denote the angles x and ^, respec- tively ; then, AOC = x — y. From any point C in OC di-aw CA and CB perpendicular to OA and OB ; also, draw BD perpendicular to 0^1, and BE perpendicular to AC produced. Since EC and BC arc perpendicular to OA and OB, the angles BCE and AOB are equal ; that is, BCE = x. We then have sin(x'-y) = ^==^^^:^ = ^-^. ^ OC OC OC 00 But and BD BD ^ OB CE CE ^ BC = ^-T-. X — r = coso; sin/y. OC BC OC Whence, sin(x — y)= sin a; cosy — cos a; sin?/. Again, ,o.{x-y) = 9A=.OD^BE^gD^BE^ ^ ^ ^^ OC OC OC OC (13) But and OD OD ^ OB oc^m'' oc= '''''' '""'y^ BE BE BC . OC BC OC Whence, cos (a; — y)= coso; cos 2/ + sin ic sin?/. (14) 44 PLANE TRIGONOMETRY. 67. The fundamental formulae of Arts. 65 and 66 are of great importance, and it is necessary to prove that they hold for all values of x and y. It is obvious that the proof of Art. 65 is not general, for we have assumed in the construction of the figure that x and y are acute angles, and that x-\-y is < 90°. Also, in Art. 66, we have taken x and y as acute angles, and x':>y. In order to prove the formulae universally, we will first show that (ll) and (12) hold for all values of x and y, and we can then give a general proof of (13) and (l4). 68. We will first prove (11) and (12) when x and y are acute, and x-\-y> 90°. D Let DOB and BOC denote the angles x and y^ respec- tively ; then, DOC = x -\-y. From any point O in 0(7 draw CB perpendicular to OB, and CA perpendicular to OD produced ; and draw BD and BE perpendicular to OD and AC. Since EC and BC are perpendicular to OD and OB, the angles BCE and DOB are equal ; that is, BCE = x. We then have, by Art. 32, But and sin DOC = BD^ OC CE OC AC^ BD-j-CE ^BD CE OC OC OC OC BD OB — — X —— = sm X cos y, OB OC ^ CE BC X — - = COS X sm y. BC OC ^ Whence, sin(a^ -f- ?/) = sinx cosy + cosa? sin 2/. Again, cos DOC = GENERAL FORMULAE. 45 OA OD-BE OD BE oc oc oc oc OD OD OB But — = X = cos X cos v, OC OB OC ^ , BE BE BC . . and = X — = sin x sin y. OC BC OC ^ Whence, cos(:(; + y) = cos x cos y — sin x sin y, 69. We have thus proved (ii) and (12) when x and y are any two acute angles; or, what is the same thing, when they are any two angles in the first quadrant. Now let a and b be any assigned vaUies of x and y for which (11) and (12) are true ; then by Art. 44, sin [90°+ (a + 6)] = cos (a -h 6) = cos a cos 6 — sin a sin 6, by (12); (a) and, cos [90°-f (a -}-/>)] = -sin («-h^) = — sin a cos 6 — cos a sin Z>, by (11). (b) But by Art. 44, cos a = sin (90°+ a) , cos 6 = sin (90°+ b) , — sin a = cos (90°+ a), — sin & = cos (90°+ 6). Hence (A) may be written in the forms, sin [(90°+ a) +6] = sin (90°+ a) cos 6 + cos (90°+ a) sin 6, sin [a + (90°+ b) ] = sin a cos (90°+ b) + cos a sin (90°+ b) , both of which are in accordance with (11) . And (b) may be written in the forms, cos [(90°+ a) + 6] = cos (90°+ a) cos b-sin (90°+ a) sin b, cos [a+ (90°+ 6)] = cos a cos (90°+ 6) -sin a sin (90°+ 6), both of which are in accordance with (12). It follows from the above that if (11) and (12) hold for any assigned values of x and y., such as a and 6, they also hold when either a or b is increased by 90°. 46 PLANE TRIGONOMETRY. But they have been proved to hold when both x and y are in the first quadrant ; hence they also hold when x is in the second quadrant and y in the first. And since they hold when X is in the second quadrant and y in the first, they also hold when x is in the thii-d quadi'ant and y in the first ; and so on. Thus (ii) and (12) are proved to hold for any values of X and y whatever, positive or negative. 70. We may now give a general proof of (13) and (14). B3' (11) and (12), we have sin {x -y) = sin [a^' + (- 2/)] = sin x cos ( — 2/) + cos x sin ( — y) — sin X cos y — cos x sin y (Art. 42) . And, cos (x — y)== cos [a? + ( — 2/) ] = cos X cos( — y) — sin x sin( — y) = cos X cos ?/ + sin a; sin y (Art. 42) . Hence (13) and (14) hold for all values of x and y, for the above proof depends on formulae which have been shown to hold universally. 71. By Art. 20, we have sin {x + y) tan (x -f- 2/) = 7 — ; — X ^ ^^ cos{x + y) sin X cos y + cos a? sin w , ^ ^ = ^-^ ;— ^, by (11) and (12) cos X COS y ~ sm x sin y Dividing each term of the fraction by cos x cos y, sin X cos y cos x sin y . cos, X cosy cos 07 cos 2/ tan (x -\-y) = : -. ^ ^ cos X cos 2/ sm X sm y cos X cos y cos x cos y tan X + tan y _ / \ "" 1 — tan X tan y GENERAL FORMULAE. 47 In like manner, we derive tan a;- tan y , . tan (X -y) = i^^ana^tan^' ^^^^ Again, by Art. 20, we have w . N ^Qs (x + y) cot (a; -\-y) = —. — 7 — rz\ cos .T cos 2/ — sin a; sin 2/ , , . j /,^v = ^^-^ by (11) and (12 ) • sin X cos y + cos x sin ?/ " ^ ' Dividing each term of the fraction by sin x sin 1/, cos X cos 2/ sin a; sin y sin X sin ?/ sin x sin ?/ cot(a;H-?/) =- sin .T cos v cos ar sin y sm X sm ?/ sin x sm ?/ cot X cot ?/ — 1 (17) cot y + cot a; In like manner, we derive cot a; cot y -\- 1 , . ^' •-'/"" cot ,y — cot a; 72. From (11), (12), (13), and (14), we obtain sin (rt -(- />) = sin a cos h + cos a sin 6, sin (a — h) — sin a cos 6 — cos a sin Z>, cos (a H- 6) = cos a cos h — sin f/ sin 6, cos (a — 6) = cos ft cos 6 + sin a sin h. Whence, by addition and snbtraction, sin (a + 6) + sin (ct — 6) = 2 sin a cos &, sin (ft + ?>) — sin (fi — 6) = 2 cos a sin 6, cos (a + &) + cos (a — 5) = 2 cos a cos 5, cos (ft + 6) — cos (« — 6) = — 2 sin a sin 6. 48 PLANE TRIGONOMETRY. Let a + h=x, and a — h = y. Then, a = |-(a; + 2/) , and 5 = ^(a; — 2/) . Substituting these values, we have sin X + sin y = 2 sin ^{x + y) cos ^{x — y)^ (l9) sin ft.' — sin 2/ = 2 cos ^{x + 2/) sin ^ (ft? — y) , (20) cosft;+cos?/= 2 cos^(ft; + ?/) cos^(ft; — ?/), (2l) cos X — cos 2/ = —2 sin |-(ft; + y) sin ^(ft; — 2/) • (22) 73. Dividing (19) b}^ (20), we obtain sin X + sin y _ 2 sin|-(a? + y) cos |- (ft.' — 2/) sin ft; — sin 2/ 2 cos |- (ft; + 2/) sin ^ (ft; — 2/) = tan|-(ft; + ?/) cot|(ft;-2/) ^ tan|-(ft; + 2/) (^^^t. 18). (23) tan ^ (ft; — 2/) FUNCTIONS OF 2 a:. 74. Putting 2/ = ftJ in (11) , we have sin 2 ft; = sin x cos x + cos x sin a; = 2 sin ft; cos x. (24) Putting 2/ = a? in (12) , cos 2 ft; = cos^ X — sin^ ft;. (25) Since cos^ a? = 1 — sin^ ft;, and sin- a; = 1 — cos^ ft? (Art. 19) , we also have cos 2 ft; = 1 — sin^ x — sin^ x =1 — 2 sin^ x, (26) and cos 2ft; = cos- ft; — (1 — cos^ft;)= 2cos^ft;— 1.(27) In like manner, from (15) and (17) , we have tan2ft;^ ^^^"^ , (28) 1 — tan- X cot 2 ft; = ^^^-^^^. (29) 2 cot X GENERAL FORMULA. 49 FUNCTIONS OF ^x. 75. From (26) and (27), we have 2 sin^ x= 1 — cos 2 .i*, and 2 cos- .i' = 1 + cos 2 x. Writing x in place of 2 x, and therefore ^ a; in place of x, 2 sin^ -^ a; = 1 — cos x, and 2 cos- ^ .i- = 1 + cos x. (A ) Dividing by 2, and extracting the square root, 8mix= JLzl£21^ (30) C09 j-a;= p +cosa; . ^. 2 Dividing (30) b}^ (31), we ol)tain tania;= jLz^2i^. (32) \l + cosa; ^ ^ Multiplying the terms of the fraction under the radical sign first by 1 + cos x, and then by 1 — cos x, we have ^x= J 1-eos-a.- tan^a; = (1 + cos a;)' -J sm^ X sm X (1 + cos a;)^ 1 + cos x (33) and tan J a; = (l-cosaQ^ \ 1 — cos- a; _ 1(1 — cos a;)^ _ 1 — cos x (34) sin- X sm X And since the cotangent is the reciprocal of the tangent, ^^+ 1 ^ 1 + cos a; sin a; , ^ cotia; = — ; :^- (35) sm a; 1 — cos x Note. The radical in each of the formulse (30), (31), and (32) is to be taken as positive or negative according to the quadrant in which the angle ^a: is situated (Art. 37). 50 PLANE TRIGONOMETRY. INVERSE TRIGONOMETRIC FUNCTIONS. 76. The expression sin"^^/? called the inverse sine of 2/, or the anti-sine of y, is used to denote the angle whose sine is equal to y. Thus the fact that the sine of the angle x is equal to y may be expressed in either of the waj^s sin X = ?/, or x = sin~-^ y. In like manner, the expression cos'^y signifies the angle whose cosine is equal to y ; tan~^ y, the angle whose tangent is equal to ?/, etc. Note. The student must be careful not to confuse this notation with the exponent —1; the —1 power of sin a: is expressed (sinx)-^, and not sin-^a:. 77. By aid of the principles of Art. 76, any relation in- volving du'ect functions may be transformed into one involv- ing inverse functions. Take for example the formula sin (^x-\-y) = sin x cos y -\- cos x sin y. (a) Let sin x = a, and sin y = b. Then by Art. 76, x= sin~^ a, and y = sin~^ b. Also by Art. 19, cos x= ^1 — siia^x, and cos y = \1 — sin^y. Putting sin x = a, and sin y = b, we have cos x= ^1 — a^, and cos y=^l — b'. Substituting these values in (a) , sin (sin~^ a + sin"'^ b) = a^l — b^ -\-b^l — a^. Whence by Art. 76, sin^^ a + sin"^ b = sin~^ (o \/l — 6^ + 6 V^l — a^) . GENERAL FORMULA. 53 EXERCISES. 1 tjiri Tj 78. 1. Prove the relation cos'2x= -^- 1 + tannic By (25) , cos '2x= cos- it* — sin^x* cos-o; — sin-a; ,. , .f.. = :, — — — r- (Art. 19). cos- a; + siQ-.^ Dividiug each term of the fraction by cos-.i*, we have , sin- X cos- a; 1 — tan- a; cos '2x = 7-:r- = :; :; — 1 -I- ^^"" ^ 1 H~ tan-a; cos- a; 2-0 4.U ^ 4-- sin 5 a.' + sin a; , „ . Prove the relation — — — ' = tan 3 a;. cos ox -j- cos X By (19) and (21), sin 5 a; + sin x _ 2 sin l(iJx-\-x) cos ^(5 a; — x) cos 5 a; + cos x 2 cos ^ (5 a; + a*) cos ^ (5 a; — x) sin 3 a' cos 3 a; Prove the following relations : « sin (a; -j-y) _ tan a- -f tan y sin (x — y) tan x — tan y = tan 3 X. 4. cos (a- -f- ?/) _ cot X cot // — 1 cos (a; — y) cot x cot y -\-l c sin (x + y) _ 1 -f cot X tan y cos [x — y) cot X -\- tan y c ' / 4-0 , \ sin y -f-cos y 6. sm(4o +?/) = — ^— ^ ^. V^2 7. tan (60° -3/)= V^-tany . 1 + V^ tan 2/ Q sin aj 4- sin ?/ , , . , . 8. — — ^ = tani(a;H-?/). cos a; -|- cos y 52 PLANE TRIGONOMETRY. g_ sinx + sinj/=_coti(iC-^). COS X — COS y 10. sm2i» = 1 + tan^aj 11 . CSC 2 a; = ^ sec x esc a;. 2 12. tan a; + cot a; = sin 2 a? 13. cot a? — tan a; = 2 cot 2 a;. 14 (l + tana;)^-(l-tana;)^^^.^^^^ (l+tana?)2+ ^i _tana;)2 15. sin {x + y) sin {x — y) = siu^x — sin^i^. 16. cos {x + y) cos (x ~y) = cos- a; — sin^?/. 17. sec^a? csc^a; = sec^a; + csc^a;. 18. cos y + cos (120° -hy) + cos (120° -y)=0. 19. sin^sin(^~(7) + sin5sin((7- J.) + sinOsin(^-5) = 0. 20. cos (A + B) cos {A-B) + cos (5 + O) cos (5 - O) + cos(C+^)cos(a-^) = cos2J[+cos25+cos2(7. rti cos a? — cos ox . ^ 21 . = tan 2 a;. sin 3 a? — sin x 22 cos 80° + cos 20° _ ^/^ sin 80° - sin 20° By putting x = x-]-y and y = z in (ii) and (12) , Art. 65, prove : 23. sin (x -\-y -\-z) = sin a? cosy cosz + cosa? sin 2/ cos;^; + cos a? cosy sinz — sina? siny sinz. 24. cos {x-\-y -{-z) = cos x cos ?/ cos z — sin a? sin y cos 2; — sin a? cosy sinz — cos a; sin 2/ sins;. GENERAL FORMULAE. 53 By putting aj=2a; and y = x in (ll), (12), and (15), prove : 25. sin 3 a; = 3 sin a^ — 4 sin^a;. 26. cos3i« = 4cos^a; — 3 cosic. oty 4. o 3 tana; — tan^a; 27. tan 3 x = 1 -3tan2a; Prove the relations : 28. sin {2x-\-y) — 2 sin x cos {x -\-y) = siny. QQ sin 3 a; _ cos 3 a; _ ,-^ sin a; cos a; 30. 1 4- cos 2 a; cos 2y=2 (sin- .r sin-?/ + cos^ajcos-?/). 31 . 1 -h tan X tan 2 x = sec 2 a;. 32. sin 4 .t = 4 sin a; cos x — S s'rn^x cos x. 33 . cos 4 a; = 8 cos^ x — 8 cos- x + 1 . 34. sin 5 a; = 5 sin a; — 20 siu'^v +16 sin' a;. 35. By putting a;= 45° and y = SO° in (13) and (14), Art. 66, prove sin 15° = ^ (v/6 - v/2) = cos 75°, cos 15° = ^ (\/6 4- V^) = sin 75°. ' 36. By putting .^' = 30° in (34) and (35), Art. 75, prove tanl5° = 2-V^ = cot75°, cot 15° = 2 + V^ = tan75°. 37. By putting a; = 45° in (30) and (31), Art. 75, prove sin22°30' = iV/2-V^, cos 22°30' = i\/2 -j- v/2. 38. By putting a; = 45° in (34) and (35), Art. 75, prove tan22°30' = V^-l, cot 22°30' = V^ + 1. 54 PLANE TRIGONOMETRY. VI. LOGARITHMS. 79. Every positive number may be expressed, exactly or approximately, as a power of 10 ; thus, 100=102; 13 = 10i"39-- ; etc. When thus expressed, the corresponding exponent is called its Logarithm to the base 10 ; thus, 2 is the logarithm of 100 to the base 10, a relation which is written logio 100 = 2, or simply log 100 = 2. And in general, if 10"= = m, then x = log m. 80. Any positive number except unity may be taken as the base of a system of logarithms; thus, if a^'^m, then a; = log^m. Logarithms to the base 10 are called Common Logarithms, and are the only ones used in numerical computation. If no base is expressed, the base 10 is understood. 81. We have by Algebra, 10°= 1, 10-' = iiff =.1, 101 = 10, 10-' = Ti7 =.01. 102=100, 10-3 = y^=. 001, etc. Whence, by the definition of Art. 79, log 1 = 0, log.l = -1 = 9-10, log 10=1, log .01= -2 = 8-10, log 100 = 2, log .001 = -3 = 7 - 10, etc. Note. The second form of the results for log.l, log .01, etc., is preferable in practice. 82. It is evident from Art. 81 that the logarithm of a number greater than 1 is positive, and that the logarithm of a number between and 1 is negative. LOGARITHMS. 55 83. If a number is not an exact power of 10, its common logarithm can only be expressed approximately ; tiie integral part of the logarithm is called the characteristic^ and the decimal part the mantissa. For example, log 13 = 1.1139. In this case the characteristic is 1, and the mantissa .1139. 84. It is evident from the first column of Art. 81 that the logarithm of any number between 1 and 10 is equal to plus a decimal ; 10 and 100 is equal to 1 plus a decimal ; 100 and 1000 is equal to 2 plus a decimal ; etc. Hence, the characteristic of the logarithm of a number with one figure to the left of its decimal point, is ; with two figures to the left of the decimal point, is 1 ; with tliree figures to the left of the decimal point, is 2 ; etc. 85. In like manner, from the second column of Art. 81, the logarithm of a decimal between 1 and .1 is equal to 9 plus a decimal — 10 ; .1 and .01 is equal to 8 plus a decimal — 10 ; .01 and .001 is equal to 7 plus a decimal — 10 ; etc. Hence, the characteristic of the logarithm of a decimal with no ciphers between its decimal point and first significant figure, is 9, with —10 after the mantissa; of a decimal with one cipher between its point and first figure, is 8, with —10 after the mantissa ; of a decimal with two ciphers between its point and first figure, is 7, with —10 after the mantissa ; etc. 86. For reasons which will be given hereafter, only the mantissa of the logarithm is given in tables of logarithms of numbers ; the characteristic must be supplied by the reader. The rules for characteristic are based on Arts. 84 and 85 : 56 PLANE TRIGONOMETRY. I. If the number is greater than 1, the characteristic is 1 less than the number of places to the left of the decimal point. II. If the number is between and 1, subtract the number of cip>hers betiveen the decimal point and first significant figure from 9, ivriting —10 after the mantissa. Thus, characteristic of log 906328.5 = 5 ; characteristic of log .007023 =7, with —10 after the mantissa. Note. Some writers, in dealing with the characteristics of negative logarithms, combine the two portions of the characteristic, writing the result as a negative characteristic before the mantissa. Thus, instead of 7.6036 — 10, the student will frequently find 3^.6036, a minus sign being written over the characteristic to denote that it alone is negative, the mantissa being always positive. PROPERTIES OF LOGARITHMS. 87. In any system., the logarithm of unity is zero. For since a° = 1, we have log^ 1 = (Art. 79). 88. In any system, the logarithm of the base itself is unity. For since a^ = a, we have log„ a = 1 . 89. In cmy system ivhose base is greater than unity, the logarithm of zero is minus infinity. For if a is > 1 , we have (x~°° = — = — = 0. Whence by Art. 79, log^ = — oo . 90. In any system, the logarithm of a product is equal to the sum of the logarithms of its factors. Assume the equations a"" = m^ , , c x = loff„ m, [ ; whence by Art. 79, ] ^" ' a^ = n } " (y =ilogan. LOGARITHMS. 57 Multiplying, we have a"" X a^ = mn^ or a"""^^ = mn. Whence, log^ 7nn = x -\-y. Substituting the values of x and y, we have log,, mn = log,, m ~\- log„ n . In like manner, the theorem may be proved for the product of three or more factors. 91. By aid of the theorem of Art. 90, the logarithm of an}' composite number may be found when the logarithms of its factors are known. 1. Given log 2 = .3010, log 3= .4771 ; find log 72. log 72 = log (2x2x2x3x3) = log 2 -f log 2 -h log 2 + log 3 + log 3 = 3 X log 2 -h 2 X log 3 = . 9030 + .!)5-i2 = 1.8572. EXAMPLES. Given log2 = .3010, log 3 = .4771, log 5 = .0990, logT = .8451 ; find the values of the following : 2. log 6. 7. log 21. 12. log 98. 17. log 135. 3. log 14. 8. log 63. 13. log 105. 18. log 168. 4. log 8. 9. log 56. 14. log 112. 19. log 147. 5. log 12. 10. log 84. 15. log 144. 20. log 375. 6. log 15. 11. log 45. 16. log 216. 21. log 343. 92. In any system., the logaritlim of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator. Assume the equations a"" = m] ( x = \oga m, >■ ; whence. a'-" = n ) (y = logrt'^i- 58 PLANE TRIGONOMETRY. Dividing, we nave _ = — , or cC^ = — cO' n n Whence, log„ — = x — y = log,, m — log„ n, 93. 1. Griven log 2 = .3010 ; find log 5. log 5 = log i5 = log 10 - log 2 = 1 - .3010 = .6990. EXAMPLES. Given log 2 = .3010, log 3 = .4771, log 7 = .8451 ; find the values of the foUowins: : 2. lopj — ^3 5. log 35. 8. 1 42 loo- "25 11. log 71. 3. 1 10 logy. 6. 1 21 log — =^16 9. log 175. 12. 1 35 loar — ° 6 4. log 31 7. log 125. 10. log 111. 13. log5f 94. In any system, the logarithm of any poiver of a quan- tity is equal to the logarithm of the quantity multiplied by the exponent of the poioer. Assume the equation cf = m ; whence, x = log^ m. Raising both members to the j?th power, we have a^"" = m^ ; whence, log„ m^ =px =p log„ m. 95. In any system, the logarithm of any root of a qnantity is equal to the logarithm of the quantity divided by the index of the root. r,— I 1 For, log„ V m = log, (m') = - log„ m (Art. 94) . 96. 1. Given log 2 = .3010; find the logarithm of 2i log 23 = 5 X log 2 = - X .3010 = .5017. Note. To multiply a logarithm by a fraction, multiply first by the numerator, and divide the result by the denominator. LOGARITHMS. 59 2. Given log 3 = .4771 ; find the logarithm of V^S. , Ao loo" 3 .4771 ,^xap log y o = " = — — = .Uoyb. EXAMPLES. Given log 2 = .3010, log 3 = .4771 , log 7 = .8451 ; find the values of the follow in oj : 3. logSi 7. log 12'. 11. log IS*^. 15. 6/- logyo. 4. log 2^ 8. log 21^. 12. logN/7. 16. log y35. 5. log7^ 9. logl4^ 13. log fd. 17. log y'98. 6. log 5^. 10. log25i 14. logV'2. 18. logV^126. 19. Find tlio logarithra of (2^ x 3'') . By Art. 90, log (2' x 3 ')= log 2' + log s' = Jl()g2+ Jlog3 = .1003 + .5964 =.6967. Find the values of the following : 20. logl'yj. 22. log(3' X2-). 24. \ogyjl 26. logy/y- 21. log-. 23. log3y^7. 25. logfl. 27. log*^. 97. In the common system, the mantissce of the logarithms of numhers having the same sequence of figures are equal. To illustrate, suppose that log 3.053 = .4847. Then, log 30.53 = log (10 X 3.053) = log 10 + log 3.053 ^l + .4847 =1.4847; log 305.3 = log (100 X 3.053) = log 100 -h log 3.053 = 2 + .4847 =2.4847; log .03053 = log (.01 X 3. 053) = log. 01 + log 3. 053 = 8 - 10 + .4847 = 8.4847 - 10 ; etc. 60 PLANE TRIGONOMETRY. It is evident from the foregoing that if a number is multi- plied or divided by any integral power of 10, thus produc- ing another number with the same sequence of figures, the mantissas of theii' logarithms will be equal. Thus, if log 3.053 = .4847, then log 30.53 = 1.4847, log .3053 = 9.4847 - 10, log 305.3 = 2.4847, log .03053 =8.4847-10, log 3053. = 3.4847, log .003053 = 7.4847 - 10, etc. Note. The reason will now be seen for the statement made in Art. 86, that only the mantissas are given in a table of logarithms of numbers. For, to find the logarithm of any number, we have only to take from the table the mantissa corresponding to its sequence of figures, and the characteristic may then be prefixed in accordance with the rules of Art. 86. This property of logarithms is only enjoyed by the common system, and constitutes its superiority over others for the purposes of numeri- cal computation, 98. 1. Given log 2=. 3010, log 3=. 4771; find log .00432. log 432 = log (2* X 3^) = 4 log 2 + 3 log 3 = 1.2040 4- 1.4313 = 2.6353. Then by Art. 97, the mantissa of the result is .6353. Whence by Art. 86, log .00432 = 7.6353 - 10. EXAMPLES. Given log 2 = .3010, log 3 = .4771, log 7 = .8451 ; find the values of the following : 2. log 1.8. 7. log .0054. 12. log 302.4. 3. log 2.25. 8. log .000315. 13. log .06174. 4. log .196. 9. log 7350. 14. log (8.1)^ 5. log .048. 10. log 4.05. 15. log \/ 9^. 6. log 38.4. 11. log .448. 16. log (22.4)i LOGARITHMS. 61 99. To prove the relation Assume the equations ce=m) f a.' = log,, 771, \ ; whence, < h" = m ) (. V = log,, 771. From the assumed equations, we have X cv'=b^, or aJ' = b. Whence, log„6 = '^ or y y loga& Substituting the values of x and y, loo" 777, log. 7?l = , , • log„6 By aid of this relation, if the logarithm of a quantity m to a certain base a is known, its logaritlnu to any otlier base b may be found by dividing by the logarithm of b to the base a. 100. To jyrove the relation logj a X log,, b= I. Putting 7?7 = a in the result of Art. 1)1), we have log,, a 1 log,a = j^ = . (Art. 88) . log, 6 \ogJj ^ ^ Whence, log,, a x log,, b= \. APPLICATIONS. 101. The value of an arithmetical quantity, in which the operations indicated involve only nuiltiplication, division, involution, or evolution, may be most conveniently found by logarithms. The utility of the process consists in the fact that addition takes the place of multiplication, subtraction of division, multiplication of involution, and division of evolution. In operations with negative characteristics the rules of Algebra must be followed. 62 PLANE TRIGONOMETRY. 102. 1. Find the value of .0631 x 7.208 x .51272. By Art. 90, log (.0631 x 7.208 x .51272) = log .0631 + log 7.208 + log .51272. log .0631 = 8.8000-10 log 7.208 = 0.8578 log .51272= 9.7099-10 Adding, .-. log of result = 19.3677 - 20 = 9.3677-10 (see Note 1) Number corresponding to 9.3677—10 = .23317. Note 1. If the sum is a negative logarithm, it should be reduced so that the negative portion of the characteristic may be — 10. Thus, 19.3677-20 is reduced to 9.3677-10. 2. Find the value of ' — 7984 By Art. 92, log ^?^ = log 336.8 - log 7984. log 336.8 = 12.5273 - 10 (see Note 2) los7984 = 3.9022 Subtracting, .-. log of result ='8.6251-10 Number corresponding = .04218. Note 2. To subtract a greater logarithm from a less, or to subtract a negative logarithm from a positive, increase the characteristic of the minuend by 10, writing —10 after the mantissa to compensate. Thus, to subtract 3.9022 from 2.5273, write the minuend in the form 12.5273 - 10 ; subtracting 3.9022 from this, the result is 8.6251 - 10. 3. Find the value of (.07396)^ By Art. 94, log (.07396)^ = 5 x log .07396. log .07396 = 8.8690 -10 5 44.3450 - 50 = 4.3450 - 10 (see Note 1) = log .000002213. LOGARITHMS. 63 4. Find the value of y. 035063. By Art. 95, log V^.035063 = ^ log. 035063. log.035063 = 8.5440 -10 20. - 20 (see Note 3) 3 )28.5449 -~30 9.5150- 10 = log .3274. Note 3. To divide a negative logarithm, add to botli parts such a multiple of 10 as will make the negative portion of the characteristic exactly divisible by the divisor, with —10 as the quotient. Thus, to divide 8.5440 — 10 by 3, add 20 to both parts of the loga- rithm, giving the result 28.5449 — 30. Dividing this by 3, the quotient is 9.5150-10. ARITHMETICAL COMPLEMENT. 103. The Arithmetical Complement of the logarithm of a number, or briefly the Colorjarithm of the number, is the loga- rithm of the reciprocal of that number. Thus, color? 409 = lojr -- = locr 1 _ W 409. "^ ^400 ^ =' log 1 = 10. - 10 (Note 2, Art. 102) log 409 =. 2.G117 .-. colog409= 7.3883-10. Again, colog .067 = log = \o^ 1 — log .067. .067 log 1 = 10. -10 log .067= 8.8261 -10 .-.colog .067= 1.1739. The following rule is evident from the above : To find the cologarithm of a number^ subtract its logarithm from 10 - 10. Note. The cologarithm may be obtained from the logarithm by subtracting the last significant figure from 10 and each of the others from 9, — 10 being written after the result in the case of a positive logarithm. 64 PLANE TRIGONOMETRY. .51384 104. Example. Find the valiie of — :: 8.709 X .0946 , .51384 . f K^oo^ sy 1 ^ _J_^ I02: = log .51o84 X X ^ 8.709 X .0946 ^ ^ ^-709 .0946; = log .51384 + log -^ h log 8.709 ^ .094(; = log .51384+colog 8.709 + colog. 0946. log .51384 = 9.7109 -10 colog 8.709 = 9.0601 - 10 colog .0946 = 1.0241 9.7951 -10 = log .6239. It is evident from the above that the logarithm of a fraction is equal to the logarithm of the numerator plus the cologa- rithm of the denominator. Or in general, to find the logarithm of a fraction whose terms are composed of factors, Add together the logarithms of the factors of the numerator, and the cologarithms of the factors of the denominator. Note. The value of the above fraction may be found without using cologarithms, by the following formula : log -^^^ = log .51384 - log (8.709 X .0946) "= 8.709 X .0946 = log .51384 - (log 8.709 + log .0946). The advantage in the use of cologarithms is that the written work of computation is exhibited in a more compact form. EXAMPLES. 105. Note. A negative quantity can have no common logarithm, as is evident from the definition of Art. 79. If negative quantities occur in computation, they may be treated as if they were positive, and the sign of the result determined irrespective of the logarithmic work. Thus, in Ex. 3, p. 65, the value of 721.3 X (-3.0528) may be ob- tained by finding the value of 721.3 X 3.0528, and putting a negative sign before the result. See also Ex. 34, p. QQ. LOGARITHMS. 65 Find by logarithms the vaUies of the following : 1. 9. 238 X. 9152. 4. (-4.3264) x(-. 050377). 2. 130.36 X .08237. 5. .27031 X .0-42809. 3. 721.3 X (-3.0528). 6. ^(-.063165) x 11.134. 52.37* ' .0.SG59 ' ' 64327* 8 LM?!. 10 9-163 ^2 -007514 10.813 .0051422 -.015822 13 3.3681 ^^ 15.008 x(-.0843) 12.853 X .6349* ' .06376 X 4.248 15. 16. (-2563) X. 03442 714.8x(-.511) * 121.6 X (-9.025) (_48.3) x3662x (-.0856) 17. (23.86)\ 22. (.8)'. 28. v/.4294. 18. (.532)«. 23. (-3.16)3. 29. v'.^2305. 19. (-1.0246)'. 24. (.021)'. 30. V^IOOO. 20. (.09323)^ 25. v/2. 31. V^-. 00951. 21. 5i 26. v^5. 32. V^.OOOlOll. 33. Find the value of 27. V-3. 2V^ 3t log ?V^ = log 2 + log ^5 + colog 3^ = log2 + |log 5 + f colog 3. log 2= .3010 log 5= .6990; divide by 3= .2330 colog 3 = 9.5229-10 ; multiply by |= 9. 6024 - 10 .1364 = logl.3691. 56 PLANE TRIGONOMETKY. 34. Find the value of Ij — ' f~ ■ 7.962 logt^55296^ai :03296_^j Q329^ ^ V 7.962 ^ ^ 7.962 ^^ ^ & ^ log .03296= 8.5180-10 log 7.962 = 0.9010 3 )27.6170-30 9.2057- 10 = log. 16059. - Kesult, -.16059. Find the values of the following 35. 2i-x3i 40. r^^^^y. 45. (^1^?^. \. 1321 J \49309 qfi 3^ ,- 46 ^-31-63Vt ^^' 3* 41. d- ^"^ 13 3 37. 5^ , 47 1003 (_10)t* 42. f^. (.7325)^ f^ „ „ 48. ^ ^^Qg^ . [j)' 43. ^^^^1 S/.0008276 39. r^¥. 44. ^x^3x^:0T. 49. (-7469)^. 38. 113y -(.2345)^ Krt V''."0073 52. (538.2x.0005969)i OU. J' (•68291)^ 53. (18.9503)iix(-.l)'^ 51. V/5.955XV61.2 54. V3734.9 x .00001108. V^298.54 55. (2.6317)* X (.71272)^ gg V -. 008193 x(.06285)t - .98342 57. V^035x y. 62667 X y. 0072103. SOLUTION OF RIGHT TRIANGLES. 67 VII. SOLUTION OF RIGHT TRIANGLES. 106. The six elements of a triangle are its three sides and its three angles. We know by Geometry that, in general, a triangle is com- pletely determined when three of its elements are known, provided one of them is a side. The solution of a triangle is the process of computing the unknown from the given elements. 107. To solve a rigJit triangle, tico elements must be given in addition to the right angle, one of which must be a side. The various cases which can occur may all be solved by aid of the following formula? : sin A = sin B = cos A= f c cos B = A ^ tan^ = Ti tan B = a Case I. When the given elements are a side and an angle. The proper formula for computing either of the remaining sides may be found by the following rule : Take that function of the angle which involves the given side and the required side. Case II, When both the given elements are sides. First calculate one of the an2;les bv aid of either of the formulae involving the given elements, and then compute the remaining side as in Case I. 68 PLANE TRIGONOMETRY. EXAMPLES. 108. 1. Given c= 203.76, 5=21° 43'. Find a and 6. In this case the formulae to be used are cos jB = -9 and sin 5 = - • c c AVhence, a-= c cos jB, and h = c sin B. By logarithms, log a = log c + log cos 5, and log h = log c + log sin B. log c = 2.3091 log c = 2.3091 log cos B = 9.9681 - 10 log sin B = 9.5682-10 log a = 2.2772 log b= 1.8773 .-. a= 189.3. .'. 6=75.38. 2. Given a = 13, A = 67° 7'. Find 6 and c. .a -, • ^ ^ In this case, tan A = ti and sm A = -' ' 6 c Whence, 6 = ^, andc=^ — -. tan A sm A By logarithms, log 6 = log a — log tan A, and log c = log a — log sin A. log a =1.1139 log a =1.1139 log tan A = 0.3746 log sin A = 9.9644-10 log b = 0.7393 log c = 1.1495 .-. & = 5.486. .-. c= 14.11. 3. Given 6= .1512, c=. 3081. Find ^ and a. b a In this case, cos A = -■> and tan A = t^ or a=b tan A. c b Whence, log cos A = log b — log c, and log a = log b -\- log tan ^. SOLUTION OF RIGHT TRIANGLES. 69 loo- h = 9.1796 log h = 9.1796 log c = 9 . 488 7 log tau A = 0.2493 log cos A = 9.6909 log a = 9.4289 .-. J. = 60° 36.4'. .-. a= .2685. Note. It is customary in practice to omit writing the — 10 after the mantissa of a negative logarithm, as illustrated in Ex. 3. 109. In the Trigonometrical solution of a triangle bj' the method of Case II., it is necessary to first find one of the angles, and the remaining side may then be calculated. It is possible however to obtain the third side directly, without first finding the angle, by Geometrical methods. Thus in Ex. 3, Art. 108, w^e have by Geometry, a- -\-b~ = c". Whence, a = ^r -b- = \/{c -{-b) (c-b). By logarithms, log a == \ [log (c -|- 6) -+- log (c — 6)] . c-}-b= .4593 ; log =9.6621-10 c^b= .1569; log =9.1956-10 2 )18.8577-20 log a = 9.4289 -10 .*. a= .2685, as before. If the given sides are a and b, the formula for c is y^a^ -f- b^t which is not adapted to logarithmic computation. In such a case it is usually shorter to proceed according to the rule of Art. 107. EXAMPLES. 110. Solve the following right triangles : 1. Given ^ = 43° 30', c=11.2. 2. Given B = 68° 50', a = 729.3. 3. Given i5= 62° 56', 6 = 47.7. 4. Given a =.624, c=.91. 70 PLANE TRIGONOMETRY. 5. Given a =5, 6=2. 6. Given ^ = 72° 7', a = 83.4. 7. Given B = 32° 10', c = .02728. 8. Given 6 = 2.887, c = 5.11. 9. Given ^ = 52° 41', 6 = 4247. 10. Given a =101, 6 = 116. 11. Given ^ = 43° 22', o = 158.3. 12. Given ^ = 58° 39', c = 35.73. 13. Given a =204.2, c= 275.3. 14. Given 5 = 30°, 6 = 1.6438. 15. Given ^=22° 14', 6=13.242. 16. Given 5=10° 51', c=.7264. 17. Given a = 638.5, 6 = 501.2. 18. Given ^=78° 17', a = 203.8. 19. Given 6 =.02497, c=. 04792. 20. Given 5=2° 19' 30", a = 1875.3. 21. Given a =24.67, 6 = 33.02. 22. Given 6=1.4367, c = 3.4653. 23. Given 5=6° 12.3', c = 37206. 24. Given ^=64° 1.3', 6 = 200.05. 25. Given a = 340.06, 6 = 231.69. 26. Given a =1.7087, c = 2.0008. 27. Given 5=21^33' 51", a =.82109. 28. Given ^= 74° 0' 18", c = 275.62. 29. Given 5=34° 14' 37", 6 = 120.22. 30. Given a = 10.107, 6 = 17.303. SOLUTION OF RIGHT TRIANGLES. 71 Solve the followiog isosceles triangles, in which A and B are the equal angles, and o, 6, and c denote the sides oppo- site the angles A^ jB, and C, respectively : 31. Given ^1=68° 57', Z> = 35.091. 32. Given 5 =27° 8', c = 3.088. 33. Given C= 80° 47', Z> = 2103.2. 34. Given a =79.24, c= 106.62. 35. Given A = 70° 19', c = .5623. 36. Given C'= 151° 28', c = 95.47. 37. A regular pentagon is inscribed in a circle wliose diameter is 24 inches. Find the length of its side. 38. At a distance of 100 feet from the base of a tower, the angle of elevation of its top is observed to be 38°. Find its height. 39. What is the angle of elevation of the sun when a tower wliose height is 103.7 feet casts a shadow 1G7.3 feet in length? 40. If tlie diameter of a circle is 3268, find the angle at the centre subtended by an arc whose chord is 1027. 41. If the diameter of the earth is 7912 miles, what is the distance of the remotest point of the surface visible from the summit of a mountain 1^ miles in height? 42. Find the length of the diagonal of a regular pentagon whose side is 7.028 inches. 43. What is the angle of elevation of a mountain-slope which rises 238 feet in a horizontal distance of one-eighth of a mile? 44. From the top of a lighthouse, 133 feet above the sea, the angle of depression of a buo}^ is observed to be 18° 25'. Required the horizontal distance of the buoy. 72 PLANE TRIGONOMETRY. 45. A ship is sailing due east at the rate of 7.8 miles an hour. A headland is observed to bear due north at 10.37 A.M., and 33° west of north at 12.43 p.m. Find the distance of the headland from each point of observation. 46. If a chord of 41.36 feet subtends an arc of 145° 37', what is the radius of the circle ? 47. The length of the side of a regular octagon is 12 inches. Find the radii of the inscribed and circumscribed circles. 48. How far from the foot of a pole 80 feet high must an observer stand, so that the angle of elevation of the toj) of the pole may be 10°? 49. If the diagonal of a regular pentagon is 32.83 inches, what is the radius of the circumscribed circle ? 50. From the top of a tower, the angle of depression of the extremity of a horizontal base line, 1000 feet in length measured from the foot of the tower, is observed to be 21° 16' 37". Find the height of the tower. o 51. If the radius of a circle is 723.29, what is the length of the chord which subtends an arc of 35° 13'? 52. A regular hexagon is circumscribed about a circle whose diameter is 10 inches. Find the length of its side. 53. From the top of a lighthouse, 200 feet above the sea, the angles of depression of two boats in line with the light- house are observed to be 14° and 32° respectively. What is the distance between the boats? 54. A ship is sailing due east at a uniform rate of speed. At 7 A.M., a lighthouse is observed bearing due north, 10.32 miles distant, and at 7.30 a.m. it bears 18° 13' west of north. Find the rate of sailing of the ship and the bearing of the lifi-hthouse at 10 a.m. SOLUTION OF RIGHT TRlAXGLES. FORMULA FOR THE AREA OF A RIGHT TRIANGLE. 111. Case I. Given the hypotenuse and an acute angle. B Denoting the area by A", we have by Geometry, 2K=ab. But by Art. L3, a = c sin A, and 6 = ceos^. Whence, 27r=c^sinYl cos^l = ^c^8in2^ (Art. 74). That is, 4 K=z r sin 2 A. (36) In like manner, 4 7r=c-sin2^. (37) Case II. Given an angle and its opposite side. We have, cot ^1 = -•» or h = a cot A. a Whence, 2 K= a • a cot A = a- cot^. (38) In like manner, 2 K=h- cot B. (39) Case III. Given an angle and its adjacent side. We have, cot A = tan B (Art. 14) . Whence by (38) , 2 7r=a-tan^. (40) In like manner, 27r=?>-tan.4. (4l) 74 PLANE TRIGONOMETRY. Case IV. Given the hypotenuse and another side. Since a^ -}- 6^ = c^, we have _:i- = a v/(c + a ) (c - a). (42) In like manner, 2/ir=5\/(c + 6)(c-6). (43) Case V. Given the tivo sides about the right angle. In this case, 2K=ab. (44) EXAMPLES. 112. 1. Given c = 10.36, B=lo° ; find the area. By (37), 4K=:rsm2B. Whence, log (4^) = 2 logc + log sin 25. lege = 1.0153 ; multiply by 2 = 2.0306 2B= 150° ; log sin = 9.6990 - 10 log(4^)= 1.7296 .-. 4.K= 53.65, and K= 13.41. Note. To find log sin 150^, take either log cos 60= or log sin 30°. (See page 10 of the explanation of the tables.) Find the areas of the followinsj trianoles : 2. Given ^=19° 36', a = 22.17. 3. Given B = 24° 7', a = .8213. 4. Given a =149.31, 5 = 76.29. 5. Given 6 =.3056, c=.6601. 6. Given ^ = 30° 56' 20", c= 192.9. 7. Given A = 58° 52', b = .05207. 8. Given a =.932, c= 2.786. 9. Given B = 72° 25', c = 27.283. 10. Given 5=29° 18' 15", 6= .33784. GENEKAL PROPERTIES OF TRIANGLES. 75 VIII. GENERAL PROPERTIES OP TRIANGLES. 113. In any triangle^ the sides are proportional to the sines of their opposite angles. There will ))e two cases, according as the angles are all acute (Fig. 1), or one of them obtuse (Fig. 2). In each case let CD be drawn perpendicular to AB, Then in either figure, we have sin A = CD In Fig. 1, sin J5 = > a and in Fig. 2, sin B = sin (180°- CBD) = sin CBD (Art. 47) ^CD a Dividing these equations, we have in either case CD sin A _ _b__ _ a sin B'~'CD~b a (45) 76 PLANE TRIGONOMETRY. In like manner, we may prove sin B _h sin C c (*6) , Sin C c /._N and - — 7=— (47) sm A a The above results may be expressed more compactly as sm A sin B sin C 114. hi any triangle^ the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. Formula (45) of Art. 113 may be put in the form a:b = sin A : sin B. Whence, by composition and division, a + 6 : a — & = sin ^ 4- sin 5 : sin ^ — sin B, a + 5 _ sin A + sin B ' a — 6 sin ^ — sin 5 But by Art. 73, sin .^ + sin J5 _ tan |(^ + B) sin A — siuB tan ^{A — B) „„ a-\-b tan-i-(^4--S) /-_x Whence, -^ = - — I) . -,,( ' (48) a — b tani(^ — ij) In like manner, we may prove or 6 + c ^ tan jrjB-j-C) 6-c~tani(J5-(7) (49) and c + a.^tani(0 + ^). (^^^ c — a tan^(C — -4) GENERAL PROPERTIES OF TRIANGLES. 77 115. Since A + B=\SO°- C, we have tani(^ + ^) =tan(90°-iC) = coti(7 (Art. 14). Thus formula (48) may be put in the form a-{-b cot ^ C a — b tan ^ (^4 — B) (51) 116. /n cmy triangle, the square of any side is equal to the sum of the squares of the other two sides, minus twice their product into the cosine of their included angle. Case I. When the included angle A is acute. There will be two cases, according as the remaining angles are both acute (Fig. 1), or one of them obtuse (Fig. 2). In each case let CD be draAvn perpendicular to AB. Then in Fig. 1, BD = c - AD, and in Fig. 2, BD = AD- c. Squaring, we have in either case, BIX = jW' + c2 _ 2c X AD. 2 Adding CD' to both members, mf + ~Clf = A& +CD^-\-(?-2cy.AD. But, BD' -{-W= a-, and AD' + CD^ = h\ Also, Therefore, cos A = 1 or AD = b cos A. b a^ = b- -\-c- — 2bc cos A. (52) 78 PLANE TlilGONOMETRY. Case II. When the included angle A is obtuse. In Fig. 3, BD = AD + c. Squaring, and adding OD' to both members, Blf+Clf = Aff 4- OS' + c^ + 2 c X AD. But, BD^ + CD" = a\ and AD" + CD" = Ir. Also, cos A = cos (180° - CAD) = - cos CAD (Art. 47) AD Whence, AD — — b cos A. Therefore, a^ =b'--^c^-2bc cos A. In like manner, we may prove b^ = c^ -\- a^ — 2 ca cos 5, and c^ = a^ -|- &2 _ 2 a6 cos C (53) (54) 117. To express the cosines of the angles of a triangle in terms of the sides of the triangle. From (52), Art. 116, a 2_ 7.2 Whence, 6^ + c^ — 2 6c cos A. or, 2 be cos J. = 6- + c^ — a^, [>2 ^ ^.- _ ^^2 cos A 2 be (55) GEXERAL PROrERtlES OF TRIANGLES. 79 Id like manner, we have cosB = — '— , (56) and cos C = — (57) 118. To express the sines, cosines, and tangents of the ha/f-angles of a triangle in terms of the sides of the triangle. From (55), Art. 117, cos A = — 26c Subtracting both members from unity, . < 1 ^" + c" — a- a-— 6- -f 2bc — c^ 1 — cos ^1=1 ■ = ■ 2 be 2 be Whence by (a), Art. 75, 2 sm- ^A = r-^ 5 ^ 2 be (a — b-\-c)(a-\-b — c) or, sin^ IA = ^ f~^ -- ^ Abe Denoting « + 6 + c by 2 s, so that s is the half-sum of the sides of the triangle, we have a - b -\- c = (a-^b -he) - 2b = 2 s-2h = 2 (s-b) , and a 4- 6 — c = {a-{-b -f-c) — 2 c= 2 s—2 c = 2 {s — c) . 4(s-6) (s-c) Hence, sin^4^ = 4 be . ■, J lis — b)(s — c) . . or, sin|-^=^^^ ^-J^ ^. (58) In like manner, we may prove sm^B-\^ — , (59) and • 1 ^ i(s — a) (s — b . ^ smiC=sj^ ^ (60) or, cos^^A = 80 PLANE TRIGONOMETRY. Again, adding both members of (55) to unity, we have l + eos^ = l + -l^^^ = Uc Whence by (a), Art. 75, (6 + c + a) (6 + c — ct) . 45c But, 64-c + a=2s, and b-{-c — a = {b-\-c + a) — 2a = 2(s — a), „ , . 4 s (s — a) Hence, cos^t^ = tt ' C0si4=JS. (61) In like manner, cosii3=Ji5HS, (62) "^ \ ca and cosiO=J^^^- (63) Dividing (58) by (ei) , we obtain be s{s — a) = >-;)(^-^) . (64) \ s{s — a) In like manner, n -r. !(S— C) (S — a) /_^N fa>"i-s=^ ^ .(;_ft) ' («=) , ^ l(s — a')(s — 6) /_->. and t^niC^^^^^^^. (66) Note. Since each angle of a triangle is less than 180'', its half is less than 90^ ; hence the positive sign must be taken before the radical in each of the formulae of Art. 118. GENERAL PROPI^RTIES OP TRIANGLES. 81 AREA OF AN OBLIQUE TRIANGLE. 119. Case I. Given two sides and their included angle. C Fig. 1. There will be two cases, according as the included angle A is acute (Fig. 1), or oljtuse (Fig. 2). In each case let CD ijc drawn perpendieuhir to ^17^. Then, denoting the area of the triangle by A", we have by Geometry, 2K=rx CD. But in Fio-. 1 , sin A = ? b and in Fig. 2, sin A = sin (180° -CAD) = sin CzlZ) (Art. 47) = CD Whence, in either figure, CZ> = 6sin^4. Therefore , 2 A" = be sin A. In like manner, 2Il = ca sin B, and 2 K= ab sin C. Case II. Given a side and all the angles. By (69) , 2 K= ab sin C. But by Art. 118, b sin B y a sin 5 or b == (67) (68) (69) a sin ^1 sin A :^Lane trigonometry. Substituting, ^K— a x X sin G sin A o? sin B sin C sin^ In like manner, (70) o -r^ IP' sin C sin ^ , , 2/1 = —, (71^ sm^ 1 c\ rp- ^ sin A sin 5 , . and 1K= ^—- (72) sni G Case III. Given the three sides. By (67), 2/r=6csin^ = 2 be sin ^A cos ^A (Art. 74) . Dividing by 2, and substituting the values of sin|^^ and cos^^ from Art. Il5, we have h'^=hr,.j (s-b)(s-c) I s(s-a) \ be \ be = V/s(s — a)(s — 6)(s — c). (73) SOLUTIOI^ OF OBLIQUE TRIANGLES. IX. SOLUTION OF OBLIQUE TRIANGLES. 120. Ill the solution of plane oblique triangles we may distinguish four cases : 1 . Given a side and any tivo angles. 2. Given two sides and their included angle. 3. Given the three sides. 4. Given t/co sides and the angle opposite to one of them. Case I. 121. Given a side and any two angles. The third angle may be found by Geometry, and then by aid of Art. 113 the remaining sides may be calculated. Tlie triangle is always possible for any values of the given elements, provided the sum of the given angles is < 180°. 1. Given b = 20, .1 = 104°, B = 19°. Find «, c, and C. C= 180° - (.1 -f- B) = l.SO° - 123° = 57°. ■D A 4. 1 1 o <^<' sin A T c sin C By Art. 113, - = , and - = b sinB b sin J5 That is, a = b sin A esc B, and c = b sin C esc B. Whence, log a = log b -f log sin A -\- log esc B, and log c = log b -j- log sin C + log esc B. log 6 =1.3010 \ogb^ 1.3010 log sin ^1= 9.9869 - 10 log sin C= 9.9236 - 10 log CSC B = 0.4874 log esc B = 0.4874 loora = 1.7753 loo: c= 1.7120 .-. a = 59.61. .-. c = 51.52. Note. To find the log cosecant of an angle, subtract the log sine from 10 — 10. To find log sin 104°, take either log cos 14° or log sin 76°. (See pages 7 and 10 of the explanation of the tables.) 84 PLANE TRIGONOMETRY. EXAMPLES. Solve the following triangles : 2. Given a =10, ^ = 38°, 5 =77° 10'. 3. Given & = .8037, 5 = 52° 20', (7 =101° 40'. 4. Given c = .032, ^ = 36° 8', 5 = 44° 27'. 5. Given 6 = 29.01, ^=87° 40', (7 = 33°15'. 6. Given ct = 5.42, 5 = 98° 22', C = 41°l'. 7. Given c = .0161, ^ = 35° 15', (7= 123° 39'. 8. Given a = 400, ^ = 54° 28', 0=60°. 9. Given 6 = 314.29, .4 = 67° 22', 5 = 57° 51'. 10. Given c = 7.86, 5 = 32° 2' 52", = 43° 25' 26". Case II. 122. Given two sides and their included angle. Since one angle is known, the sum of the remaining angles may be found, and then their difference may be calculated b}^ aid of Art. 114. Knowing the sum and difference of the angles, the angles themselves may be obtained, and then the remaining side may be computed as in Case I. The triangle is possible for any values of the data. 1. Given a = 167, c = 82, 5 = 98°. Find A, O, and 6. By Geometry, ^ + O = 180° - 5 = 82°. By Art. 114, o + c^ tanJC^ + 0)^ a — c tan^(^— O) or, tanl-(^-0)=-^^ tan 1(^ + 0). Whence, log tan |(^ — O) = log {a — c) -\- colog (a -f- c) + logtani(^ + 0). SOLUTION OF OBLIQUE TRIANGLES. 85 a-c = 85 log =1.9294 a + c = 249 colog = 7.6038 i(A + C) = 41° log tan = 9.9392 logtan^(^-C)= 9.4724 .-. i{A-C)=l6°Sl.T. Hence, A = ^{A + C) + \{A -C) = 57° 31.7', and C = i{A -\- C) - ^{A- C)= 24° 28.3'. To find the remaining side, we have by Art 113, , a sin B ■ n a h z=z = a sin B CSC A. sin A Whence, log h = log a + log sin B + log esc A. loga = 2.2227 log sin 2^ =9.9958 log CSC ^1 = 0.0739 log 6 = 2.2924 .-. 6=196.05. EXAMPLES. Solve the following triangles : 2. Given c/ = 27, c=15, J5 = 46°. 3. Given a = 486, 6 = 347, 0=51° 36'. 4. Given 6 = 2.302, c = 3.567, ^=62°. 5. Given a =.3, 6 = .363, 0= 124° 56'. 6. Given 5=1192.1, c = 356.3, ^ = 26° 16'. 7. Givena=7.4, c = 11.439, 5 = 82° 26'. 8. Given a = 53.27, 6 = 41.61, (7= 78° 33'. 9. Given 6 = .02668, c=. 05092, ^=115° 47'. 10. Given a = 51.38,* c= 67.94, ^=79° 12' 34". 86 PLANE TKIGONOMETRY. Case III. 123. Given the three sides. The angles might be calculated by the formulae of Art. 117; but as these are not adapted to logarithmic computa- tion, it is more convenient to use the formulae of Art. 118. Each of the three angles should be computed trigonometri- call}^ as we then have a check on the work, since their sum should be 180°. If all the angles are to be computed, the tangent formulae are the most convenient, as only four different logarithms are required. If but one angle is required, the cosine for- mulae will be found to involve the least work. The triangle is possible for any values of the data, pro- vided no side is greater than the sum of the other two. If all the angles are required, and the tangent formulae are used, they may be conveniently modified as follows : ^ . ^ -.J \(s — a)(s — b)( s —c) By Art. 118, tani^=^^^ s\s-ay l_ Ki^a){s-b){s-c) s — a\ s ^ . \(s — a)(s — b) (s — c) , Denotmg lI- — — by r, we have tan^^ = -. s — a In like manner, tan ^B = , and tan |- (7 = s—b s—c 1. Given a = 2.5, 6 = 2.79, c = 2.33 ; find A, B, and G. In this case, 2s = a -f- 5 + c = 7.62, and s= 3.81. Whence, s — a= 1.31, s — b = 1.02, and s — c= 1.48. By logarithms, we have log r = -2-[log (s — a)-]- log (s — 6) + log (s — c) + colog s] . SOLUTION OF OBLIQUE TRIANGLES. 87 Also, log tan ^A = log r — log (s — a) , log tan ^B = log r — log (s — 6) , log tan ^C = log r — log (s — c) . log (s - a) = 0.11 73 log r = 9.8576 - 10 log (s - h) = 0.0086 log (s - 6) = 0.0086 log (s - c) = 0.1703 log tan 4^i?= 9.8490 - 10 cologs = 9.4191 -10 ^5=35° 14.1' 2 )9.7153-10 .'.B= 70° 28.2'. .•.logr= 9.8576 -10 log r = 9.8576 - 10 log r = 9.8576 - 10 log (s - a) = 0.1173 log (s - c) = 0.1703 log tan ^A = 9.7403 - 10 log tan ^C= 9.6873 ^ 10 iA = 28° 48.3' i C= 25° 57.2' .-. A = 57° 36.6'. .-.•0= 51° 54.4'. Check, ^14-J5 + C=179°59.2'. 2. Given a = 7, 6 = 11 , c = 9.6 ; find B. Ms-b) By Art. 118, cos^^ Or, log cos^B = 4-[log s + log {s — b)-\- colog c + colog a], In this case, 2s = a -|- 6 + c = 27.6. Whence, s=13.8, and s-b = 2.8. logs =1.1399 log (s - 6) = 0.4472 colog c = 9.0177 -10 colog a = 9. 1549 — 10 2 )19.7597-20 log cos 1^=9.8798 -10 1^=40° 41.8' r.B = 8r 23.6'. 88 PLANE TRIGONOMETRY. EXAMPLES. Solve the following triangles : 3. Given a = 2, & = 3, c = 4. 4. Given a = 4, 5=7, c=6. 5. Given a = 5. 6, 5 = 4.3, c = 4.9. 6. Given a=. 23, 5=.26, c=.198. 7. Given a =79.3, 6=94.2, c = 66.9. 8. Given a = 321, 6 = 361, c = 402. 9. Given «=. 641, 6=.529, c=.702. 10. Given a = 3.019, 6 = 6.731, c = 4.228. Case IY. 124. Given two sides and the angle opposite to one of them. It was stated in Art. 106 that a triangle is in general com- pletely determined when three of its elements are known, provided one of them is a side. The only exceptions occur in Case IV. To illustrate, let us consider the following : 1. Given a = 52.1, 6 = 61.2, ^ = 31° 26'. Required B, (7, and c. -n A ^ -. 1 o sin B i> . ^ 6 sin ^ By Art. 113, - — - — —> or sinB = Whence, log sin B = log 6 + colog a + log sin A. log 6 =1.7868 cologa = 8.2832 -10 log sin J. = 9.7173 -10 log sin 5 = 9.7873 -10 .-. ^ = 37° 47.5', from the table. SOLUTION OF OBLIQUE TRIAXGLES. 89 But in determining the angle corresponding, attention must be paid to the fact that an angle and its supplement have the same sine (Art. 47). Therefore another value of 5 tvIII be 180° — 37° 47.5', or 142° 12.5' ; and calling these values Bi and ^o. we have ^1=37° 47.5', and 5,= 142° 12.5'. Note. The reason for tliis ani})iguity is at once apparent when we attempt to construct the triangle from the data. Bi D We first lay off the angle DAF equal to 31° 2(V, and on AF take ^C— 61.2. With C as a centre, and a radius equal to 52.1, describe an arc cutting AD at Bi and B.^. Then either of the triangles AB^C or vlC^C satisfies the given conditions. Tlie two values of B which were obtained are the values of the angles AB^C and AB.,C respectively; and it is evident geometrically that these angles are supplementary. To complete the solution, denote the angles ACB^ and ACBo by C\ and 02, and the sides ABi and xlB., b}- q and Cg. Then, Ci= 180°- (.1 + 7?,) = 180°- 69° 13.5'= 110° 46.5', and O2=180°-(.l + i?,,) = 180°-173°38.5'= 6° 21.5'. . . c, sin C, Agam, -L = -^- ^ a sm A and "Whence, c, = a sin d csc^, log a =1.7168 log sin ^1 = 9.9708 log csc^ = 0.2827 ^2. a C2 sin Co sin .4 and C2 = a sin C^ esc A. log a =1.7168 log sin a = 9.0443 log csc^ = 0.2827 logCi= 1.9703 .-. Ci=93.4. log Co =1.0438 .-. C2= 11,06. 90 PLANE TEIGOJs^OMETRY. 125. AYhenever an angle of an oblique triangle is deter- mined fr.om its sine, both tiie acute and obtuse values must be retained as solutions, unless one of them can be shown by other considerations to be inadmissible ; and hence there may sometimes be two solutions, sometimes only one, and sometimes none, in an example under Case IV. I. Let the data be a, 6, and A, and suppose b<.ct- Since, by Geometry, B must be < A, only the acute value of B can be taken ; in this case there is but one solution. II. Let the data be a, b, and A, and suppose b > a. Since B must be > A, the triangle is impossible unless A is acute. . . . sin B b 17-^ -Tj-^-.* Agam, smce = -5 and is > a, smB is > sin J.. sin A a Hence both the acute and obtuse values of B are > A, and there are tico solutions, except in the following cases : If the data are such as to make log sin -5 = 0, then sin 5= 1 (Art. 87) and B= 90°, and the triangle is a right triangle ; if log sin 5 is positive, then sin 5 is > 1, and the triangle is impossible. 126. The results of Art. 125 may be stated as follows : If, of the given sides, that adjacent to the given angle is the less, there is but one solution, corresponding to the acute value of the opposite angle. If the side adjacent to the given angle is the greater, there are two solutions unless the log sine of the opposite angle is or positive ; in which cases there are one solution (a right triangle), and no solution, respectively. 127. We will illustrate the above points by examples : 2. Given a =7.42, 6 = 3.39, ^=105° 13'; find 5. Since ?> is < o, there is but one solution, corresponding to the acute value of B. SOLUTIOX OF OBLIQUE TRIANGLES. 91 Air V. • 13 ^sin^ We nave, sin B = • a log 6 = 0.5302 colog a = 9.1296 log sin .4 = 9.9845 log sin 5 = 9.6443 .-. i3=2G° 9.0'. 3. Given & = 3, c=2, C=100°; find J5. Since h is > c, and C is obtnse, the triangle is impossible. 4. Given rt= 22.764, c = 50, ^1 = 27° 4.8'; find C. We have, sin C = — logc= 1.G990 colog a = 8.G428 log sin .4 = 9.6582 log sin C'= 0.0000 .-. sin C= 1, and C=90°. Here there is bnt one solntion ; a right triangle. 5. Given a = .83, 6 =.715, 7^=61° 47'; find ^. WT \^ A ft sin B We have, sin A = h log rt = 9.9191 colog & = 0.1457 W sin 5= 9.9451 log sin .1 = 0.0099 Since log sin A is positive, the triangle is impossible. EXAMPLES. Solve the followinor triansiles : 6. Given a = 5.08, 5 = 3.59, ^=63° 50'. 7. Given 5 = 74.8, c = 02.2, 0=27° 18'. 92 PLANE TRIGONOMETRY. 8. Given 6 = .2337, c = .1982, 5 = 109°. 9. Given a =1.07, c = 1.71, = 31° 53'. 10. Given a =.1864, 5 = .17, 5 = 63° 40'. 11. Given a = 50, c = 66, ^=123°11'. 12. Given 6 = 50.3, c = 66.8, = 32° 49'. 13. Given a = 8.656, c = 10, ^=59° 57'. 14. Given 6 = 5.161, c = 6.84, J5=44°3'. 15. Given a = 214.56, 6 = 284.79, 5 =104° 20'. 16. Given 6 = 3069, c = 1223, = 55° 52'. 17. Given a = .7097, c = .5112, ^ = 35° 11'. 18. Given a =106. 85, 6 = 166.21, ^ = 40°0'21". 19. Given a=. 3216, c = .2708, 0=52°24'. 20. Given 6 = 811.3, c = 606.4, jB = 126° 5' 20". AREA OF AN OBLIQUE TRIANGLE. 128. 1. Given a =18.063, ^ = 96° 30', 5=35°; find A". _ . ^ ^ ^^ a^ sin B sin O By Art. 119, 2A = -. — -. -^ ' sm ^ = a^ sin B sin esc A. Whence, log (2 K) = 2 log a + log sin B + log sin O + log esc A. From the data, = 180° -{A + B) = 48° 30'. log a = 1.2568 ; multiply by 2 = 2.5136 log sin 5 = 9.7586 log sin = 9.8745 log CSC ^=0.0028 log (2Ar) = 2.1495 .-.2^=141.1, and ^=70.55. SOLUTION OF OBLIQUE TRIANGLES. 93 EXAMPLES. Find the areas of the following triangles : 2. Given rt = 38, c=G1.2, 5 = 67° 56'. 3. Given a =5, 5 = 7, c = 6. 4. Given 6 = 2.07, ^=70°, 5 = 36° 23'. 5. Given /j = 116.1, c = 100, ^ = 118°16'. 6. Given a =79, b = \)-i, c=67. 7. Given a = 3.123, ^ = 53M1', 5=13°57'. 8. Given /j = .439, ^1 = 76° 38', (7= 40° 35'. 9. Given a = 23.1, 6 = 19.7, c=25.2. 10. Given « = . 3228, c = .9082, i^ = 60°lG'. 11. Given r =80.25, 7? =100° 5', 0=31° 44'. 12. Given (< = .010168, 6 = .018225, C = ll°18'26". 13. Given (t = 5.82, 6 = 6, c=4.26. MISCELLANEOUS EXAMPLES. 129. 1. From a point in the same horizontal plane with the base of a tower, the angle of elevation of its top is 52° 39', and from a point 100 feet fin-ther away it is 35° 16'. Required the height of the tower, and its distance from each of the points of observation. 2. In a field ABCD, the sides AB, BC, CD, and DA are 155, 236, 252, and 105 rods, respectively, and the length of a line from xi to C is 311 rods. Find the area of the field. 3. From the top of a bluff, the angles of depression of two posts in the plain below, in line with the observer and 1000 feet apart, are found to be 27° 40' and 9° 33', respec- tively. AYhat is the height of the bluff above the plain? 4. Two yachts start at the same time from 'the same point, and sail, one due north at the rate of 10.44 miles an 94 PLAXE TRIGONOMETRY. hour, and the other due northeast at the rate of 7.71 miles an hour. How far apart are they at the end of 40 minutes ? 5. A ship is sailing due southwest at the rate of 8 miles an hour. At 10.30 a.m., a lighthouse is observed to bear 30° west of north, and at 12.15 p.m., it is observed to bear 15° east of north. Find the distance of the lighthouse from each position of the ship. 6. Wishing to find the distance of an inaccessible object A from a position B, I measure a line BC\ 208.3 feet in length. The angles ABC and ACB are measured, and found to be 126° 35' and 31° 48', respectively. Required the distance AB. 7. A flagpole 40 feet in height stands on the top of a tower. From a position near the base of the tower, the angles of elevation of the top and bottom of the pole are 38° 53' and 20° 18', respectively. Required the distance and height of the tower. 8. A surveyor observes that his position A is exactly in line with two inaccessible objects B and C. He measures a line AD, 500 feet in length, making the angle BAD = 60°, and at D observes the angles ADB and BDC to be 40° and 60°, respectively. Required the distance BC. 9. To find the distance between two buoys A and 5, I measure a base line CD on the shore, 150 feet in length. At the point C the angles ACD and BCD are measured and found to be 95° and 70°, respectively ; and at D the angles BDC and ADC are found to be 83° and 30°. What is the distance between the buoys ? 10. The sides of a field ABCD are AB=^1, BC=63, and DA = 20, and the diagonals AC and BD are 75 and 42, respectively. Required the area of the field. Paet II. SPHERICAL TPJGOXOMETRY. 3>*iOO- X. GEOMETRICAL DEFINITIONS AND PRINCIPLES. 130. If a triedral angle is formed with its vertex at the centre of a sphere, it intercepts on the surface a spherical triangle. 131. The triangle is bounded by three arcs of great circles called its sides, which measure the face angles of the triedral angle. The angles of the spherical triangle are the diedral angles of the triedral angle ; and by (ieometry, each is measured by the angle between two straight lines drawn, one in each face, and perpendicular to the edge at the same point. 132. The sides of a spherical triangle, being arcs, are usually expressed in degrees. If the length of a side in terms of some linear unit is desired, it ma}^ be obtained by finding the ratio of its arc to 360°, and multiplying the result by the length of the circum- ference of a great circle. 133. Spherical Trigonometry treats of the trigonometrical relations between the elements of a spherical triangle ; or what is the same thing, between the face and diedral angles of the triedral angle which intercepts it. 96 SPHERICAL TRIGONOMETRY. 134. The face and diedral angles are not altered in mag- nitude by varj^ing the radius of the sphere, and hence the relations between the sides and angles of a spherical triangle are independent of the length of the radius. 135. We shall limit ourselves in this work to snch tri- angles as are considered in Geometry, where each angle is less than two right angles, and each side less than the semi- circumference of a great circle ; that is, where each element is less than 180°. 136. The proofs of the following properties of spherical triangles may be found in any treatise on Solid Geometrj' : (a) P^ither side of a spherical triangle is less than the sum of the other two sides. (5) If two sides of a spherical triangle are unequal, the angles opposite them are unequal, and the greater angle lies opposite the greater side ; and conversely. (c) The sum of the sides of a spherical triangle is less than 360°. (d) The sum of the angles of a spherical triangle is greater than 1 90°. c B Complete the lune ABA'C. 98 SPHERICAL TRIGOXOMETRY. Then in the right triangle A'BC, A'C= 180" - b. That is, the sides a and A'C of the triangle A'BC are each < 90°; and by I., A'B and the angles A' and A'BC are each < 90°. But, c = 180° - A'B, A = A', and B = 180° - A'BC. ^ "Whence, c is > 90°, A < 90°, and B > 90°. In like maimer, if a is > 90° and & < 90°, then c is > 90°, ^>90°, and5< 90°, III. Suppose a > 90° and b > 90°. Complete the lune ACBC. Then in the right triangle ABC, AC =180'' -b, and BC = 180° -a. That is, the sides AC and BC of the triangle ABC are each < 90° ; and by I., AB and the angles BAC and ABC are each < 90°. But, A = 180° - BAC, and B = 180° - ABC. Whence, c is < 90°, A > 90°, and B > 90°. Hence, in any spherical right triangle : 1. If the sides including the right angle are in the same quadrant, the hypotenuse is < 90° ; if they are in different quadrants, the hypotenuse is > 90°. 2. An angle is in the same quadrant as its opp)Osite side. SPHERICAL RIGHT TRIANGLES. 99 XI. SPHERICAL RIGHT TRIANGLES. 139. Let C be the right anole of the spherical right tri- angle ABC\ and the centre of the sphere. Join OA, OB, and OC. At any })oint A' of OA draw A'B' and A'C perpendicular to OA, and join B'C Then hj' Art. L'31, the sides «, 6, and c measure the angles BOC\ COA, and AOB, respectively, and the angle B'A'C is equal to the angle A of the spherical triangle. Since OA is perpendicular to A'B' and A'C, it is perpen- dicular to the plane A' B'C. AVhence, since each of tlie })lanes A'B'C and OBC is perpendicular to the plane OAC\ their intersection B'C is perpendicular to OAC. Therefore B'C is perpendicular to A'C and OC. In the right triangle OA'B', we have cos c = cos A' OB' = OA' OB' OC OA' OB' ^ OC' But in the right triangles OB'C and OCA\ OC -, OA' , = cos 0, and = cos h. OB' OC Whence, cos c = cos a cos 6. (74) 100 SPHERICAL TRlGOXOMETiiY. Again, smA=smB'A'C' = ^r^, = -jf^, = ^r^^, (75) 'OB' A'C and co^ A = Qo&B A C^ = -rr^. = ——■.— (76) A'B' AB' tan c In like manner we have, . -r, sin 5 / \ smB = 5 (77) sin c -t -r-» Tian 0/ ^ V and cosi3=^ -• (78) tan c 140. From (75) and (76), we obtain , J sin A sin a tan c sin a tan ^ = = X cos A sin c tan 6 cos c tan 6 Whence by (74) , , 4 sin a tan a , ^ tan^ = — = (79) cos a cos b tan 6 sin b Similarly, , T) tan b , X tan^ = iSO) sin a 141. Since sin a = cos a tana (Art. 20), (75) may be ^^•^t*^^ tana . cos a tan a tan c sm ^ = : = cos c tan c cos c cos a Whence by (74) and (78), sin Az= cos B COS b Similarh^, sin B = COS A COS a (81) (82) SPHERICAL RIGHT TRIANGLES. 101 142. From (74), (8l), and (82), we have cos c = cos a cos h _ cos A cos B sin B sin A = cot A cot B. (83) 143. The proofs of Art. 139 cannot be regarded as gen- eral, for in the construction of the figure we have assumed a and 6, and therefore c and A (Art. 138), to be less than 90°. To prove formulfe (74) to (73) universally, it is necessary to consider two additional cases : Case I. When one of the sides a and b is < 90°, and the other > 90°. In the right triangle ABC, let a be < 90° and b > 90°. Complete the lune ABA'C; thou in the triangle A'BC, A'B = 180° -c, A' = A, A'C = 180° - b, A'BC = 180° - B. But by Art. 138, c is > 90°, A < 90°, and B > 90°. Hence each element, except the right angle, of the right triangle A'BC is < 90°, and we have by Art. 139, cos A'B= cos a cos A'C, A , sni a sm A' = ? sin A'B tan A'C cos A' = At Tin Sin A'C sm A'BC = » sin A'B tan a tan A'B cos A'BC tan A'B 102 SPHERICAL TRIGONOMETRY. Putting for A'B, A'C, A', and A'BC their values, we have cos (180° -c)= cos a cos (180°-&), sin^ = ?ia^ , sin(180°-B) = ?iBll551=L^, sin (180° -c) ^ ^ sin (180° -c) cos^^^^"(^^Q°-^), cos(180°-5) = ^^^ tan (180° -c) \ ^ tan(180°-c) Whence by Art. 47, — cos c = cos a ( — cos 5) , .^ A sin a . -D sin & sin^=: ^ sm5=i ■) sin c sin c ^^^ /f — tan 6 -D tan a cos^ = 5 — eosi5 = - ; — tan c — tan c and we obtain the formulae (74) to (78) as before. In like manner, the formulae may be proved in the case where a is > 90° and 6 < 90°. Case II. When both a and h are > 90°. a R In the right triangle ABC\ let a and 6 be > 90°. Complete the lime ACBC By Art. 138, c is < 90°, A > 90°, and B > 90°. Hence each element, except the right angle, of the right triangle ABC is < 90°, and we have by Art. 139, cos c = cos ^C cos J5C • T) int s'lnBC . ATyrn sin^C" sm^^u' = 9 sm ABC = ? sin c sin c TyAnt tan J.0' Annt tun BC cos BAC = —9 cos ABC =^ tan c tan c SPHERICAL RIGHT TRIANGLES. 103 Putting for AC\ BC, BAC, and ABC their values, we have cos c = cos (180° - a) cos (180° - 5) , sin(180°-^) = ^iHn^5!z:i^, sin(180°-5) = ^Hil^5!zL^, sin c sin c cos(180°-.l) = '^"(''^"°-^-\cos(180°-^)=t""('«^°-'''>. tan c tan c Whence by Art. 47, cos c = ( — COS a) ( — cos b) , ^ sin a . -n sin 6 sm^= 1 sin 5 = ? sin c sin c .^r, A — t^" ^ r. — tan a — cos ^ = •) — cos B = ; tan c tan c and we obtain the formuki} (74) to (78) as before. . 144. The forinuLT of Arts. 139 to 142 are collected below for the convenience of the student : cose = cos a cos b. i sin a sin A = — • r> sin 6 sill B = sin c sin c C0Su4= • tanc cos B = • tanc tan A = *•'•"«. , n tan b tan B = — — sin b sin a . ^ cos B sin^= siuB-Oos^. cos b cos a cosc = cot A cot ^. By comparing the formula? for the sines, cosines, and tan- gents of A and B with the corresponding forms for plane triangles as given in Arts. 10 and 14, no difficulty will be found in retaining them in the memory. 104 SPHERICAL TRIGONOMETRY. NAPIER'S RULES OF CIRCULAR PARTS. 145. These are two artificial rules which include all the formulse of the preceding article. In any spherical right triangle, the elements a and &, and the complements of the elements A^ B, and c (written in abbreviated form, co. A, co. jB, and co. c), are called the cir- cular' parts. CO. B CO. If we suppose them arranged in the order in which the letters occur in the triangle, any one of the five may be taken and called the middle par^; the two immediately adjacent are called the adjacent parts, and the remaining two the opposite parts. Then Napier's rules are : I. The sine of the middle part is equal to the product of the tangents of the adjacent p)arts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 146. Napier's rules may be proved by taking each of the circular parts in succession as the middle part, and showing that the results agree with the formulae of Art. 144. 1. If a is the middle part, 6 and qo.B are the adjacent parts, and co. c and co. A the opposite parts. Then the rules give sin a — tan h tan (co. B) , and sin a = cos (co. c) cos (co. A) ; or, by Art. 14, sina = tan b cotB, and sin a = sin c sin ^ ; which agree with (80) and (75). SPHERICAL RIGHT TRIANGLES. 105 2. If 6 is the middle part, a and co. A are tlie adjacent parts, and co. c and co. B the opposite parts. Then, sin b = tan a tan (co. ^4) = tana cot ^4, and sin b = cos (co. c) cos (co. B) = sine sini^ ; which agree with (79) and (77). 3. If CO. c is the middle part, co. ^ and co. B are the adjacent parts, and a and b the opposite parts. Then, sin (co. c) = tan (co. A) tan (co. B), or cos c = cot A cot B ; and sin (co. c) = cos a cos 6, or cos c = cos a cos b; which agree with (83) and (74). 4. If CO. A is the middle part, b and co.c are the adjacent parts, and a and co.^ the opposite parts. Then, sin (co. ^4) = tan?; tan (co. c) , or cos^l = tan b cote, and sin (co. ^4) = cos a cos (co. B) , or cos^l = cos a sin B ; which agree with (76) and (82). 5. If CO. B is ti)e middle part, a and co.c are the adjacent parts, and b and co.A the opposite parts. Then, sin ( CO. J5) == tana tan (co.c), or cos i^== tana cote, and sin (co. B) = cosb cos (co. A) , or cos^ = cos 6 sin^l ; which agree with (78) and (81). 147. Writers on Trigonometry differ as to the practical value of Napier's rules ; but in the opinion of the highest authorities, it seems to be regarded as preferable to attempt to remember the formulae by comparing them with the anal- ogous forms for plane triangles, as stated in Art. 144. 106 SPHERICAL TRIGONOMETRY. SOLUTION OF SPHERICAL RIGHT TRIANGLES. 148. To solve a spherical right triangle, two elements must be given in addition to the right angle. There may be six cases : 1. Given the hypotenuse and an adjacent angle. 2. Given an angle and its opposite side. 3. Given an angle and its adjacent side. 4. Given the hypotenuse and another side. 5. Given the tivo sides a and b. 6. Given the tivo angles A and B. 149. Either of the above maj^ be solved b}^ aid of Art. 144. The formula for computing either of the remaining elements when any two are given may be found b}' the following rule : Take that forniida which involves the given parts and the required part. If all the remaining elements are required, the following rule may be found convenient in selecting the formulae : Take the three formuke ivhich involve the given parts. 150. It is convenient in the solution to have a check on the logarithmic work, which maj- be done in every case with- out the necessity of looking out any new logarithms. Examples of this will be found in Art. 153. The check formula for any particular case may be selected from the set in Art. 144 by the following rule : Take that formula ivhich involves the three required parts. Note. If Napier's rules are used, the following rule will indicate which of the circular parts corresponding to the given elements and any required element is to be regarded as the middle part : If these three circular parts are adjacent, take the middle one as the mid- dle part, and the others are then adjacent parts. Ij' they are not adjacent, take the part ivhich is not adjacent to either oj the others as the middle part, and the others are then the opposite parts. For the check formula, proceed as above with the circular parts correspond- ing to the three required elements. SPHERICAL RIGHT TRIAXGLES. 107 Thus if c and A are the given elements, 1. To find a, consider the circuhir parts a, co.c, and co.^; of tliese, a is the middle part, and co.c and co.^ are opposite parts. Then, by Napier's rules, sin a = cos (co.c) cos (co.^1) = sin c sin^. 2. To find h, tlie circular parts are b, co.c, and co.A; in this case CO. ^ is the middle part, and b and co.c are adjacent parts. Then, sin (co.^1) = tan 6 tan (co.c), or cos ^1 = tan 6 cote. 3. To find 13, tlie circular parts are co.B, co.c, and co.^; co.c is the middle part, and co. .1 and co. D are adjacent i)arts. Tlien, sin (co.c)= tan (co. ^1) tan (co. B), or cose = cot.l cot^. 4. For the check formula, the circular parts are a, b, and co. B; a is the middle part, and b and co. B are adjacent parts. Then, sin a — tan b tan (co. B) = tan b cot B. 151. lu solving spherical triangles, cnreful attention mnst be paid to the ah/ehraic signs of the functions ; the cosines, tangents, and cotangents of angles greater than 1)0° being taken negative. It is convenient to place the sign of each function just above or below it. as illustrated in the examples of Art. lo3 ; the sisfn of the function in the first member beino- then deter- mined in accordance willi the i>rinciple that like signs pro- duce -h, and unlike signs produce — . Note. In the examples after tlie first of Art. 153, the signs are omitted in every case wliere both factors of the second member are +. 152. In finding the angles corresponding, if the function is a cosine, tangent, or cotangent, its sign determines whether the angle is less or greater than 1)0° ; that is, if it is +, the angle is < 90° ; and if it is — , the angle is > 90°, and the supplement of the acute angle obtained from the tables must be taken (Art. 47). If the function is a sine, since the sine of an angle is equal to the sine of its supplement, both the acnte angle obtained from the tables and its supplement mnst be retained as solu- tions, unless the ambiguit}^ can be removed by the principles of Art. 138. 108 SPHERICAL TRIGONOMETRY. EXAMPLES. 153. 1. Given 5= 33° 50', a = 108° ; find A, 6, and c. B}' the rule of Art. 149, the formulae from Art. 144 are, . ^ cos A ^ tan b _ tan a sm5= ? tanij = -; — -^ cosB = - cos a sm a tan c That is, _ ~. - +_ + ."'" +^ ~ tanci cos A = cos a sin -B, tan o = sm a tan 5, tan c = cos-B + Hence, log cos ^ = log cos a + log sin ^, log tan b = log sin a -\- log tan B, log tan c = log tan a — log cos B. Since cos JL and tanc are negative, the supplements of the angles obtained from the tables most be taken (Art. 152). Note. When the supplement of the angle obtained from the tables is to be taken, it is convenient to write 180° minus the element in the first member, as shown below in the cases of A and c. By the rule of Art. 150, the check formula for this case is cos A = -^^—^ or log cos A = log tan b — log tan c. tanc The values of log tan b and log tan c may be taken from the first part of the work, and their difference should be equal to the result previously found for log cos ^. log cos o = 9.4900 - 10 log tan a = 0.4882 W sin5 = 9.7457 - 10 log cos B = 9.9194 - 10 log cos ^=9.2357 -10 log tan c =0.5688 180°- ^=80° 5.5' 180° -c = 74° 53.8' .-. ^=99°54.5'. .-. c =105°6.2'. log sin a = 9.9782 — 10 Check. log tan 5 =9.8263 -10 log tan 5 = 9.8045-10 log tan b =9.8045-10 log tan c = 0.5688 b = 32° 31.1'. lo^ cos A = 9.2357 - 10 sin A = sin a sine That is, SPHERICAL RIGHT TRIAXGLES. 109 2. Given c = 70° 30', A = 100° ; find a, 6, and B. By Art. 149, the three formulae are, cos A =z , cos c = cot A cot B. tan c - + - - + - sin a = sin c sin A, tan b = tan c cos A, cot 5 = cos c tan ^. The side a is determined from its sine ; but the ambiguity is removed by the principles of Art. 138 ; for a and A must be in the same quadrant. Therefore a is > 90°, and the sup- plement of the angle obtained from the table must be taken. By Art. 150, the check formula is tan B = — , or sin a = tan b cot B, sin a Note 1. The check formula should always be expressed in terras of the functions used in determining the required parts; thus, in the case above, the check formula is transformed so as to involve cot^ instead of tan B. log sin c = 9.9743 log cos c = 9.5235 log sin A = 9.9934 log tan .1 = 0. 7537 log sin a =9.9677 log cot ^ = 0.2772 180°- a = 68° 10' 180° - J5 = 27° 50.6' .-. rt=lll°50'. .'. J5 = 152°9.4'. log tan c = 0.4509 log cos ^ = 9 . 2 3 9 7 Check. log tan b = 9.6906 ^^S ^^^ ^ = ^-^^^^ lSO°-b = 26° 7.5' l^g ^^^ ^ = Q-^^72 .-. b= 153° 52.5'. log sin a = 9.9678 Note 2. We observe here a difference of .0001 in the two values of log sin a. This does not necessarily indicate an error in the work, for such a small difference might easily be due to the fact that the logarithms are only approximate! t/ correct to the fourth decimal place. 110 SPHERICAL TRIGONOMETRY. 3. Given a = 132° 6', b = IT bV ; find A, B, and c. In this case the formulae are, — , tan a , ^ tan b ~~ "" "^ , tan A = . , ; tan B = ~. — ? cos c = cos a cos b. sm b sm a + The check formula is cos c = cot A cot 5, or cos c tan -4 tan B= 1. That is, log cos c + log tan A + log tan -B = log 1 = log tan a = 0.0440 log cos a = 9.8263 log sin b =9.9901 log cos b =9.3232 log tan ^ = 0.0539 log cos c =9.1495 180° -A = 48° 32.8' 180° - c = 81° 53.4' .-. ^ = 131°'27.2'. .-. c= 98° 6.6'. log tan b =0.6670 Check. loo- sin a = 9.8704 log cos c = 9.1495 log tan J5 = 0.7966 loo- tan ^=0.0539 log 1 = 0.0000 4. Given A = 105° 59', a = 128° 33' ; find 6, 5, and c. The formulae are, sin a sm c = -• sm-4 + sin b ■■ tana tan^ + sinjB = cos J. cos a The check formula is sin 5 = sin& sm c In this example, each of the required parts is determined from its sine ; and as the ambiguit}^ cannot be removed by Art. 138, both the acute angle obtained from the tables and its supplement must be retained in each case. SPHERICAL RIGHT TRIANGLES. HI log tau a = 0.0986 lo^ sin a = 9.8932 l02 tan A = 0.5430 looj sin .4 = 9.9828 log sin b =i}Jj3dC) log sin c =9.9104 .-. b = -2r3.d', .-. c = 54° 26.7', or 158° 56.1'. or 125° 33.3'. log cos ^ = 9.4399 Check. loor COS a = 9.7946 I02: sin b = 9.5556 log sin B = 9.6453 log sin c =9.9104 .-. jB=26°13.5', log sin 5=9.6452 or 153° 46.5'. It does not follow, however, that these values can be com- bined promiscuously ; for by Art. 138, since a is > 90°, with the vahie of b less than 90° must be taken the value of c greater than 90°, and the value of B less than 90° ; while with the value of b greater than 90° must be taken the value of c less than 90°, and the value of B greater than 90°. Thus the only solutions of the example are : 1. ^=21°3.9', c= 125° 33.3', 15 =26° 13.5'. 2. 6= i:)8°56.1', c = 54°2(;.7', i5= 153° 46.5'. Note. The figure shows geometrically why there are two solutions in tliis case. B For if AB and J. Care produced to A^, forming the lune ABA^C, the triangle A'BC has the side a and the angle A^ equal, respectively, to the side a and the angle A of the triangle ABC, and both triangles are right-angled at C. It is evident that the sides A^B and A^C and the angle A^BC are the supplements of the sides c and h and the angle ABC, respectively. 112 SPHERICAL TRIGONOMETRY. Solve the following spherical right triangles : 5. Given a = 159% c = 137° 20'. 6. Given J. = 50° 20', 5= 122° 40'. 7. Given a = 160°, b = 38° 30'. 8. Given 5= 80°, b= 67° 40'. 9. Given 5 = 112°, c= 81° 50'. 10. Given a =61°, B = 123° 40 . 11. Given a = 61° 40', 6 = 144° 10'. 12. Given xl= 99° 50', a = 112°. 13. Given b = 15°, c = 152° 20'. 14. Given ^= 62° 59', B= 37° 4'. 15. Given ^= 73° 7', c = 114°32'. 16. Given 5= 144° 54', 5 = 146° 32'. 17. Given 5= 68° 18', c= 47° 34'. 18. Given ^=161° 52', 6=131° 8'. 19. Given a = 113° 25', 6 = 110° 47'. 20. Given a = 137° 9', J5= 74° 51'. 21. Given ^=144° 54', 5 =101° 14'. 22. Given a = 69° 18', c= 84° 27'. SOLUTION OF QUADRANTAL TRIANGLES. 154. A spherical triangle is called quadrantal when it has one side equal to a quadrant. By Art. 136, (/), the polar triangle of a quadi-antal til- angle is a right triangle. Therefore to solve a quadrantal triangle we have only to solve its polar triangle, and take the supplements of the parts obtained by the calculation. 1. Given c=90°, a = 67° 38', & = 48°50'; find A, B, and C. SPHERICAL RIGHT TRIANGLES. 113 Denoting the polar triangle bv A'B'C'^ we have by Art. 136, (/) : C = 90°, A' = 112° 22\ B' = 131° 10' ; to find a', 5', and c'. By Art. 144, the formula for the solution are ~ , cos A' ~ ,, cos B' '^ , ~ i, ~ -r>, cosa' = ^ — — , cos 6' = ,, cosc'=cot^l'cot5'. sin B' sin A' + + Tlie check formula is cos o' = cos a' cos 6'. log cos xV = 0.5804 log cot ^' = 9.6143 log sin JB' = 9.8767 logcot 5' = 9.9417 log cos ((' = 9.7037 log cos c' = 9.5560 .-. 180° - «' = 59° 38.2'. .-. c' = 68° 54.8'. log cos ii' = 9.8184 Check. log sin.1' = 9.9660 log cos a' = 9.7037 log cos h' = 9.8524 l^g ^'^s ^' = j^f524 .-. 180° -6' = 44° 36.7'. log cos c' =9.5561 Then in the given quadrantal triangle, we have ^=180°-a'= 59° 38.2', B = \so°-b'= 44° 36.7', C=180°- c' = lll° 5.2'. EXAMPLES. Solve the following quadrantal triangles : 2. Given ^=139°, 6 = 143°. 3. Given ^= 45° 30', JB= 139° 20'. 4. Given a = 30° 20', C = 42° 40'. 5. Given 5= 70° 12', (7 = 106° 25'. 6. Given A = 105° 53', a = 104° 54'. 114 SPHERICAL TRIGONOMETRY, XII. SPHERICAL OBLIQUE TRIANGLES. GENERAL PROPERTIES OF SPHERICAL TRIANGLES. 155. In any sjjJiericcd triangle, the sines of the sides are proportional to the sines of their opposite angles* G Fig. 1. -D Let ABC be any spherical triangle, and draw the arc CD perpendicular to AB. There will be two cases according as CD falls upon AB (Fig. 1), or upon ^J5 produced (Fig. 2). In the right triangle ACD, in either figure, we have by Art. 144, sin CD sin A = sin 6 Also, in Fig. 1, . T5 sin CD sin B = ? sm a and in Fig. 2, sin B = sin (180°- CBD) = sin CBD (Art. 47) sin CD sin a Dividing these equations, we have in either case s'm CD sin A sin b sin a sin B sin CD sin b (84) gm a SPHERICAL OBLIQUE TKIAXGLES. 115 In like manner we have, sin B _ sin b sin C sin c (85) , sin C sin c /„_x and - = (86) sin A sin a Tiie above results may be expressed more compactl}' as follows : sin a _ sin b _ sin c sin A sin B sin C 156. /?i any spherical triangle, the cosine of either side is equal to the product of the cosines of the other two sides, plus the continued p>t'oduct of their sines and the cosine of their included angle. In the right triangle BCD, in Fig. 1 of the preceding article, we have by Art. Ill, cos a = cos BD cos CD = cos {c- AD) cos CD. And in Fig. 2, cos a = cos BD cos CD = cos (AD — c) cos CD. Whence, in either case, cos a = cos c cos AD cos CD -f sin c sin AD cos CD. But in the right triangle ACD, by Art. 144, cos AD cos CD = cos b. Also, cos b sin AD cos CD= sin AD -— = cos b tan AD cos AD = sin6-^ = sin 6 cos J. (Art. 141). tan 6 ^ ^ Whence, cos a = cos b cos c + sin b sin c cos A. (87) 116 SPHEPJCAL TRIGONOMETRY. Id like manner we have, cos 6 = cos c cos a + sin c sin a cos B, (88) and cos c = cos a cos h + sin a sin h cos (7. (89) 157. Let ABC and A'B'C be a pair of polar triangles. Applying the theorem of Art. 156 to the side a' of the triangle A'B'C ^ we obtain cos a' = cos b' cos c'-\- sin b' sin c' cos -4'. Putting for a', 6', c', and ^' their values as given in Art. 136, (/), we have cos(180°-^)=cos(180°-5)cos(180°-e) + sin (180°- B) sin (180°-^) cos (180°- a). Whence by Art. 47, — cos J. = (— cos 5) (— cos(7) + sin5sin(7(— cosa). That is, cos A = — cos B cos O + sin ^ sin C cos a. (90) In like manner, cos B = — cos C cos A + sin C sin A cos &, (91) and cos C= — cos >4 cos 5 + sin ^ sin B cos c. (92) The above proof illustrates a very important application of the theory of polar triangles in Spherical Trigonometry^ If any relation has been found between the elements of a triangle, an analogous relation may be at once derived from it, in which each side or angle is replaced by the opposite angle or side, with suitable modifications in the algebraic signs. SPHERICAL OBLIQUE TRIAXGLES. 117 158. To express the sines, cosines, and tangents of the half- angles of a spherical triangle in terms of the sides of the triangle. From (87), Art. 156, we obtain sin b sin c cos A = cos a — cos b cos c, J cos a — cos 6 cos c .^. or, C0S^ = : — ; — : (BJ sm sin c Subtracting both members from unity, ^ . 1 cos a — cos b cos c 1 — cos A = l sin 6 sin c _ cos b cos c + sin 6 sin c — cos a sin b sin c Whence b}' (A), Art. 75, ^ . ., , J cos (b — c) — Gos a 2 snr -kA = ^^ sin b sin c But by Art. 72, cos y — cos X = 2 sin 4- {x -j- y) sin 4- {x — y) . Whence, 2sini[a+(6-0]sini[a-(6-c)] Z sin 7>-xt = ; ^ ; ■ sni b sm c . , , . sin l{a-\-b — or, sm-^^ = ^ ~ , . sm b sm c Denoting a -|- 6 + c by 2 s, so that ,9 is the half -sum of the sides of the triangle, we have a + /> - c = (a + & + o) - 2c = 2s - 2c = 2 (.9 - c) , and a - & + c = (a + 5 + c) - 2 6 = 2 s — 2 6 = 2 (s — 6) . Whence, . _ . . sin (s — b) sin (s — c) sm^ \A— ^^ — . , . — ^ -^ sm sm c , ... Isin (s — 6) sin (s — c) ^ . and sm \A = d ^ — . / • (93) \ smo sm c ^ ^ 2 1 /I _ ®^" T (<^ + ^ — c) sin ^(a — b-\-c) 118 SPHERICAL TRIGONOMETRY. In like manner we have, sm^B= \ ^sia (s — c) sin {s — a) SI sin c sin a cU. 1 ri — > I'sin {s — a) sin {s — &) (94) and sin|-C=V ^ — ■ ^^T (95) -^ \ sua a sin Again, adding both members of (B) to unity, we have cos a — cos b cos c 1 + cos ^ = 1 + sin b sin c _ cos a — (cos 6 cos c — sin & sin c) sin 6 sin c „ 2 1 /< cos a — cos (5 + c) or, 2cos^t^= ^-, — r^ — — sm b sui c Whence, as in the proof of (93), „, , sin -l-(6 H-C + a) sin i- (5 + c — a) cos^ iA = — — . i .- — -• "^ sin b sin c But, 6 -f c + a = 2 s, and 6 + c — a = 2 (s — a) . Whence, „ , , sin s sin (s — a) cos^ I A = . , . -1 ^ sin sin c , , sin s sin (s — a) , , and cosi^=d ^^-A -• (96) "^ ^ sin 6 sin c ^ In like manner , _ sin s sin (s — b) , . cos 1-5= V' ^ ^ -' (97) "^ \ sin c sm a ^ , ^ sin s sm (s — c) , . and cos^C=J -. \ . • (98) \ sin a sm b Dividing (93) by (96) , we obtain Isin (8 — 6) sin(s — c) ( sin 6 sin c tan ^A = V ^ . I . sJ- r—. ^ "^ \ sm b sm c \ sin s sm [s—a) ^J sin (s— &) sin (s — c) , , -4 . , ^ ' • (99) sm s sin (s — a) SPHERICAL OBLIQUE TRIANGLES. 119 In like manner, ,^ Isin (.s — c) sin (s — a) tan i ^ = V ^^ ^/^S:^-^' (100) ^ \ sin s sm (8 — h) ^ ' , ^, Isin (s — a) sin (s — 6) and taniC=0 \ — /, , - (lOl) \ sins sin (s — c) ^ 159. To express the sines, cosines, and tangents of the half- sides of a spherical triangle in terms of the angles of the triangle. From (90), Art. 157, we obtain sin B sin C cos a = cos A -\- cos B cos (7, cos A + cos B cos C z^x or, cos a = ^ . (C) sm B sm C rr., 1 . cos yl + cos B cos C Then, 1 — cos a = 1 ' — ^ sin B sin C o . 21 — (cos ^ cos (7— sin ^ sin C) — cos ^ or, 2 sin^ | a = — ^ ^—~ ^—- ^ sm ^ sin C = cos(^4-<7) + cos J. sin B sin O Therefore by Art. 72, 2 sin^ ^ a = - 2cosi(^4-e+^)cos^(^+0-^) sin B sin C Denoting A -\- B + C by 2 6', so that S is the half-sum of the angles of the triangle, we have B+C — A=2(S — A). Whence, smH«= _ <:<^^S cosjS - A) ^ sin ^ sin C 1 -1 cos >S^ cos ( 6 — ^) /,««N and sin I a = i — ■ ; — ^r^"^ — ^. — " (102) \ sm B sin C In like manner, . ,7 I COS tS cos ( aS — B) /^««\ and sin i c = L COS'S' cos (5-:C)_ .^^^. \ sin ^4 sin B 120 SPHERICAL TRIGONOMETRY. Again, adding both members of (C) to unity, we have -, , -, , cos A + cos B cos O 1 + cos a = 1 H . ^ . ^ sin B sm C _ cos A + cos B cos C + sin B sin C sin B sin G Then, 2cos^ia = 55^ii+^^5^%l^ sin ^ sin C _ 2cos|[^+^-0]cosi[^-(^-a)] "" sin B sin (7 _ cosK^ + ^-C) cosl-M-^+C) or, cos^ A- a .= ^-^ . -1 . --^ ! ^• ' "^ smB sm O But ^4-5-C=2(^-0),and^-S + (7=2(5'-5). Whence ' .os2 1 a - cos (.S--^) cos (^-C) tub 2 t* — : — . — — ? sm B sm C and eosia = J ""^<^-:^^'^"^(f-^) - (105) \ sin B sin C In like manner. cos 1 5 ^ J cQs (^ - <^) cos (>S - ^) ^ ^^Qg^ \ sin C sin ^ 1^^ / cos(^-^) COS(/S-^) \ sin A sin Dividing (102) by (105), we obtain and cosic = .J """^"~"":^"""^" ""^ - (107) \ sin A sin B tania= J cos S cos (S - A) ^3^ ^ SI cos(/^-j5)cos(/S-0) ^ ^ • In like manner. and tan ^ 5 = k - cos ^ cos (S — B) coBiS-C)cos{S-Ay (109) tan 1 c = yl - cos iS COS {S — C) (110) oas(S — /l^ ons(S— B) SPHERICAL OBLIQUE TRIANGLES. 121 NAPIER'S ANALOGIES. 160. Dividing (99), Art. 158, hy (100), we have tanjj4_ /sin (s — 6) sin(s — c) I sin 6* sin (s — 5) tan ^ B \ sin § gin (s — a) ^ sin (s — c) sin (s — a) sm^AQOs^B_ I sin" (.9 — h) _ sin(s — &) cos ^ ^ sin 4^ i^ V sin^ (s — a) sin (s — o) Whence, by composition and division, sin 7}^ cos j-jB-f-cos j-xlsin^-B sin (g — 5) + sin (s — a) sin^^cos^J3 — cos-^ J. sin|-^~~ sin (s — b) — sin (s — a) That is, by Arts. Go, 6G, and 73, sin (iA-{-iB) _ tan j-[.s - ?> + (g - a)] sin (^ J. — I--B) ~ tan ^[s — 6 — (s -^ a)] But s— ?/ -h s— rt = 2 s — (I — b = c. sin4^(^ + J5) tan^c Hence, . T , . J, = 1 — ,, ^ ,, • (ill) sin-^(^ — -B) tanj(a — 6) ^ 161. Multiplying (99) by (lOO), we have , ^ ,^ Isin (s — 6) sin (s— c) Isin (.s— c)sin(6' — a) tanl^tanii3=: V — -^^ / ' ^ , \ !^ . , ^ ,, S "^ '^ \ snis sin(.s — a) \ snissm(s— 0) sin ^^ sin ^^ |sin-(s — c) sin (s — c) cos^^cos^-B V sin-s ~ sins Whence, by composition and division, cos ^yl cos ^5 — sin ^^4 sin ^B sin s — sin {s — c) cos^yl (tos^B -\- sin-^^^l sin-^-^ ~ sin s -j- sin (s — c)' or, by Art. 73, ^o^i^^ + ^B) _ tani[s - (.9 - c)] ^ -" cos (i^ -.1-5) tani[s + (s-c)] But s + s — c=2s — c = a + 6. TT cos^(^4-f-jB) tan^c Hence, yt~a ^ = z — tt-^t-tt- (112) QO^^{A — B) tan J(aH- 6) ^ 122 SPHERICAL TRIGONOMETRY. 162. Proceeding as in Art. 157, and applying (ill) to the triangle A'B'C, we obtain, sin i(^' - B') ~ tan i(a'- b') ' But, i(^'+ B') = i-(180°- a + 180°- 6) = 180°-i(a + &), i{A'- B') = i(180°- a - 180°+ 6) = |( -a + 6), ic'=i(180°-C) = 90°-iO, and i(a'-b')=:i(^80°-A-180°-{-B)=i{~'A-\-B). Whence, sin[180°-i(a + 5)] _ tan(90°-ia) sin-^( — a + 6) "tan^( — ^ + 5)* Therefore, by Arts. 42, 46, and 47, sin^(a-|-6) cot ^(7 — sin ^{a — b) —tsin^{A — B) sin-i-(a + 5) _ cotja ^^' sini(a-6)~tani(^-jB)' ^ ^ In like manner, from (112) , we obtain cosj-(^'+^') _ tan|c^ cosi(^'-5')~tan.i(a'+6')* But, -|(a'+ b') = 180°- i(v4 + B). Whence, cos [180°- i(a + &)] _ tan (90°- j C) cos i(- a + &) "~ tan [180°- i(^ + ^)]' Therefore, by Arts. 42, 46, and 47, — cos|-(a + 6)_ eot^C cosi(a — 5) ~ — tan ^{A + -B) ' cosi-(a + 6) _ cot JO ^^' cosi(a-6)~tani(^ + ^)' ^ ^ SPHERICAL OBLIQUE TRIANGLES. 123 163. The formulae exemplified in Arts. 160, 161, and 162, are known as Napiefs Analogies. In each case there may be other forms, according as other elements are used. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 164. In the solution of spherical obliqne triangles, we may distinguish six cases : 1 . Given a side and tivo adjacent angles. 2. Given two sides and their included angle. 3. Given the three sides. 4. Given the three angles. 5. Given two sides and the angle opposite to one of them. G. Given two angles and the side opposite to one of them. 165. By application of the principles of Art. 136, (/), the solution of any example under Cases 2, 4, and 6 may be made to depend upon the solution of .'mother example under Cases 1, 3, and 5, respectively ; and vice versa. Thus it is not essential to consider more than three cases in the solution of spherical oblique triangles. 166. The student must carefully bear in mind the remarks made in Arts. 151 and 152. Case I. 167. Given a side and two adjacent angles. 1. Given ^=70°, 5 = 131° 20', c'=116°; find «, h, and C. By Napier's Analogies (Arts. 160, 161), we have sini(5 + -'l) _ tan|c sin-2-(5 — ^) tan 1^(6 — a)' and eosi(^ + ^) ^ tan^c cos ^{B — A) tan |(6 + a) * 124 SPHERICAL TRIGONOMETRY. Whence, tan i(5 — a) = sin ^-{B — A) esc ^(B-\-A) tan |-c, + _ + and tsin^{b -\-a) = cos^{B — A) sec^{B-{- A) tsiuic. From the data, 1(5-^) = 30° 40', i(^ + ^)=100°40', ic = 58°. log sin i{B -A)= 9.7076 log cos ^{B- A) = 9.9316 log CSC ^{B-\- A) = 0.0076 log sec i{B + A) = 0.7326 log tan|c = 0.2042 log tan|-c = 0.2042 log tan i{b-a) =9.9194 log tan i(6 + a) =0.8714 .-. i{b -a)= 39° 42.8'. 180° - ^(b + a) = 82° 20.5' .-. i(b-j-a) =97° 39.5'. Then, a= i(b-j- a) - ^(b- a) = 57° 56.7', and b = ^(5+ a) + ^{b- a) = 137° 22.3'. To find C, we have by Art. 162, , _, sin 4-(6 +«) . .^ = sin |^(& + a) CSC -^-(^ — a) tan •2-(-S— ^). log sin i{b + a) = 9.9961 log GSGi(b - a) = 0.1946 log tan i{B -A)= 9.7730 log cot 10=9.9637 .-. i(7= 47° 23.6', and 0= 94° 47.2'. Note 1. The value of C may also be determined by the formula Note 2. The triangle is always possible for any values of the given elements. SPHERICAL OBLIQUE TRIANGLES. 125 EXAM PLES. Solve the followino; trianojles : 2. Given ^= 78°, B= 41% c=108°. 3. Given 5 = 135°, 0= 50°, a = 70° 20'. 4. Given ^= 31° 40', C'=122°20', h= 40° 40'. 5. Given J. = 108° 12', i^=145°4G', c=12G°32'. Case II. 168. Given tico sides and their included angle. 1. Given b = 137° 20', c = 110°, .4 = 70° ; find B, C, and a. By Napier's Analogies (Art. 1G2), sin ^{b -\-c) _ cot 4^ A sin ^(6-0 ~tan^(7i-C')' T cos4-(^ + c) cot I J and '^ — ■ — - = — ^ cos H^ - c) tan i v-^ + ^') Whence, tani(7^- C)= s\n ^{b-c) csci(&+c) cotj^, - + - + and tan^(Z?+ C) = cos^(/> — c) seci(b -f- c) cot^^. From the data, ^{b-c) = 10° 40', |(/j + c) = 12G°40', 4-^4 = 35°. From which we find, ^(B-C)= 18° 14.5', i (5 + C) = 113° 2.9'. Then, B = ^ (B -{- C) -{-^ {B - C)= 131° 17.4', and C=^{B-\-C)-^(B-C)= 94° 48.4'. To find rt, we have by Art. IGO, tan i a = 2H±i|+^ tan i (i - 0) . sm ^ (is — C ) From which we obtain a = 57° 56.6'. 126 SPHERICAL TRIGONOMETRY. Note. The triangle is always possible for any values of the given elements. EXAMPLES. Solve the following triangles : 2. Given a = 72°, 6= 47°, C= 33°. 3. Given a = 98°, c= 60°, J5 = 110°. 4. Given 5 = 120° 20', c= 70° 40', A= 50°. 5. Given a =125° 10', 6 = 153° 50', C = 140° 20'. Case III. 169. Given the three sides. 1. Givena = 60°, 5=137°20', c = 116°; find ^, £', and 0. By Art. 158, tani^= j Bin(.-5)sin(. ;^^ ^ sin s sm (s — a) 1 •„ . sin (s — c) sin (s— a) tan i 5 = V^ ^; i— — ^— - — ^5 ^ sin s sm (s— o) tan * C= >i" (« - «) ^i" ( ZEg. ^ sin s sin (s — c) From the data, 2s = ft + 6 + c = 313°20', or s=156°40'. AYhence, s-a= 96° 40', s - 5 = 19° 20', s - c = 40° 40'. log sin (s- 6) =9.5199 log sin (s-c)= 9.8140 log CSC s = 0.4022 log CSC (s- a) =0.0029 2 )9.7390 logtan 1^=9.8695 0-. 1^ = 36° 31.2', and ^=73° 2.4'. and SPHERICAL OBLIQUE TRIANGLES. 127 In like manner we find, 5 =131° 32.2', and (7= 96° 55.4'. The values of A, B, and C may also be obtained by aid of the sine or cosine formnlae of Art. 158. If all the angles are to be computed, the tangent formulae are the most convenient, as only four different logarithms are required. If but one angle is required, the cosine formula will be found to involve the least work. Note. The triangle is always possible for any values of the given elements which satisfy the conditions of Art. 138, (a) and (c) ; that is, if a + 6 + c is < 360°, and no side is greater than the sum of the other two. EXAMPLES. Solve the following triangles : 2. Given a = 38°, 6 = 51°, c = 42°. 3. Given a = 101°, b = 4 = 3'.)°, c = 92°. 5. Given a = 62° 20', ?/=54°10', c = 97°50'. Case W. 170. Given the three angles. 1. Given ^ = 70°, 5= 131° 10', 0=94° 50'; find a, 6, and c. By Art. 159, tania=v/- ^os .S cos ( ^ - ^) ^ cos {S - B) cos {S - C) f.nn|7>-J C0S>SC0S(^-^) ^ COS {S - C) cos {S - A) and tan 1 c = J cos ^- cosT^- C ) ^ cos {S - A) cos {S - B) 128 SPHERICAL TRIGONOMETRY. From the data, S-A=78°, S-B=16° 50', S-C=53°10'. Note 1. Since cos 5 is— (Art. 37), while the cosines of S — A, S — B, and S— C are +, the quantities under the radical signs are essentially positive, and hence no attention need be paid to the nega- tive signs in the formula9. log COS ^S'^ 9.9284 log cos {S- A) = 9.3179 log sec (/S - 5) = 0.0190 log sec (^- C)= 0.2222 2 )9.4875 logtania= 9.7437 .•.|a= 28° 59.7', and a = 57° 59.4'. In like manner we find, 6 =137° 11.8', and c= 115° 55.8'. The values of a, 5, and c may also be obtained by aid of the sine or cosine formulae of Art. 159. If all the sides are to be computed, the tangent formulae are the most convenient, as only four different logarithms are required. If but one angle is required, the sine formulae will be found to involve the least work. Note 2. The triangle is always possible for any values of the given elements, provided S is between 90° and 270°, and each of the quantities S-A, S-B, sind S - C between 90° and -90° (Art. 37). EXAMPLES. Solve the following triangles : 2. Given .4= 75°, B= 82°, 0=61°. 3. Given ^=120°, 5=130°, 0=80°. 4. Given ^= 91° 10', B= 85° 40', 0=72° 30'. 5. Given ^=138° 16', B= 31° 11', 0=35° 53'. SPHERICAL OBLIQUE TRIANGLES. 129 Case Y. 171. Given two sides and the angle opposite to one of them. 1. Given a = 58°, 6 = 137° 20', 5 = 131' 20' ; find A, (7, and c. o i *. 1 - - sin ^1 sin a sin B sm h or, sin ^ = sin a CSC 6 sin 5. log sin rt 3=9.9284 log CSC h = 0.1689 logsini? = 9.8756 log sin^l = 9.9729 .-. .1 = 69° 58', or 110° 2' (Art. 152). To find C and c, we have by Arts. 100 and 102, cotiO=^!lii;^ta„K£-^), and tan |- c = - — r ;-^ -. ' tan 1 ( 6 - a ) , sin ^(5 — ^) ^^ ^ Using the first value of .4, we have i(i5 + ^l)=100°39', ^(i3-.4)=30°41'. Also, ^{h+a)= 97° 40', ^ (6 - «)= 39°40'. From which we obtain C=94°41.6', and c= 115° 53. 6'. Using the second value of ^4, we have i(i3 + .4)=120°41', ^(5 -.4) = 10° 39'. From w4nch we obtain (7= 147° 26.4', and c = 150° 56.8'. Thus the two solutions are : 1. A= 69° 58', C= 94° 41.6', c= 115° 53.6'. 2. ^=110° 2', (7=147°26.4', c= 150° 56. 8'. 130 SPHERICAL TRIGONOMETRY. As in the corresponding case in the solution of plane oblique triangles (compare Arts. 124 to 126), there may sometimes be two solutions, sometimes only one, and some- times none, in an example under Case V. After the two values of A have been obtained, the number of solutions may be readily determined by inspection ; for by Art. 136, (5), if ais b, A must be > B. That is, only those values of A can be retained which are greater or less than B according as a is greater or less than b. Thus, in Ex. 1, a is given < b ; and since both values of A, 69° 58' and 110° 2', are < B, we have two solutions. Also if the data are such as to make log sin A positive, there will be no solution corresponding. 2. Given a = 58°, c = 116°, 0=94° 50'; find ^, 5, and 6. sin A sin a In this case, sin C sin c or, sin ^ = sin a esc c sin C. log sin a =9.9284 log CSC c =0.0463 lo2 sin = 9.9985 log sin^= 9.9732 ... ^==70° 5', or 109° 55'. Since a is given < c, only values of A which are < O can be retained ; hence there is but one solution, corresponding to the acute value of A. To find B and 6, we have by Arts, 160 and 162, ,^ sin-Kc + a)- , ,^ .. cot J5 = ■ I) 4 tan 1(0 - A) , ,, sin 1(0 + ^), ,, and tan 2^ = ^^^^.^ _ ^. t^rii(G- a). SPHERICAL OBLIQUE TPJAXGLES. ISl Using the first value of A, we have i(C + .4)==82°27.o', ^{C-A)= 12° 22.5'; also, ^(c-\-a) = 87% i{c-a) = 2r. From which we obtain i5= 131° 21.8', and b= 137° 23.6'. Thus the only solution is ^=70° 5', ^=131° 21.8', 6 = 137° 23.6'. 3. Given b = 126°, c = 70°, i3 = 56° ; find C. T .1 . sin C sin c In this case, — = -> sin B sin b or, sin C = sin c esc 6 sin 5. log sin c = 0.0730 log CSC b =0.0020 log sin ii= 0.0186 log sin (7=0.0836 .-. 0=74° 20', or 105° 40'. Since both values of C are>J5, while c is given, there is no solution. EXAMPLES. Solve the following triano-les : 4. Given 6 = 99° 40', c = 64° 20', 5 =95° 40'. 5. Given a = 40°, & = 118°20', ^ = 20° 40'. 6. Given a =115° 20', c=146°20', O=141°10'. 7. Given a = 109° 20', c = 82°, ^=107° 40'. 8. Given 6 = 108° 30', c = 40°50', O=39°50'. 9. Given a = 162° 20', 6 =15° 40', 5=125°. 10. Given a = 55°, c = 138° 10', A = 42° 30'. 132 SPHERICAL TRIGONOMETKY. Case VI. 172. Given two angles and the side opposite to one of them. 1. Given J. = 110°, jB=131°20', 5 = 137° 20'; find a, c, and C. -r , , . sin a sin A In this case, = •> sin b sin B or, sin a = sin A esc B sin b. log sin ^ = 9.9730 log csc5= 0.1244 log sin b =9.8311 log sin a = 9.9285 .-. a = 58° 1.2', or 121° 58.8'. To find c and (7, we have by Arts. 160 and 162, ^ , sm^(B-i-A) ^ ,.^ . tan -i-c = -. — ri^^ aT tan f(b — a) , ^.^ sin +(5+ a), ,,^ .. and cot i C = . ,,. : tan ^(B-A). sinf(o — a) ^ ^ Using the first value of a, we have c= 150° 53.8', and (7= 147° 23'; and using the second value of a, c=64°7.8', and = 85° 17.6'. Thus the two solutions are : 1. a=58°1.2', c= 150° 53.8', (7= 147° 23'. 2. a =121° 58.8', c=64°7.8', (7= 85° 17.6'. In Case VI., as well as in Case V., there are sometimes two solutions, sometimes only one, and sometimes none ; and it may be shown, exactly as in Art. 171, that only those values of a can be retained tohich are greater or less than b according as A is greater or less than B. SPHERICAL OBLIQUE TRIAXGLES. 133 Also if log sin a is positive, the triangle is impossible. EXAMPLES. Solve the followinsj triangles : 2. Given ^ = 11P/, (7=80% c = 84°. 3. Given yl = 132°, ^=140°, 6 = 127°. 4. Given .4 = 62°, (7=102°, a = 64° 30'. 5. Given ^1=133° 50', 73=66° 30', a = 81° 10'. 6. Given i5 = 22° 20', 0=146° 40', c = 138° 20'. 7. Given .4 =61° 40', (7= 140° 20', c = 150°20'. 8. Given 15 = 73°, (7= 81° 20', 6 = 122° 40'. APPLICATIONS. 173. In questions concerning geodesy or navigation, the earth may be regarded as a sphere. Tiie shorteat imth between any two points is the arc of a great circle which joins them, and the angles between this arc and the meridians of the points determine the bearings of the points from each other. Thus, if Q and Q' are the points, and PQ and PQ' their meridians, the angle PQQ' determines the bearing of Q' from Q, and the angle PQ'Q determines the bearing of Q from Q'. 134 SPHERICAL TRIGONOMETRY. Jf the latitudes and loDgitudes of Q and Q' are known, the arc QQ' and the angles PQQ' and PQ'Q may be determined by the sohition of a spherical triangle. For if EE' is the equator, and PG the meridian of Green- wich, we have angle QPQ'= angle Q'PG- angle QPG = longitude of Q'— longitude of Q. Also, PQ = PE-QE = 90°- latitude of Q, and PQ'= PE'-^ Q'E'= 90° + latitude of Q'. Thus, in the triangle PQQ', two sides and their included angle are known, and the remaining elements ma}' be com- puted. Note. When QQ' has been found in angular measure, its length in miles may be calculated by the method of Art. 132. In the following problems the diameter of the earth is taken as 7912 miles. 1. Boston lies in lat. 42° 21' N., longitude 71° 4' W. ; and the latitude of Greenwich is 51° 29' N. Find the shortest distance in miles between the places, and the bearing of each place from the other. 2. Calcutta lies in lat. 22° 33' N., Ion. 88° 19' E. ; and Valparaiso lies in lat. 33° 2' S., Ion. 71° 42' W. Find the shortest distance in miles between the places, and the bear- ing of each place from the other. 3. Sandy Hook lies in lat. 40° 28' N., Ion. 74° 1' W. ; and Queenstown lies in lat. 51° 50' N., Ion. 8° 19' W. In what latitude does a great circle course from Sandy Hook to Queenstown cross the meridian of 50° W. ? If the latitude of a place is known, and the altitude and declination of the sun, the solution of a spherical triangle serves to determine the hour of the day at the time and place of observation. SPHERICAL OBLIQUE TRIANGLES. 135 Thus let be the position of the observer ; P the celestial north pole ; EE' the celestial equator ; ////' the horizon ; Zthe zenith ; S the sun's position ; PSM a meridian passing through the sun's position ; and ZSN a great circle passing throngh Z and *S'. Then SM is tlie sun's declination, SJSf its altitude, and EZ the latitude of the plac^e of observation. Then in the spherical triangle SPZ, we have SP= PM- SM=\nf- the sun's declination, SZ= ZN- 6'.Y=- 1)0"- the sun's altitude, and PZ= EP - EZ = 'J0°- the latitude of the place. That is, the three sides of the triangle SPZ are known, and the angle SPZ may be computed. If 24 hours is multiplied by the ratio of this angle to 360°, we have the time required for the sun to move from S to the meridian EP. Hence, if this time is subtracted from 12 o'clock, if the observation is made in the morning, or added, if made in the afternoon, we obtain the hour of the da}' at the time and place of observation. If the Greenwich time of the observation is noted on a chronometer, the difference between this and the local time as calculated above serves to determine the longitude of the place of observation. 136 SPHERICAL TRIGON^OMETRY. In reducing time to longitude, it should be borne in mind that 24 hours of time correspond to 360° of longitude; that is, one hour of time corresponds to 15° of longitude, one minute to 15', and one second to 15''. 4. A mariner observes the altitude of the sun to be 14° 18', its decUnation beiug 18° 36' N. If the latitude of the vessel is 50° 13' N., and the observation is made in the morn- ing, find the hour of the day. If the observation is taken at 9 A.M., Greenwich time, what is the longitude of the vessel? 5. What will be the altitude of the sun at 4 p.m. in San Francisco, lat. 37° 48' N., its declination being 12° S.? 6. In Melbourne, lat. 37° 49' S., the altitude of the sun is observed to be 25° 46'. If the sun's declination is 3° S., and the observation is made in the morning, find the hour of the day. 7. At what hour will the sun rise in Boston, lat. 42° 21' N., when its declination is 15° N. ? Note. At sunrise the sun's altitude is 0, so that the arc SZ be- comes 90°. FORMULA. 137 FORMULA. PLANE TRIGONOMETRY. Art. 10. sm^l = --) c tan A = -1 b sec^ = -1 b A ^ cos^ = -, c cot A = -1 a A ^ CSC A = -' a sin jB = - , c tanB = -? a secB = -1 a cos B = -■) c cot B =-^ b CSC 5=-. b Art. 18. sin^ 1 . = 7? tan . 1 A = •> sec A- 1 , (1) (2) CSC A cot^ COS ^1 _1 sin^ cos^ = ? cot^l = , csc^l = sec A tan A Art. 19. sin^ A + cos' .1=1. Art. 20. , J sin A ,^^ . . cos^ tan A = 7- (5) cot A sin^l csc-^= 1 +cot-^. cos A Art. 21. sec2^=lH-tanM. (7) Art. 42. sin (— ^) = — sin^, cos(— .4)= cos^, tan (—.4)= —tan J., cot (—^) = —cot J., sec( — ^)= sec^, csc(— ^)= — csc^. Art. 44. sin (90° + ^)= cosu4, cos (90°+^)= -sin^, "] tan (90°+ 4)= -cot ^, cot (90°+^) = -tan ^, f sec (90°+^)= -CSC ^, csc(90°+^)= secX J (3) W (6) (8) (9) (10) 138 FORMULA. Art. 65. sin (x-\-y) = sin x cos y + cos x sin y. (ll) cos {x -\-y) = cos X cos y — sin x sin 2/. (12) Art. 66. sin (a; — 2/) = sin ic cos 2/ — cos a; sin 2/. (13) cos (ic — 2/) = cos X cos 2/ + sin x sin 2/. (14) tan a; + tan V /__x Art. 71. tan(^ + y)=^-^^^^. (15) tan X — tan 2/ /, g\ ^ '^^~ 1 -}-t&.nxtsiny cot a; cot 2/ — 1 /^ n\ ^ cot 2/ + cot X ^, - cota;cot2/ + l /,«>. cot (.T — 2/) = — 7 ; — CJ-8) ^ cot 2/ — cot X Art. 72. sin a; + sin 2/= 2sin ^(a; + 2/) cos|-(cc— 2/). (19) sin x — siny=z 2 cos ^{x + v/) sin v;(x—y) . (20) cosaj + cos2/= 2cos|-(a;H-2y) cos-^(a;— 2/). (21) cos a; — cos 2/= —2 sin i(aj + y) sin i(^— 2/)- (22) Art. 73. sin x + sin 2/ tan ^{x-\-y) ^ .^^. sin ic — sin y ~~ tan ^ {x — 2/) Art. 74. sin 2. ^• = 2 sin a? cos a;. (24) cos 2 a; = cos^ a? — sin^ x. (25) cos 2 a.' = 1 -2sin2a;. (26) cos 2 a; = 2 cos^ a; — 1. (27) . J. 2 tan a; /-o\ tan 2 a; = — (28) 1 — tan-" X C0t2a;=: ^^^'^~^ - (29) 2 cot a; FORMULA. 139 Art. 75. sinia;=^ (30) cos|-a; = %|--^— (31) tsini-x = \l (32) tan*a;= (33) taDia; = (34) 1 + cos X sin X . , 1 4- cos a; sin x ^__x cot^a; = — = (35) sin X 1 — cos X Art. 111. 4/r=c-sin2^. (36) 4/r=c-sin25. (37) 2K=d-Q0iA. (38) 2 K=b- cot B. (39) 2/f=a-tanJ5. (40) 2A^=?;-tan^. . (41) 2K=a^(c + a){c-a). (42) 2 A^= ?> V(c-f /j) (c-6). (43) 2K=ab. (44) Art. 113. ^^ = ± (45) ?i^=?'. (46) 5iB^=l (47) slnZ^ 6 ^ ^ sinC c ^ ^ sin^ a ^ ^ Art. 114. cj±b^ tani(A + B) ^ ^^3^ a-b tain {A -B) 6 4-c ^ tau^(J3+0) 6-c~tani(J5-(7)' c + ^-\--"> . (65) \ s{s — h) tan iC=JBi£Hl. (66) ^ \ s (s — c) 2^=5csin^. (67) 2^= -, — (70) 2K=casmB. (68) „ h^ sinC smA ,__n ZK = : — vvi; 2^=a6sina. (69) ^ • a - -n ^j^^c'smAsmB^ (72) sinO K=>Js{s-a){s-h){s-c). (73) FORMULA. 141 SPHERICAL TRIGONOMETRY. Art. 139. cos c = COS a cos b. (74) 4 sin a .„^s . ^ sin 6 ^ ^ smA = - (75) sin^ = ^ (77) sine smc ^ ^ A tan 6 /„^x ^ tan ft , . COS^ = ■' (76) C0S5 = 7 -• (78) tanc tanc ^ Art. 140. tan^ = ^-^^. (79) tan5 = *^. (80) sin b sin a Art. 141. J cos B /„, X . -D cos^ /__K sin A = (81) sm B = (82) cos 6 cos a Art. 142. cos c = cot A cot B. (83) Art. 155. §^ = ^- (84) (85) ... (86) Sin A sin a Art. 156. cos ct = COS b cos c + sin b sin c cos.^^ (87) cos b = cos c cos a + sin c sintt cos 5. (88) cos c = cos a cos b + sin a sin b cos C. (89) Art. 157. cos A= — cos B cos (7 + sin 5 sin O cos a. (90) cos B= — cos C cos ^ + sin C sin ^ cos b. (91) cos (7= —cos A cos 5 + sin A sin 5 cos c. (92) sin^ sin 6 sin B sin 6 sin C sine sin C sine 142 FORMULA. . , , /sin(s — 6) sin(s — c) ^.-n Art. 158. sini^=v— ^ — . ' • (^3) ^ \ sin sm c ^. lsm(«-c)sin(.-g_ (3^) ^ \ sm c sm (X sin * O = /'"(«-'^)^'"(«- J). (95) \ sm a sm o cosi^= >'°^.t^^~'^^ - (96) "= \ sm sm c "^ \ sm c sm a ,^ lsmssin(s — c) /qqx cos 1^(7=. : \ , ' - (98) \ sm a sm o tan i ^ = , Ni" («-?;) sin (a -^ . (99) \ sm s sm (s — a) tan J B = , p(>-c) sin (.- _«). ^^^^^ \ sm s sm (s — o) tan 1 = , | sin(«-")^i"(«^ . (101) \ sms sm (s—c) A-i. -icn -1 I cos /S cos (/S' — ^) /inoN Art. 159. sm4-a=. -, — — ^ — - — ^- (102) \ sm B sm (J . 1, I cos >i^ cos (/S — 5) /ino\ \ sm C sm A . 1 I COSTS' cos (S— C) /TnA\ ^^■'*^=V sin ^ sin .B ' ^'°*^ eos i a = l ooHS-B)oos(S^ Vl_ ^ (,05^ \ sm B sm G 1, IcOS (O — U ■) COS ^O — ^) /TrtCN COS i 6 = i ^ ■ ri - A (^°^) \ sm C sm ^ ' cos(^-C)cos(^^-^) FORMULA. 143 "^ \ smA sin B tan^a-y^ cos (6' -5) cos (.S - O) ^ ^ tan i 6 = J ^^^(^Z:^. (109) ^ \ cos(^— C) cos(aS-^) tan i c = J- ^^f^^l^S^ CJ^ (,,0) "^ \ COS {S — A) cos (>S^ — B) Art. 160. sin ^ (^4 + ^) _ tan j^ c (111) sm^(^-^) tan ^ (a -6)* Art. 161. cos^(.l-ii) tan ^ (a + 6) Art. 162. sin i- (ff + ft) ^ cot i (7 sin i (a -ft) tan^(^-J5)' cos 4^ (g H- ft) _ cot ^ (7 cos i(a-b) tan i (^ + 5) (113) (114) AI^SWEES Art. 9; page 3. 9. 28° 38' 52.4". 14. 114° 35' 29.6". 13. 42° 58' 18.6". 15. 100° 54' 5.1". 16. 30° 40' 33.8". Art. 91; page 57. 2. .7781. 7. 1.3222. 12. 1.9912. 17. 2.1303. 3. 1.1461. 8. 1.7993. 13. 2.0212. 18. 2.2252. 4. .9030. 9. 1.7481. 14. 2.0491. 19. 2.1673. 5. 1.0791. 10. 1.9242. 15. 2.1582. 20. 2. 5741. 6. 1.1761. 11. 1.6532. 16. 2.3343. 21. 2.5353. Art. 93; page 58. 2. .3680. 5. 1.5441. 8. .2252. 11. .8539. 3. .1549. 6. .1182. 9. 2.2431. 12. .7660. 4. .5229. 7. 2.0970. 10. 1.0458. 13. .7360. Art. 96; page 59. 3. .2863. 9. 4.5844. 15. .1165. 22. .2601. 4. 2.7090. 10. 3.2620. 16. .3860. 23. .6884. 5. 4.2255. 11. .9801. 17. .2212. 24. .1840. 6. .1398. 12. .4225. 18. .1750. 25. .2215. 7. .7194. 13. .1590. 20. 2.6145. 26. .2494. 8. .6611. 14. .0430. 21. .1678. 27. .1449. ANSWERS. 14£ Art. 98; page 60. 2. .2552. 7. 7.7323-10. 12. 2.4804. 3. .3522. 8. 6.4983-10. 13. 8.7905-10 4. 9.2922-10. 9. 3.8663. 14. 6.3588. 5. 8.6811-10. 10. .6074. 15. .1964. 6. 1.5841. 11. 9.6511-10. 16. .1688. Art. 105: ; pages 65 and 66. 1. 8.454. 19. -1.184. 40. .5010. 2. 10.73. 20. .000007038. 41. 1.062. 3. — 2202. 21. 2.924. 42. -.9102. 4. .2179. 22. .9146. 43. 1.093. 5. .01157. 23. 4.638. 44. .7035.' 6. -.7032. 24. .0000639. 45. .5807. 7. 7.G72. 25. 1.414. 46. -.6313. 8. .6688. 26. 1.495. 47. 24.62. 9. -3.908. 27. -1.246. 48. .2979. 10. 1782. 28. .6553. 49. 98.50. 11. .3500. 29. .2846. 50. 1.660. 12. - .4748. 30. 2.372. 51. 3.076. 13. .4127. 31. -.5142. 52. .8678. 14. -4.671. 32. .1588. 53. 1.134. 15. .2415. 35. 5.883. 54. .5881. 16. -.0725. 36. .7885. 55. 1.805. 17. 13587. 37. 1.195. 56. .003229. 18. .006415. 38. 39. .6803. .6443. 57. .03344. Art. 110; pages 69 to 72. 1. a = 7.708, 5 = 8.124. 3. a = 24.37, c = 53.56. 2. 6 = 1883, c = 2019.5. 4. ^ = 43° 17.9', 6 = .6622. 146 ANSWERS. 5. ^=68° 12.2', c = 5.385. 6. b =26.91, c = 87.64. 7. a =.02309, 6 = .01452. 31. c = 25.206. 8. ^ = 55° 36.1', a = 4.216. 32. a =1.735. 9. a =5571, c=7007. 33. c = 2725.6. 10. A = 4:1° 2 A', c = 153.8. 34. ^ = 47° 42.9'. 11. b =167.6, c = 230.5. 35. a =.8346. 12. a =30.51, 5 = 18.59. 36. a =49.25. 13. ^ = 47° 52.5', 6 = 184.7. 37. 14.106 in. 14. a =2.847, c = 3.287. 38. 78.12ft. 15. a =5.4125, c=14.306. 39. 31° 47.1'. 16. a =.7133, 5 =.1367. 40. 36° 37.9'. 17. ^=51° 51.9', c = 811.7. 41. 99.45 miles. 18.6=42.27, c=208.15. 42. 11.371 in. "^ 19. ^ = 58° 35.7', a = .0409. 43. 19° 50'. 20. 6=76.13, c=1877. 44. 399.5ft. 21. ^ = 36° 45.9', c = 41.22. 45. First, 25.22 miles; 22. A = 65° 30', a = 3.153. second, 30.07 miles. 23. a =36992, 6 = 4021. 46. 21.65 ft. 24. a =410.5, c = 456.7. 47. 14.487 in. ; 15.682 in. 25. ^ = 55° 44.1', c = 411.5. 48. 453.7 ft. 26. ^ = 58° 40', 6=1.0405. 49. 17.26 in. 27. 6 =.3245, c=.8828. 50. 389.4ft. 28. a =264.9, 6=75.95. 51. 437.6. 29. a =176.64, c = 213.65. 52. 5.773 in. 30. ^ = 30° 17.2', c = 20.04. 53. 481.9 ft. 54. Eate, 6.792 miles an hour; bearing, 63° 8.4' west of north. Art. 112 ; page 74. 2. 69.03. 5. .08938. 8. 1.223. 3. .151. 6. 8208. 9. 107.2. 4. 5695. 7. .002245. 10. .1017. ANSWERS. 147 Art. 121 ; page 84. 2. 5=15.837, c = 14.703. 7. a = .011162, 5=.006962. 3. a =.445, c=.9942. 4. a=. 01913, ?;=. 02272. 5. rt= 33.78, c = 18.54. 6. 6 = 8.24, c = 5.464. 8. 6 = 447.4, c = 425.7. 9. a = 342.6, c = 303.3. 10. a =11.067, 6=6.067. Art. 122; page 85. 2. ^ = 100^56.7', 6 = 19.78. 3. A= 83° 14.7', c = 383.5. 4. B= 39° 15.8', a = 3. 211. 5. A= 24°41.8', c=.5886. 6. J5=143°29.1', a = 886.4. 7. ^=35° 1.9', 6=12.78. 8. ^=59° 16.3', c=60.74. 9. 7^ = 21° 1.2', a=.06699. 10. .4=40°52' 19", 6=77.14. Art. 123; page 88. 3. yl=28°57.4', B= 46° 34.2', C= 104° 28.4'. 4. J = 34° 46.4', B= 86° 24.8', C= 58° 49'. 5. .4 =74° 40', B= 47° 46.4', C= 57° 33.4'. 6. .4 = 58° 26', B= 74° 23.8', C= 47° 10.6'. 7. .4 = 55° 55.4', B= 79° 43.8', C= 44° 20'. 8. .4 = 49° 24.2', B= 58° 38', C= 71° 57.8'. 9. .4 = 60° 50.8', B= 46° 6.2', (7= 73° 1.6'. 10. ^=18° 12.4', 5 =135° 50.6', C= 25° 56.6'. Art. 127; pages 91 and 92. 6. B= 39° 21.3', c = 5.511. 7. J5= 33° 28.4', a =118.33; or, 5 =146° 31.6', a =14.58. 8. 0= 53° 18.9', a =.07508. 9. A= 19° 18.1', 6 = 2.522. 148 ANSWERS. 10. A= 79° 20', G=. 11416; or, ^ = 100° 40', c=. 05121. 11. Impossible. 12. B= 24° 5.4', a =103.3. 13. 0= 90°, & = 5.007. 14. C= 67°10', a = 6.918; or, 0=112° 50', a = 2.913. 15. A= 46° 53.3', c= 141.48. 16. Impossible. 17. 0= 24° 31.4', & = 1.0637. 18. ^= 90°, c= 127.32. 19. A= 70° 12', 6 = .2879; or, ^=109° 48', &=.1045. 20. C= 37° 10', a = 289.3. Art. 128 ; page 93. 2. 1077.9. 6. 2604. 10. .1273. 3. 14.697. 7. 1.353. 11. 4491. 4. 3.257. 8. .06858. 12. .00001817. 5. 5114. 9. 215.9. 13. 11.732. Art. 129 ; pages 93 and 94. 1. Height of tower, 153.64 ft. ; distance from first point, 117.27 ft. ; from second, 217.27 ft. 2. 29800 square rods. 3. 247.7 ft. 4. 4.927 miles. 5. From first position, 9.9 miles ; from second, 19.122 miles. 6. 298 ft. 7. Distance, 91.66 ft. ; height, 33.9 ft. 8. 1113.1 ft. 9. 248 ft. 10. 1569.6. ANSWERS. 149 Art. 153; page 112. 5. ^=148° 5', B= 65° 23.2', 6= 38° 2'. 6. a = 40° 41.8', 5 == 134° 30.8', c = 122° 7.5'. 7. J. = 140° 41', S= 66° 43.8', c = 137° 20'. 8. ^= 27° 11.6', a= 25° 25.2', c= 69° 54' ; or, ^=152° 48.4', a = 154° 34.8', c=110°6'. 9.' ^ = 109° 22.5', a = 110° 57.6', 5 =113° 22'. 10. ^= 66° 12.1', ?> =127° 17.4', c = 107° 5.1'. 11. .1= 72° 28.9', ^=140° 38.1', c = 112° 37.7'. 12. B= '2ri.2\ h= 25° 24.4', c = 109° 46' ; or, 5 = 152° 52.8', 6 = 154° 35.6', c = 70° 14'. 13. .4= 120° 44.3', a = 156° 30', J3 = 33° 52.6'. 14. a= 41° 5.5', h = 26° 25', c= 47° 32.1'. 15. a= 60° 31.4', 2^ =143° 50', 6 =147° 32.1'. 16. .1= 78° 46.7', a= 70° 10', c = 106° 27.5'; or, ^1 = 101° 13.3', « = 109°50', c= 73° 32.5'. 17. .1= 30° 32.1', a= 22° 1.1', h= 43° 17.9'. 18. a=166°8.7', 5=101° 48.9', c= 50° 18.4'. 19. ^1=112° 2.5', i5=109°12', c= 81° 53.6'. 20. yl=135°2.5', h= 68°17.3', c = 105° 44'. 21. a =146° 32.5', Z. = 109° 48.1', c= 73° 35'. 22. A= 70°, i^= 75° 6.2', 6= 74° 7.3'. Art. 154; page 113. 2. a=117°1.2', J5=153°41.9', C=132°34'. 3. a = 57° 22.1', 6 =129° 41.6', 0=57° 52.5'. 4. A= 20° 0.9', 5= 141° 29.6', 6 =113° 17.1'. 5. ^= 33° 27.5', a= 35° 4.4', 6= 78° 46.7'. 6. 5= 69° 16', 5= 70°, 0= 84° 30' ; or, 5 = 110° 44', h =110°, C= 95° 30'. 150 ANSWERS. Art. 167; page 125. 2. a= 95° 37.8', 6= 41° 52.2', (7= 110° 48.8'. 3. 6 = 120° 16.6', c= 69° 19.6', A= 50° 26.2'. 4. a= 34° 3', c= 64° 19.4', B= 37° 39.6'. 5. a= 69° 4', 6 = 146° 25.6', 0= 125° 12.2'. Art. 168; page 126. 2. ^ = 121° 32.8', J5= 40° 56.8', c= 37° 25.8'. 3. ^= 86°59.7', C= 60°50.9', & = 111°17'. 4. 5 =134° 57.3', 0= 50° 41.1', a= 69° 8.8'. ' 5. ^ = 147° 29', 5= 163° 8.6', c= 76° 8.4'. Art. 169; page 127. 2. A= 51° 58.2', B= 83° 54.4', C= 58° 53.2'. 3. ^=142° 32.8', B= 27° 52.6', (7= 32° 27.2'. 4. J.= 35° 31', 5= 24° 42.6', 0=138° 24.8'. 6. A= 47° 21.2', 5= 42° 19.4', 0=124° 38'. Art. 170; page 128. 2. a= 67° 51.8', 6= 71° 44.4', c= 57°. 3. a = 144° 9.8', 6 = 148° 48.6', c= 41° 43.6'. 4. a= 89° 51.2', b= 85° 49.2', c= 72° 31.8'. 5. a =100° 4.6', 5= 49° 58.8', c= 60° 6'. Art. 171; page 131. 4. 0= 65° 30', A= 97° 20', a = 100° 44.6'. 5. B= 42° 40', = 159° 54', c = 153° 30.2'; or, 5 = 137° 20', 0= 50° 20.6', c= 90° 9.6'. 6. Impossible. 7. 0= 90°, ^ = 113° 36.9', 6 = 114° 51.9'. ANSWERS. 151 8. ^= 68° 18', .4 =132° 33.8', a =131° 15.8'; or, 5 = 111° 42', A= 77° 4.6', a= 95° 50'. 9. Impossible. 10. = 146° 37.9', B= 55° 0.6', 6= 96° 34.4'. Art. 172; page 133. 2. 5 = 114° 50', A= 79° 20', a= 82° 56'. 3. a= G7°24', 0=164° 6.4', c = 160° 6.4' ; or, a = 112° 36', 0=128° 20.6', c = 103° 2.4'. 4. c= 90°, B= 63° 42.7', b= 66° 26.2'. 5. Impossible. 6. b= 27° 22.1', A= 47° 21.2', a =117° 9.2'. 7. a= 43° 3.1', B= 89° 23.8', & = 129° 8.4'; or, a =136° 56.9', B= 26°58.6', 6= 20°35.8'. 8. Impossible. Art. 173; pages 134 and 136. 1. Distance, 3275 miles; bearing of Greenwich from Boston, N. 53° 7' E. ; of Boston from Greenwich, N. 71° 39.4' W. 2. Distance, 11010 miles ; bearing of Calcutta from Val- paraiso, S. 64° 20.4' E. ; of Valparaiso from Calcutta, S. 54° 54.6' W. 3. Latitude 49° 58.4' N. 4. 6 h., m., 40 s., a.m. ; longitude 44° 50' W. 5. 15° 0.8'. 6. 8 h., 2 m., 40 s., a.m. 7. 5 h., 3 m., 26.4 s., a.m. 1 Date Due /Oh ^/ro 1 / r T ,..>.-.- on ' ' ,\ L. U / / / f For Reference Not to be taken from this room BOSTON COLLEGE 3 9031 01550243 8 t. BOSTON COLLEGE LIBRARY UNIVERSITY HEIGHTS CHESTNUT HILL, MASS. Books may be kept for two weeks and may be renewed for the same period, unless reserved. Two cents a day is charged for each book kept overtime. If you cannot find what you want, ask the Librarian who will be glad to help you. The borrower is responsible for books drawn on his card and for all fines accruing on the same. 9 BOSrOR CmiEGE 19433