Digitized by the Internet Arcinive in 2010 with funding from Boston Library Consortium IVIember Libraries http://www.archive.org/details/planesphericaltrOOwent eHC^ 1 WENTWORTH'S SERIES OF MATHEMATICS First Steps in Number. Primary Arithmetic. Grammar School Arithmetic High School Arithmetic. Exercises in Arithmetic. Shorter Course in Algebra. Elements of Algebra. Complete Algebra. College Algebra. Exercises in Algebra. Plane Geometry. Plane and Solid Geometry. Exercises in Geometry. PI. and Sol. Geometry and PI, Trigonometry. Plane Trigonometry and Tables. Plane and Spherical Trigonometry, Surveying. PI. and Sph. Trigonometry, Surveying, and Tables Trigonometry, Surveying, and Navigation. Trigonometry Formulas. Logarithmic and Trigonometric Tables (Seven). Log. and Trig. Tables (Complete Edition). Analytic Geometry, Special Terms and Circular on Application, PLANE AND SPHEEICAL TRIGONOMETRY, AND SURVEYING BY G. A. WENTWORTH, A.M., ^ PBOFEBSOR OF JE^^MATICS IN PHILLIPS EXETER ACADEJttT. 2Ceacl)ers* lEtiitton* BOSTON COLLEGE PHYSICS DEPT. BOSTON, U.S.A.: GINN & COMPANY, PTTBLISHEES. 1891. D/^ -y Jj t X Entered, according to the Act of Congress, in the year 1886, by G, A. WENTWORTH, in the Office of the Librarian of Congress, at Washington. h4 Typography by J. S. Gushing & Co., Boston, U.S.A. Prksswork. by Ginn & Co., Boston, U.S.A. JUN19 / PREFACE. THIS edition is intended for teachers, and for them only. The publishers will under no circumstances sell the book except to teachers of Wentworth's Trigonometry ; and every teacher must con- sider himself in honor bound not to leave his copy where pupils can have access to it, and not to sell his copy except to the publishers, Messrs. Ginn & Company, It is hoped that young teachers will derive great advantage from studying the systematic arrangement of the work, and that all teach- ers who are pressed for time will find great relief by not being obliged to work out every problem in the Trigonometry and Sur- veying. G. A. WENTWORTH. ^\ TEIGONOMETET. Exercise I. Page 5. 1. What are the functions of the other acute angle B of the triangle ABC{Yig. 2)? sin B = -1 c cos 5 = tan B = -, a cot 5 = sec B = -, a CSC B = 2. Prove that if two angles, A and B, are complements of each other (i,e., ifA + B = 90°), then sin A = cos B, cos A = sin B, tan A = cot B, sec A = esc B, Bin A = -y c COS A^-> c tan^== -, cot A=-, a cot A = tan B, CSC A = sec B. (1-) cos B sec A = esc J. = — a sin B=~, c cot B = -, tan B = -, a CSC iJ = -, sec 5 = 3. Find the values of the func- tions of A, if a, b, c respectively have the following values : (i.) 3, 4, 5. (iv.) 9, 40, 41. (ii.) 5, 12, 13. (v.) 3.9, 8, 8.9. (iii.) 8, 15, 17. (vi.) 1.19, 1.20, 1.69. (lii. (v.) sin A cos A ■■ tan^ cot^ = sec A -■ CSC A -- ) sin ^ : cos A = tan.A = cot J.- sec J. = CSC A = sin A -- tan^ = sec A = 3 — ) 5 4 5' 3 4 4 3' 5 4 5^ 3 17 15 15' 15 ' 8' 17 15' 17 .39 "89' 39 80' 89 80' (ii ) sin^ cos A tan^ cot -4 sec A CSC 4 (iv .) sin A COS, A tan^ cot J. sec A CSC A- cos A COt^: CSC A ■■ 13 = 1? 13' __5^ ^12 _12 " 5' ^13 12 13 " 5* 9 ^41' 40 Ti' _^ '40' 40 ^T 41 ^40' .41 " 9' 80 89' 80 39' 89 39" TRIGONOMETRY. , ., . , 119 ( VI.) Sin A = — 1 ^ ^ 169 tan A = - — ) 120 A 169 sec A = > 120 A 120 cos A = . 169 , , 120 cot A = . 119 CSC A 169 119" 4. What condition must be ful- filled by the lengths of the three lines a, b, c (Fig. 2) in order to make them the sides of a right tri- angle? Is this condition fulfilled in Example 3 ? a^ + b^ = c^. 5. Find the values of the func- tions of A, if a, h, c respectively have the following values : (i.) 2mn, m^ — n'^, mn? -\- n^. (ii.) _^, x-vy, x-y x'^ + y'^ x-y (iii.) pqr, qrs, rsp. /■ s mn mv nr rsp p qrs s cot J. = = -' pqr p rsp s CSC A = — = -• 2W q (iv.) sm J. = — X =^- = — , pq nr qr mv ps mpv cos A = — X ^— = ^-- nr sq nqr . n sq ns tan J. = — X -^ = --. pq mv pv mn sq ns , sq nr sec A == -^ X — nqr mv ps mpv ,€se Ti^n ^ ps ms TEACHERS EDITION. 6. Prove that the values of a, h, c, in (i.) and (ii.), Example 5, satisfy the condition necessary to make them the sides of a right triangle. (i-) a2 _^ 52 ^ g2^ (2 mnf + (m^ — n^Y = (m'^ + n^)^, 4 rn^n^ + m* — 2 m?v? + n* = m* + 2 m^v? + n*, 2xy\ (li.) 3^ +(aj + 2/)' = x-yj x^ + 3/^ •2 o - + x2 + 2a;2/+2/2 x^ — Axy ■\- y^ _ a;*+2a;V+.V^ a;'' — 2 a;?/ + 2/2 4a;y + a;*-2a;y + 2/* = a;* + 2cey + 2/*, a* + 2a;y + 2/* = a;* + 2a;y + 2/*. 7. What equations of condition must be satisfied by the values of a, 5, c, in (iii.) and (iv.), Example 5, in order that the values may repre- sent the sides of a right triangle. (iii.) ^2g2^2 _|_ g2j,2g2 _ 7'2g2p2^ or Jp^-)2 2/1 12 1 or m^n's^ + m^p^v^ = n^(fr^. 8. Compute the fun<;tions of J. and B when a = 24, & = 143. c = \/(24)2 + (143)2 = \/21025 = 145. 24 sin J. = — = cos B, 145 J 143 . „ cos A = — = sm B, 145 94 tan J. = — = cot B, 143 cot J. = — = tan B, 24 A 145 o sec A = — • = CSC B, 143 A 145 r> CSC -4 = — = sec B. 24 9. Compute the functions of A and B when a = 0.264, c = 0.265. ^2 = g2 _ 0^2 = 0.070225-0.069696 = 0.000529. .-. 6 = 0.023. sin^ = ^ = ?^ = cos5, c 265 cos ^ = - = — = sin B, c 265 tanJl = ? = ^ = cot5, 6 23 cot .4 = - = — =tan5, a 264 sec^ = ^ = ^ = csc5, 6 23 csc^ = ^ = ?^ = sec.g. a 264 10. Compute the functions oi A and B when 6 = 9.5, c = 19.3. a^ = c^ — 52 = 372.49 - 90.25 = 282.24. .-. a = 16.8. sin.4=^=l^ = cos5, c 193 TRIGONOMETEY. . 5 95 .J, cos A= - = — = sm ij, c 193 tan A = - = — ■ = cot B, b 95 cot ^ = - a 95 168 tan 5, sec ^ = - = — = CSC is, b 95 . c 193 D CSC ^ = - = — = sec B. a 168 11. Compute the functions of A and B when a = Vp^ + g^, b = V2pq p^ + 2pg' + ^-^ = c'^. p + q = c. a sin A = cos A = - = c p + q p + q \/2pq cos 5, sin B, V2pq ^2pq cot J. = - = ^ y/p"^ + q^ o y/2pq = tan B, = esc B, CSC ^ c _ p -\- q _ ^ y/p^ + (^ = sec 5, 12. Compute the functions of A and B when a=Vp'^ +pq, c=p + q, b^ = c^ — Q? = (f -{-pq. .'. b = Vq^ + pq. ■ A ^ Vp2 +pq ^ sm A = -= —^ i-2 = cos B, c P + q cos . b y/f +pq . A = - = — i i-i = sm B, c P + 2 " Vq^ + pq « \/pi -j-pq A ^ P + Q T> sec A = j- = f ^ - = CSC B, " Vq^ + pq CSC A = c _ p + q = sec B. ^ Vp^ + pq 13. Compute the functions of A and B when b = 2->/pq, c =p + q. a2 + 52 = c\ a? + ipq =p'^ + 2pq + q^, a =p — q' sm tI = - = ^ — ^ = cos B, c p + q . b 2Vpq . „ cos J. = - = — i-i = sm B, c p + q tan -4 cot J. = - = a p pj-q ^y/pq 2V^q sec J. == ^ CSC A = - a 9. P + 9. pq . P + 9 p-q cot B, tan 5, CSC B, = sec B. ( 14. Compute the functions of A when a = 2 6. a = 2&, a2 + 62 = c2, 462 + 62_c2, 562 = c2, C-6V5. TEACHERS EDITION. sm -d. = - = CO? A = a _2b_ = f V5 = 0.89443, bV5 2. b b cotA = - = h a 2 b b cscA = '- = ^'=iV5. a 2b ^ 15. Compute the functions of J. when a = I c. a = fc, 52 ^ ^2 _ 0^2^ 6 = Vc^ — a^ 16. Compute the functions of A when a + b = ^c. a2 + J2 _ g2^ a2 + 62 + 2a5 = ffc2, a2-2a6 + 62 = _7_c2, 2 sm ^ = - = — p 3 2 - = -, a 3' & 2^ cosul = - = -T — = ^\/5, tan J. = - = ^ ^V5 cot^ = ^ = :^. fa = fV5, sec ^ = - 2 csc^ = ^ = ?. a 2 ^"' =fV5, a — b = 1^- a + b = =i«. 26 = = fo-|V7, 6 = = 1-1 V7, 2a = 4 a = = fc+|V7, c a 8 5+ V? 8a c = sm ^ = - = c a 5+ V7 5 + V7 8a 6 + V7 c 6._£ V7 cos ^ = - = c , . a 5+v^ tan J. = T = ^' ^ 6 -a/7 cot^ = 5 = 5-^ 5-V7 a 5+^7 sec -4 = - = b 5-V7 esc ^ = - = a 5+y^ .'.^ 6 TRIGONOMETRY. 17. Compute the functions of J. when a — 6 = — ' 4 a2-2a5 + Z.2 16 + 52 _ c2 2ab 15c^ 16 + &2 a2 + 2a6 + 62 = a + 6 = - Vsl, 4 31 c^ 16 I. I' a — = -, 4 2a = -V31+-- 4 4 .. a = -(\/31 + l). 2 6 = - VSI - -. 4 4 .-. 6 = |(V3l-l). ^(V3l + 1) CSC ^ = - = « V3I + 1 18. Find a if sin J. = | and G = 20.5. sin^ = ^ = ^. c 5 g ^3 20.5 5' 5a = 61.5, a =12.3. 19. Find h if cos A = = 0.44 and c = 3.5 h c h 3.5 .-. h = 0.44, = 0.44. = 1.54. Sin J. = - = c h 8 ^(V31-l) cos A = — = c \/3l + l V31-1 tanyJ=g_ V31+l , ^ V31-1 , , h V31-1 cot A = - = 1 « Vsl + i sec A = - = V31-1 20. Find a if tan J. = J^ and /> — 2-5- '^" 11 a _ _a_ _ 11 6~"2^~ a" . lla_ll "27 3' a = 9. 21. Find 5 if cot J. = 4 and a = 17. 5 = A = 4 a 17 .-. & = 68. 22. Find c if sec J. == 2 and & = 20. ^ = — =2 6 20 ■ .-. c = 40. . TEACHERS EDITIOISr. 6.45 and 23. Find c if esc J. a = 35.6 . CSC J. = - = -^ = 6.45. a 35.6 .-. c = 229.62. 24. Construct a right triangle ; given c = Q, tan -4 = f . tan^ = -. .*. a = 3 and & = 2. Draw AB -= 2, and .5C ± to ^5 = 3 ; join Cand A. Prolong vie to D, making AD = 6. Draw DE 1. to AB produced. Rt. A ADE will be similar to rt. ^ACB. .'. ADE is the rt. A required. 25. Construct a right triangle ; given a = 3.5, cos A = ^. Construct A A^B'C so that y=l, c^= 2. Then cos A = >}. Construct A ABC similar to A^B'C^, and having a = 3.5. 26. Construct a right triangle ; given 5 = 2, sin J. = 0.6. Construct rt. A A^B^C^, m'aking a^= 6, and c = 10. Then sin A^= j%. Construct A ABC similar to A^B'C^, and having 6 = 2. 27. Construct a right triangle; given & = 4, cscA = 4. Construct rt. A A^B^C'', having c''= 4 and a^= 1. Then construct A J.5(7 similar to A A^B'C, and having 6 = 4. 28. In a right triangle, c = 2.5 miles, sin J. = 0.6, cos A = 0.8; com- pute the legs. • A ^ sm A= — e A ^ cos A-= — c a = c sin A. .'. b = c cos A a =1.5. .-. 6 = 2. 30. Find, by means of the table, the legs of a right triangle ifA= 20°, c = 1 ; also, if ^ = 20°, c = 4. A = 20°, c = 1. sm A =-. c cos ^ .•. a = c sin ^. .*. 6 = c cos J.. .-. a = 0.342. .-. 6 = 0.940. ^ = 20°, c = 4. .-. a = 4x0.342 .-. 6 = 4x0.940 = 1.368. =3.760. 31. In a right triangle, given a = 3 and c = 5 ; find the h3^pote- nuse of a similar triangle in which a = 240,000 miles. a: C:: 240,000 : x, 3:5:: 240,000 : x. .-. a; = 400,000. 32. By dividing the length of a vertical rod by the length of its horizontal shadow, the tangent of the angle of elevation of the sun at the time of observation was found to be 0.82. How high is a tower, if the length of its horizontal shadow at the same time is 174.3 yards ? tan^ = - = 0.82. 6 .-. a = 0.82 6. b = 174.3 yards. .-. « = 0.82 of 174.3 yards = 142.926. TRIGONOMETRY. Exercise II. Page 8. 1. Eepresent by lines the func- tions of a larger angle than that shown in Fig. 3. Fig. 1. 2. Show that sinx is less than tana;. In Fig. 1, OM: PM::OA: AS, but 0M< OA. .'. PM< AS. 3. Show that sec x is greater than tana;. 0/S'=sec, AS=id.n. In rt. A OAS, Hyp. OaS'> side AS .'. sec > tan. 4. Show that esc x is greater than cot>r. OT=csc, BT=cot In A BOT, Hyp. 0T> side BT. .'. CSC > cot. 5. Construct the angle x if tan x Let O BAM he a unit circle, with centre 0; then construct AT tan- gent to the circle at J. = 3 OA ; then J. or is required angle. Construct the angle x if esc x Let O ABM be a unit circle, with centre 0; construct -ST tangent to the circle at B = 2 OA ; connect OT; then J. OT is required angle. 7. Construct the angle a; if cos x — 2^- Take 0M= I radius OA. At M erect a ± to meet the circumference at P. Draw OP. Then is POif the angle required. 8. Construct the angle a; if sin x = cos X. Let PM= sin x and 0M= cos x. But, by hypothesis, PM--= OM. .•. by Geometry, x = 45°. Hence, construct an Z 45°. 9. Construct the angle a; if sin x = 2 cos X. Construct rt. Z PMO, making PM= 2 OM. Draw OP. Then POM\b the angle required. if 10. Construct the angle 4 sin X = tan x. Take J of radius OA to M. At J/ erect a _L to meet the circumfer- ence at P. Draw OP. Then POil/ is the required angle. 11. Show that the sine of an angle is equal to one-half the chord of twice the angle. Have given Z POA. Construct POB = 2 POA. Draw chord PB. Then it is ± to OA • and PM, its half, is the sine of POA. .•. sin a; = 2 chord 2a;. TEACHEES EDITION. 9 12. Find x if since is equal to one-half the side of a regular in- scribed decagon. Let AC\)Q a side of a decagon. 360° Then 10 = 36° or^Oa Draw OB bisecting AO. Then Z ^0(7 will be bisected, and Z ^05 = 18°. But the sine oiAOB=^\AQ. .'. X or AOB = 18°. 13. Given x and y {x + y being less than 90°) ; construct the value of sin {x + y) — sin x. Let AB = sin (a; + y) in a circle whose centre is 0, and CD = sin x. Then, with a radius equal to CI), describe an arc from B, as centre, cutting AB &t E. Then BJA will be the constructed value of sin (x + y) — sin x. 14. Given x and y (x + y being less than 90°) ; construct the value of tan (x + y) — sin {x + y) + tan x — sin X. Let AB = sin {x + y), and CI) = sin a; ; also EF = tan {x + 3/), and 0F=i2,xxx. From F with a radius = AB take From ^with a radius = GF add HI. From / with a radius = CD take Then ^^will be the constructed value of tan {x + y) — sin {x + y) + tan X — sin x. 15. Given an angle x ; construct an angle y such that siny = 2 sin a?. Let AB be the sine of the Z a; in a circle whose centre is 0. Draw AC perpendicular to the vertical diameter. Then CO = AB. Take CF on vertical diameter = CO. Draw FD perpendicular to vertical diameter, and meeting cir- cumference at D. Draw DE perpendicular to OB and draw OD. 0F= 2 CO by construction. ED=FO; FO being the projec- tion of the radius OD. .-. DE = 2 AB, and DOB = &nglQ required. 16. Given an angle x ; construct an angle y such that cos 3/ = J cos a;. Let OB = cos AOB. Erect a ± CD at C, the middle point of OB, and meeting the cir- cumference at D. Draw DO. Then DOB is the angle required. 17. Given an angle x ; construct an angle y such that tan 3/ = 3 tan a;. Let AB be the tangent of x. Prolong AB to C, making AC= 3 AB, and draw OC from 0, the centre of the circle. COA is the required angle. 18. Given an angle x ; construct an angle y such that sec y = esc a;. Since sec = esc, £ — £ b a .'. a = b. Hence, construct an isosceles right triangle. The required angle will be 45°. 10 TRIGONOMETRY. 19. Show by construction that 2 sin J. > sin 2 A. Construct Z BOC and Z COA each equal to the given Z A. Then AB = 2 sin A, and AD, the J_ let fall from A to OB, = sin 2 ^. But AB > AD. Hence 2 sin J. > sin 2 A. 20. Given two angles A and B {A^ B being less than 90°), show that sin {A + B) < sin A + sin B. Construct HOK = Z ^, and COH = ZB. Then sin (A + B) = CP, sin A = UK, sin ^ = CD. Now CPDE. .-. (7P< CD+HK. .-. sin ( J. + P) < sin ^ + sin B. 21. Given sin a; in a unit circle ; find the length of a line corre^^^'^ ^ ing in position to sin a; 11 .„uio whose radius is r. 1 : r : : sin x : required line. .•. length of line required = r sin x. 22. In a right triangle, given the hypotenuse c, and also sin A = 7?i, cos A = n ; find the legs. sm ^ = - = m. c .'. a = cm. cos A — - = n. c .'. b = en. Exercise III. Page 11. 1. Express the following func- tions as functions of the comple- mentary angle : sin 30°. CSC 18° 10^ cos 45°. cos37°24^ tan 89°. cot82° 19^ cot 15°. CSC 54° 46^ sin 30° = cos (90° - 30°) = cos 60°. cos 45° = sin (90° -45°) = sin 45°. tan 89° = cot (90° -89°) = cot 1°. cot 15° = tan (90° -15°) = tan 75°. esc 18° 10^= sec (90° - 18° 10^ = sec 71° 50^. cos 37° 24^= sin (90° - 37° 24^ = sin 52° 36^. cot 82° 19^= tan (90° - 82° 19^ = tan 7° 41^ esc 54° 46^= sec (90° - 54° 46^ = sec 35° 14^ 2. Express the following func- tions as functions of an angle less than 45° : sin 60°. CSC 69° 2^. cos 75°. cos 85° 39^ tan 57°. cot 89° 59^ cot 84°. CSC 45° V. sin 60° = cos (90° - 60°) = cos 30°. cos 75° = sin (90° - 75°) = sin 15°. tan 57° = cot (90° - 57°) = cot 33°. cot 84° = tan (90° - 84°) = tan 6°, esc 69° 2^ = sec (90° - 69° 2') = sec 20° 58^. cos 85° 39^= sin (90° - 85° 39^ = sin 4° 21^. cot 89° 59^= tan (90° - 89° 590 = tanO°l^ esc 45° V = sec (90° - 45° 1^ = sec 44° 59^ TEACHEES EDITION. 11 3. Given tan 30° = ^ V3 ; find cot 60°. tan30° = cot(90°-30°) = cot 60°. .•.cot60° = J\/3, 4. Given tan A = cot A ; find A. tan A = cot (90° - A), 90° - J. = A, 2^ = 90°. .-. A = 45°. 5. Given cos ^ = sin 2 J. ; find J.. cos A = sin (90° - A), dO°-A = 2A, 3^ = 90°. .-. A = 30°. 6. Given sin ^ = cos 2 .4 ; find A. sin A = cos (90° - A\ 90°-A = 2A, 3 A = 90°. /. A = 30°. 7. Given cos J. = sin (-15° — J ^ ; find J.. cos A = sin (90° - A), 90°- J. = 45° -J J., 180° -2^ = 90° -A .-. A = 90°. 8. Given cot ^ J. = tan A ; find A. tan A = cot (90° - A), J^ = 90°-^, ^ = 180° -2^, 3^ = 180°. .-. A = 60°. 9. Given tan (45° + ^) = cot ^ ; find A cot A = tan (90° - A), tan (90° -A) = tan (45° + A), 90° - J. = 45° + A, 2 J. = 45°. .'. A = 22° 30^ 10. Find A if sin ^ = cos 4 A. sin A = cos (90° - A), dO°-A = 4:A, 5 A = 90°. .'. A = 18°. 11. Find ^ if cot ^ = tan 8^. cot A = tan (90° - A), 8A = 90°-A, 9 A = 90°. .'. A = 10°. 12. Find ^ if cot J. = tan nA. cot A = tan (90° - A), 90°- A = nA, 90° = A{n+ 1). .:A = ^^. n + 1 Exercise IV. Page 12. 1. Prove Formulas [1] - [3], using for the functions the line values in unit circle given in § 3, [1]. sin^^ + cos^yl = 1. [2]. tan^=^-HL4. cos J. 12 TEIGONOMETRY. [3]. sin AxcscA = l, cos AxsQc A = l, tan J.X cot J. = l. Fig. 2. [1]. DC= sin J., BD = cos J., DC^ + BD^ = (7^2 . but CB^ = 1. .-. BC^ + BD"" = 1. .'. sin^A + cosM = 1. [2] BC= sin A, BB = cos A, EF=iQ.nA. A FBE and ^Ci) are similar. .-. FE : BE:: CB : BB. FE_CB_. BE BB' but BE==1. CB BB sin A Or. FE = tan J.= cos A [3]. CB = sin J., 55"= CSC A. In similar /i^ HGB and (75i), ^S" : QB :: BC: CB. . BH^BC ^ ■ * (?5 CX) ' but (?5 = 1, BC^l. BH=—. CB BHxCB = l, CSC J. X sin J. = 1. cos A = BB, sec J. = BF. In similar A 5i^^ and BCB, BF: BE:: BC : BB. BF _BC . BE BB' but BE=1, BC^l. ,'. BF=^ ~ BB BFxBB = l, sec A X cos J. = 1. tan A = EF, cotA=GR. In similar A GSB and FEB, OH: OB:: BE: FE, GH_BE. GB FE' but GB = 1. BE==1. 1 FE GBxFE=l. cot A X tan A = l. 2. Prove that 1 +tan2J. = sec'A tan^ = 7, sec^ = -. b c a-' + b^ = c\ Dividing all the terms by 5^, GH- TEACHERS EDITION. 13 a^ 52 ^ g2 Substituting for — and — their values idiU^A and sec^J., we have tan^^ + 1 = sec^J.. 3. Prove that 1 + cot^^ - csc^^. cot^ =-', a CSC A = — a 0,2 _j. 52 ^ c^. Dividing all the terms by o?, O^ 52 ^ g2 a? a? 0? Substitutins for — and — their ^ o? a? values cot^J. and csc^J., we have 1 + cot2^ = csc^J.. 4. Prove that cot A = cot^ cos A sin A sin A = a — ~i c cos J. = h c Substituting, _ . h a a c c .'. cot J.= cos A sin A ExEECisE V. Page 14. 1. Find the values of the other functions when sin A = -^f. sin^J. + cos^J. = 1, m2 13 cosM = 1 cos A =v-(iy 25 169' •. cos J. = — • 13 tan^=^ cos 12 5" cot A is reciprocal of tan A. .'. cot^ = A. 12 sec A is reciprocal of cos A. .-. sec A = ^. CSC A is reciprocal of sin A. 13 12* .-. CSC A = ^^ 2. Find the values of the other functions when sin A = 0.8. sin^J. + cos^J. = 1, cos2^ = 1 - (0.8)2, cos A = Vl-0.64. .-. cos A = 0.6. tan^ = ^-iB = 0^. cos 0.6 .-. tan^= 1.3333. 0.6 cot^ = 0.8 14 TRIGONOMETKY. cot ^ = 0.75. sec A = — 0.6 sec A = 1.6667. J_ 0.8' CSC A = 1.25, CSC A = 3. Find the values of the other functions when cos A = f f . sin^ + cos^ = 1, sm 1 3600 J 121 ^'3721 3721 sin 11 cos 60 cot 1 tan _60 ll" sec 1 cos _61 60' 1 61 5. Find the values of the other functions when tan A = ^. GSC = — - = --• sm 11 4. Find the values of the other functions when cos ^= 0.28. sin^ + cos^ = 1. sin = Vl - (0.28)2 _ VO.9216. = 0.96. sin 0.96 tan^ cos 0.28 1 = 3.4285. tan 3.4285 = 0.29167. sec 1- = — = 3.5714. 0.28 cos 1 esc =-^=-'-= 1.04167. sin 0.96 tan J.= ^3 . cot J. = 3 ^4 tan J.= _ sin A cos A 4 sin A 3 Wl- sin ^A 3 sin ^ = 4VT~ smM, 9 sin^A = 16-16 sin^^, 25 sin2^ = 16, 5 sin J. = 4. sin A = — 5 cos A = sec A = CSC A = sin A tan^ 1 cos A 1 sin A 6. Find the values of the other functions when cot J. = 1. cot J. = 1. .'. tan J. = l. sin A tan -4 = 1 = cos A sin A Vl - sin'' A sin A = Vl — sinM, sm '^ = 1 - sin2^, 2 sin2^ = 1, sm '■A=^^. TEACHERS EDITION. 15 COS J.= tan J. sec A^ 1 CSC ^ = COS A I V2 1 sin A \y/2 = V2. = V2. 7. Find the values of the other functions when cot J. = 0.5. 0.5 tan -4 = tan^ = cot A sin A 2. cos -4 2 cos A = sin A. 4 cos^ J. — sin^^ = (squaring) cosM + sin^^ = 1 5 cos^ J. = 1 cos J. = a/- = 0.45. 4cos2J.4- 4sin2J. = 4 4cos2^— sin2^ = 5sin2J. = 4 sin A = -^- = 0.90 sec A = CSC A = '5 1 cos A 1 sin A = 2.22. = 1.11. 8. Find the values of the other functions when sec -4 = 2. cos A = = -, sec A 2 sin A = Vl — cos^^ -^l^h4 .-. sin A = iVS. tan A = . = ^—— = Vs. cos A 1 2 1 1 cot^ = CSC tan^ ^ sin-4 iV3. 9. Find the values of the other functions when esc A = \/2. sin J.= — = 1V2, V2 cos ^ = Vl - (^ V2)2 = vT^ = v|=jA tan^=i^ = l, ^V2 cot ^ = 1 = 1, 1 ' sec A = r-^ = V2. iV2 10. Find the values of the other functions when sin A = 7n. cos A = Vl — sinM = Vl — m^, sin ^ m tan -4 = cot^ cos A Vl — TO^ w Vl — m'^ \ — w? 1 1 —7n? tan yl mVl — m'^ 16 TRIGONOMETRY. sec A = CSC A = 1 1 cos A ^i _ ^2 1 ^ 1 sin A m 11. Find the values of the other ,■ ^ ■ A 2m Oill yi — ■ 1 +m2 cos A == Vl - - sin^. .'. cos A = a/i - 4m2 l + 2m2 + m* =Ji^ -2m2 + m* ^1 + 2m'^ + m* 1 — m^ l + m2 taii^ = sin cos 2m l-m^ cot J.= 1 tan l-m2 2m sec A = 1 1+m^ cos l-m^ rsr, y1 = 1 l+m2 sin 2m 12. Find the values of the other 2'mn functions when cos A = m^ + n^ sin A=Vl^ COS'' 4 m^n^ m* + 2 m^w^ + ?i* m^ — 2m''n'-^ + n* m* + 2 m^Ti^ + n* m^ — n' sm m'' ,,1 ^ 10 2 tan A = ^^^^ = ^ cos 2 mil , . 1 2m?i cot A = - — = —^ ; tan . 1 m? ^- V? sec A = •= —^ cos 2 mn esc A = sm m^ + n^ Tin? — n^ 13. Given tan 45° = 1 ; find the other functions of 45°. (1) sin 45° cos 45° sin 45° tan 45°. = 1. cos 45° (2) sin'^ + cos^ = 1. By (1), sin 45° = cos 45°. By (2), cos2 45° + cos2 45° = 1. 2cos2 45° = l, cos2 45°=-, 2 cos 45° = -ytt = |\/2. sin 45° = JV2. cot 45° - ^ -^ -1 tan 45° 1 sec 45° -^1 -V2. iV2 CSC 45° iV2 14. Given sin 30° = | ; find the other functions of 30°. sin'^ + cos^ = 1.- TEACHERS EDITION. 17 cos 30° =V-i = A? = iV3. tan 30° cot 30° = 1 .V3. iV3 sec 30° = J^ = JV3. ^V3 CSC 30° = 1 = 2. J 15. Given csc 60° = f V3 the other functions of 60°. find sm = ■ — , CSC )° = -^=-iV3. sin 60^ = == -2 fV3 cos 60° = Vl - sin2, COS 60° = Vl - (i V3)2 \ A 9. 4 2 JrVS tan 60° = ^^ = Vs. cot 60° = — = I Vs. V3 sec 60° = i = 2. 16. Given tan 15° = 2 -VS the other functions of 15°. find sin 15° cos 15° = 2 -Vs. sin2 15° + cos2 15° = 1. sin 15° = cos 15° (2 -VS), [cos(2-V3)]2 + cos2 = l, cos2 (4 - 4V3 + 3) + cos2 = 1, cos2(8-4V3) = l. 1 2+V3 cos2 15° = 4(2 -Vs) 4 cos i5° = y2+V3^,y^;;^ sin^ = 1 — cos^. sinn5° = l-2iV3^2-V3 4 4 in 15° = aV2-V3. sin cot 15° = tan 15° 2-V3 = 2+V3. 17. Given cot 22° 30^= V2 + 1 ; find the other functions of 22° SO''. tan = — = 1 cot V2 + 1 =V2-1. sm , — == tan, cos cos^ + sin^ = 1. (1) (2) From (]), cos tan = sin. Squaring, cos^tan^ = sin'^ From (2), cos^ = — sin^ + 1 Add, cos^tan^ + cos^ = 1 cos^(tan2 + 1) = 1, cos2(4-2V2) = l, cosV4-2V2 = l. 18 TRIGONOMETRY. .*. COS = V4-2V2 =v 4 + 2\/2 = aV2+V2. sm = ^1_2±V2 \ '4-2-V2 =.^'2z:^=iV2:vI. 18. Given siB(F = 0; find the other fanctions of 0°. cos = Vl — sm^ = Vl -0. .*. cos = 1. tan=?^=? = 0. cos 1 cot = = - = 00. tan 1 1 1 sec = — = - = 1. cos 1 1 1 CSC = = _ = 00. sin 19. Given sin 90°= 1; find the other functions of 90°. sin 90° = 1. cos = Vl — sin^ = 0. . sin 1 _ tan = — - = - = CO. cos cot = — = — = 0. tan CO 1 1 sec = — = - = CO. COS 1 1 1 CSC= -r- = 7= i. sm 1 20, Given tan 90° = 00 ; find the other functions of 90°. tan 90°= 00. tan 00 cot = — = — = 0. sm cos sin^ = CO cos^ sin^ + cos^ = 1 — COS^ = CO cos^ — 1 CO cos^ = 1. \ CO COS = -xi — = 0. sin = 00. • sin = 1. 1 sec = - = 00. CSC = 1. 21. Express the values of all the other fanctions in terms of sin A. By formnlae on pages 11 and 12, sin A = sin A, cos A = Vl — sin^^, , . sin ^ tan^ = cot J.= sec A = CSC A = Vl — sin'^^l Vl - sin^^ sin A 1 Vl - sinM 1 sin^ 22. Express the values of all the other fanctions in terms of cos A. TEACHEE-S EDITION. 19 By formula on pages 11 and 12, sin A = Vi- COS 'A cos A = ces A, tan J.= Vi- COS 'A cos A cotA = COf A 1 Vi- cos 'A sec A = 1 COS J. , CSC A = 1 ' Vi- cos 'A 23. Express the values of all the other functions in terms of tan J.. cot^ tan^ - = tan A. b a? + h'' = 1. a = h tan A. a? = 6^ tan^ J.. a2-62tan2^ =0 a2 + 62 =1 h-' (i + tan^.l) = 1 1 + tan^^ sin A = Vi — cos'^^ V = Al- 1 + tan^^ 4 1 - 1 + tan^J. 1 + tan2 J. tan J. Vi + tan^ J. sec A = — = VrTtanlJ. cos 4 1 Vi + tan'-^J. CSC ^ = — = sm tan^l 24. Express the values of all the other functions in terms of cot A. = tan A. cot^ sin A = tan A. cos A Let X = sin, y = cos. x__l y cot A X cot A = y, x^ cot^A = y"^. x^co\?A-y'^ = ^ x^ +3/^ = 1 a;2(l+cot2J.) = l 1 1 + C0t2 J. sin A = cos Vi + cot^^ ^ = Vl — sin'-^J. V = Al- =v 1 + cot^J. 1 + cot^A - 1 1 + C0t2 J. cot J. VTTcoFl sec A = cos -4 CSC \/\ + COt''^ J. cot^ A = ^^ = VlTcot^Z. sin A 25. Given 2 sin ^ = cos A ; find sin A and cos A. 20 TRIGONOMETRY. sin^J. + cos^J. = 1. sin^ J. + 4 sin^ J. = 1. 5 siu^A = 1, sinM = i- 5 sin A ' n iV5. ". cos J. = |V5. 26. Given 4 sin ^ = tan ^ ; find sin A and tan J.. J, sin ^ But cos A tan J.= = 4 sin A. 4 sin J. = sin A cos A 4 sin A ; K cos A = sin ^. cos A sin ^ 1 4sinA 4 sin^J. + cos^J. = 1. sin A >'16 tan J. sin A cos A 15 27. If sin ^ : cos J. = 9 : 40, find sin A and cos A. 40 sin J. = 9 cos A. (sq.) 1600 sin^A - 81 cosM. 1600 sin2^ - 81 cosM - 0. But sin2^ + cos2 J. = 1. Multiplying bj ' 81 and adding, 1681 sin 2^1 = = 81, .'. 41 sin A = 9. sin A = _ 9 '4l" sin^J. + cc A== A = )S2A = = 1. cos Vi- - smM. .:. cos v^- { ^ Y 40 '41 28. Transform the quantity tan^ J. + cot^-^ — sin^J. — cos'^J. into a form containing only cos A. idX^A = coi^A = sinM _ 1 — cos^^ cos^A cos'-^J. cos^J. cos^^ sin^A 1 — cos^^ cos2^ 1 — cos^tI cos^ J. 1 — COS^A — 1 + cos^yl — COS^J. 1— 2 cos^ J-+2 COS* J.— cosM+cos^^ COS^^ — COS* J. _ 1 - 3 cos^^ + 3 cos* J. cos^ A — cos*^ 29. Prove that sin J. + cos J. = (1 + tan A) cos A. ^i5_4=.tanA cos A sin A = tan J. cos ^. sin A + cos A = tan ^ cos J. + cos A = (1 + tan A) cos A. 30. Prove that tan^ + cot^ = sec A X CSC A. TEACHERS EDITION. 21 tan A cot J.= _ sin A cos A cos A sin A tan^ + cot^=^HL^ + ^-^^ cos A sin A EXEECISE VI. 1, In Case II. give another way of finding c, after h has been found. cos A = -, c h = c cos A, h c = ■ COS A 2. In Case III. give another way of finding c, after a has been found. ■ A « c c&va. A = a, a c = sin A 3. In Case IV. give another way of finding 6, after the angles have been found. cos A = -y c 6 = c cos A. 4. In Case V. give another way of finding c, after the angles have been found. A ^ cos A = —, G c cos J. = h, c = cos J. sin^^ + cos^^ cos A sin A ^va?A + cos^J. = 1. tan J. + cot A- cos ^sin^ = sec A Xcsc A Page 20. 5 . Given sir B A vA a and c; = (90° a c = c sin find A, A, a, h cos -4 b c h = C COS A. 6. Given B and h ; find A, a, c. A = (90° - B), tan J.= a h a = h tan^ sin B = h — 1 c b = c sin B^ c = h sin 5 7. Given B and a ; find A, b, c. A = (90° - B), tan J. = -) b tan A = a, tan J. 22 TRIGONOMETEY. sin A c &'m A = a, c = sin J. 8. Given b and c ; find A, b, a. cos J. = -. B = = (90° -A). sin A = a c a = = c sin A. 9. Given a = 6, c = 12; required ^ = 30°, B = 60°, b = 10.392. sin A A a 1 • = - = - = sm c 2 = 30°. 30°, £ = (90° -A) = = 60°. cos A b b ~ c = c cos A. log cos A = 9.93753 log 12 = 1.07918 log b = 1.01671 b = 10.392. 10. Given ^ = 60°, 6 = 4; re- quired B = 30°, c = 8, a = 6.9282. Since A = 60° and 6 = 4, B = (90° - 60°) = 30°, and c = 8. (By Geometry.) c^ = a^ + b^. .•.c2-62 = a2 = 48. log48 = loga2= 1.68124, log a = 0.84062, a = 6.9282. 11. Given ^ = 30°, a = 3; re- quired B = 60°, c = 6, 6 = 5.1961. Since A = 30° and a = 3, B = (90° - 30°) = 60°, and c = Q. c'- = a? -\- b^. .-. c2-a2 = 62 = 27. log27 = log 62 = 1.43136, log 6 =0.71568, 6 =5.1961. 12. Given a = 4, 6 = 4; required J. = ^=45°, c = 5.6568. Since a and 6 each = 4, the A is an isosceles A, and the A A and B are equal. :.A = ^ of 90° = 45°, 5 = J of 90° = 45°. c2 = a2 + 62 = 32. log 32 = log c2= 1.50515, log c = 0.75257, c =5.6568. 13. Given a = 2, c = 2.82843 ; required A=^B = 45°, 6 = 2. 6 = \/^ = ■\J{c-[-a){c — a), log 62 = log {c + a) + log (c— a). log {c^a)= 0.68381 log (c - a) = 9.91826-10 log 62 - = 0.60207 TEACHEES EDITION". 23 log h = 0.30103, h =2. .*. the A is an isosceles rt. A. .-. ^ = ^ = 45°. 14. Given c = 627, A = 23^ 30^ ; required B = 66° 30^ a = 250.02, b = 575.0. B = (90° - ^) = 66"" 30^. a = c sm A. log a = log c + log sin A. log c = 2.79727 log sin A = 9.60070 log a = 2.39797 a = 250.02. 6 = c cos A. log 6 = log c + log cos A. log c - 2.79727 log cos A = 9.96240 log 6 =2.75967 h = 575. 15. Given c = 2280, A = 28° 5^ ; required B = 61° 55^, a = 1073.3, 6 = 2011.6. B = 61° 55^ a = c sin A. log « = log c + log sin ^. log c = 3.35793 log sin A = 9.67280 log a = 3.03073 a = 1073.3. 5 = c cos A. log 5 = log c + log cos A. log c = 3.35793 log cos A = 9.94560 log b = 3.30353 b =2011.6 16. Given c = 72. 15, A = 39° 34^; required B = 50° 26^ a = 45.958, b = 55.620. ^ = 50° 26^ a = c sin J., log a = log c + log sin A, log c = 1.85824 log sin A = 9.80412 log a = 1.66236 a = 45.958. 6 = c cos A. log 6 = log c + log cos A. log c = 1.85824 log cos A = 9.88699 log b = 1.74523 b = 55.620. 17. Given g=-1, A = 36°; required 5 = 54°, a = 0.58779, 5 = 0.80902. B = 54°. sin A a _ a c = c sin A. log a = log c + log sin A log c = 0.00000 log sin A = 9.76922 log a = 9.76922 a = 0.58779. cos A- c 24 TRIGONOMETRY. b = c cos A. log 6 = log c + log cos A. log c = 0.00000 log cos A = 9.90796 log b = 9.90796 - 10 b = 0.80902. 18. Given c = 200, B = 21° 47^ ; required ^ = 68° 13^ a = 185.72, b = 74.219. 13^ sm -d = — c a = c sin ^. log a = log c + log sin A. log c = 2.30103 log sin A = 9.96783 log a = 2.26886 a = 185.73. cos .4 = -• c & = c COS .4. log 6 = log c + log COS A. log G = 2.30103 log cos A = 9.56949 log b = 1.87052 6 = 74.22. 19. Given c = 93.4, B = 76° 25^ ; required A = 13° 35^ a = 21.936, 6 = 90.788. A = 13° 35^ a = c sin A. log a = log c + log sin A. logc = 1.97035 log sin A = 9.37081 log a = 1.34116 a = 21.936. b = a cot A. log b = log a + log cot A log a = = 1.34116 log cot A = = 0.61687 log b = = 1.95803 b = = 90.788. 20. Given a = 637, A = 4° 35^ ; required B = 85° 25^ b = 7946, c = 7971.5. B^ = 85° 25^ b = = a cot A. lo gb- = log a + log cot A. log a = = 2.80414 log cot A = = 1.09601 log b = = 3.90015 b = = 7946. logc = = log a + colog sin A. log a = = 2.80414 colog sin A = = 1.09740 log c = 3.90154 c = 7971.5. 21. Given a = 48.532, ^ = 36° 44^; required B = 53° 16^ b = 65.033, c = 81.144. ^ = 90° - .4 = 90° - 36° 44^ = 53° 16^ • A « sm A = — c TEACHERS EDITION. 25 c sin J. logc = log a + colog sin log a = 1.68603 colog sin A = 0.22323 log c = 1.90926 c = 81.144. cos A b b = c cos A. log b = log c + log cos A. log c = 1.90926 log cos A = 9.90386 log b = 1.81312 b = 65.031. 22. Given a = 0.0008, ^ = 86°; required B = ^°,b = 0.0000559, c = 0.000802. B = dO°-A = 90° - 86° = 4°. sin^ = -- sin J. log c = log a + colog sin A. log a = 6.90309 - 10 colog sin A = 0.00106 log c = 6.90415 - 10 c = 0.000802. cos A = ~ c b = c cos A. log 6 = log c + log cos A. log c = 6.90415 - 10 log cos A = 8.84358 log b = 5.74773 - 10 b = 0.0000559. 23. Given 6 = 50.937, 5 = 43° 48^; required ^ = 46° 12^ a = 53.116, c = 73.59. A = 46° 12^ tan^ = -. b a = b tan A. log 5 = 1.70703 log tan A = 0.01820 log a = 1.72523 a = 53.116. sin A a c a sin A log a = 1.72523 colog sin A = 0.14161 log c = 1.86684 c = 73.593. 24. Given 5 = 2, 5 = 3° 38^ ; re- quired A = 86° 22^, a = 31.497, c = 31.560. A = 86° 22^ tan .4 = -. 6 a = 5 tan A. log 6 = 0.30103 log tan^ = 1.19723 log a = 1.49826 a = 31.496. 26 TRIGONOMETRY. sin A a c c a sin J. log a = 1.49826 colog sin A = 0.00087 log c = 1.49913 G = 31.560. log a = 1.86332 colog sin A = 0.44305 25. Given a = 992, B = 76° 19^ ; A = 13° 41^ b = 4074.45, c = 4193.55. ^ = 90° - 76° 19^ = 13° 41^ sm A = — c log c = log a + eolog sin A. log a = 2.99651 colog sin A = 0.62607 log c = 3.62258 c = 4193.6. sin B = — c log 6 = log c + log sin B. log c = 3.62258 log sin B = 9.98750 log h = 3.61008 h = 4074.5. 26. Given a = 73, 5 = 68° 52^ ; required A = 21° 8^ h = 188.86, c = 202.47. ^ = 90° - .5 = 21° 8^ • A ^ sm A = — c log c = log a + colog sin A. log c = '- 2.30637 c = = 202.47. sin B = 6 c log h = = log c + log sin B. log c = = 2.30637 log sin B = = 9.96976 log 5 = = 2.27613 & = = 188.86. 27. Given a = 2.189, B = 45° 25^; required yl = 44° 35^ h = 2.2211, c = 3.1185. ^ = = 90° - 45° 25^ = 44° 35^ sin A = c- a c a sin^ lo^ yc- = log a + colog sin A. log a = = 0.34025 colog sin ^ = = 0.15370 log G : = 0.49395 c = 3.1185. cos A. h- b c = c cos A. lo^ l^ = log c + log cos A. log c = 0.49395 log cos A = 9.85262 log 6 b = 0.34657 = 2.2211. 28. Given 6 =4, ^ = 37°56^ re- quired 5=52°4^ a =3.1176, c = 5.0714. teachers' edition. 27 B = 52° ¥. h cos A = c h = c cos ^. h c = COS J. log G = log h + Colog COS A. log 5 = 0.60206 colog COS A = 0.10307 log c = 0.70513 c = 5.0714. tan A = -- a = b tan ^. log a = log & + log tan A. log h = 0.60206 log tan A = 9.89177 log a = 0.49383 a = 3.1176. 29. Given c = 8590, a = 4476 required ^ 6 = 7332.8. 31° 24^ B = 58° 36^ Sin A = — log sin J. = log a + colog c. log a = 3.65089 colog c = 6.06601 log sin A = 9.71690 A = 31° 24^. B = 58° 36^ cot A = — a log h = log a + log cot A. log a = 3.65089 log cot A = 10.21438 log b = 3.86527 b = 7332.8. 30. Given c = 86.53, a = 71.78 ; required A = 56° 3^ B = 33° 57^ b = 48.324. log sin A = log a + colog c. log a = 1.85600 colog c = 8.06283 log sin A = 9.91883 A = 56° 3^ .5 = 33° 57^ log 6 = log a + log cot A. log a = 1.85600 log cot J. = 9.82817 log b = 1.68417 & = 48.324. 31. Given c = 9.35, a = 8.49 ; required A = 65° 14^ B = 24° 46^ 6 = 3.917. A <^ sm J. = — c .5 = 90° - .4. colog c = 9.02919 log a = 0.92891 log sin A = 9.95810 A = 65° 14^ B = 24° 46^ cos A = — c 6 = c COS A. log c = 0.97081 log cos A = 9.62214 log b = 0.59295 b = 3.917. 32. Given c=2194, 6 = 1312.7; required ^=53° 15', B = 36° 45^ a = 1758. 28 TRIGONOMETRY. COS A = log 6 = 3.11816 colog c = 6.65876 log COS A = 9.77692 A = 53° 15^ B = (90° - A) = 36° W, Sin A = -' c a = c sin A. log c = 3.34124 log sin A = 9.90377 log a = 3.24501 a = 1758. 33. Given c = 30.69, b = 18.256 ; required A = 53° 30^ B = 36° 30^ a = 24.67. A ^ cos A = — c log cos J. = log b + colog c. log 6 = 1.26140 colog c = 8.51300 log cos ^ = 9.77440 A = 53° 30^. £ = 36° 30^ tan A = -- log a = log tan A + log 6. log tanJ. = 10.13079 log b = 1.26140 log a = 1.39219 a = 24.671. 34. Given a = 38.313, b = 19.522 ; required A = 63°, .B = 27°, c = 43. tan A = -- b log tan A = log a + colog h. log a = 1.58335 colog b = 8.70948 log tan J. = 10.29283 A = 63°. B = 27°. sin -4 a logc = log a + colog sin A log a = 1.58335 colog sin A = 0.05012 log c = 1.63347 c = 43. 35. Given a = 1.2291, 6 = 14.950; required J. = 4° 42^ B = 85° 18^, c = 15. tan^ = ?. b log a = 0.08959 colog b = 8.82536 - 10 log tan^ = 8.91495 A = 4°42^ B' = 85° 18'. sin A = a a = c sin A. a sm-d log a = 0.08959 colog sin A = 1.08651 log G = 1.17610 c -15. TEACHERS EDITION. 29 36. Given a = 415.38, & = 62.080; required ^ = 81° 30^ .S = 8° 30^ c = 420. tan A = -■ log a = 2.61845 colog b = 8.20705 - 10 log tan A = 10.82550 A = 81° 30^ B = 8°30^ • A ^ Bin A = -- a = c sin A. c a sin J. log a = 2.61845 colog sin A = 0.00480 logc = 2.62325 c = 420. 37. Given a = 13.690, 6 = 16.926; required A = 38° 58^ B = 51° 2', c = 21.77. tan A = -' log tan A = log a + colog b. log a = 1.13640 colog b = 8.77144 - 10 log tan7l = 9.90784 A = 38° 58^ B =51° 2\ A Of c = sin J. log c = log a + colog sin A. log a = 1.13640 colog sin A = 0.20144 log c = 1.33784 c = 21,769. 38. Given e = 91.92, a = 2.19; required J. = 1° 21^ 55^^ ^ = 88° 38^ 5^^ b = 91.894. • A ^ sin ^ = — c log sin -4 = log a + colog c. log a = 0.34044 colog c = 8.03659 - 10 " log sin ^ = 8.37703 A =1° 21^ 55^^. B =88° 38^ 5^^ cos A = -' c 6 = c cos -4. log b = log c + log cos A. log c = 1.96341 log cos A = 9.99988 log 6 = 1.96329 b = 91.894. 39. Compute the unknown parts and also the area, having given a = 5, 6 = 6. tan A = — b log tan A = log a + colog 6. log a colog 6 = 0.69897 = 9.22185-10 log tan J. = 9.92082 A = 39° 48^ B ^ 50° 12'. TRIGONOMETRY. sin A = — c sin A log c = log a + colog sin A. log a = 0.69897 colog sin A = 0.19375 log c = 0.89272 c = 7.8112. axh 30 F 15. 40. Compute the unknown parts and also the area, having given a = 0.615, c = 70. I . . a sm A^-- G log sin J. = log a + colog c. log a =9.78888-10 colog c = 8.15490-10 log sin A = 7.94378 A = 30^ 12^^. £ = 89° 29^ 48'^ tan A^-- log & = log a + colog tan A. log a = 9.78888 - 10 colog tan A = 2.05626 log b = 1.84514 b = 70.007. F= --lab. log« = 9.78888- -10 log 6 = 1.84514 colog 2 = 9.69897 - -10 41. Compute the unknown parts and also the area, having given b = ^2. c=\/3. n/3 = cos^ = 1.25991. 1.73205. b c log COS A = -- log b + colog c. log b = 0.10034 colog c = 9.76144- -10 log F = 1.33291 F = 21.528. log cos A = 9.86178 A = 43° 20^ B - 46° 40^ . a sm A = — c a = c sin A. log a — log c + log sin A. log c = 0.23856 log sin A = 9.83648 log a = 0.07504 a = 1.1886. J'^ J ab. log a = 0.07504 log b = 0.10034 colog 2 = 9.69897 - 10 log F = 9.87435 - 10 F = 0.74876. 42. Compute the unknown parts and also the area, having given a =7, il = 18°14^ sm A = — c c = sin A TEACHERS EDITION. 31 log c = log a + colog sin A. log a = 0.84510 colog sin A = 0.50461 log c = 1.34971 c = 22.372. tan J. = -• tan J. log h = log a + colog tan ^. log a - 0.84510 colog tan A = 0.48224 - 10 log b = 1.32734 b = 21.249. log a log 6 colog 2 logi^ F=^ab. = 0.84510 = 1.32734 = 9.69897-10 = 1.87141 " ■ = 74.371. 43. Compute the unknown parts and also the area, having given 6 = 12, ^=29°8^ A = 29° 8^. B = 60° 52^ COS A = — c h c = COS A log c = log b + colog COS A. log b = 1.07918 colog COS A = 0.05874 log c = 1.13792 c = 13.738. sm ^ = -. c a = c sin^. log a = log c + log sin A. log c = 1.13792 log sin A = 9.68739 log a = 0.82531 a = 6.6882. F= I ab. log F= log a + log 5 + colog 2. log a = 0.82531 = 1.07918 = 9.69897-10 log 5 colog 2 logF F = 1.60346 = 40.129. 44. Compute the unknown parts and also the area, having given c = 68, ^ = 69°54^ A = 69° 54^ B = 20° 6^ sm A = -' c a = c sin A. log a = log c + log sin A. log c = 1.83251 log sin A = 9.97271 log a =1.80522 a = 63.859. cos A = -. c 6 = c cos ^. log 6 = log c + log cos A. 32 TRIGONOMETKY. log c = 1.83251 log cos A = 9.53613 log b = 1.36864 b = 23.369. F= J ab. log a = 1.80522 log b = 1.36864 colog 2 = 9.69897-10 log i^ = 2.87283 F =746.15. 45. Compute the unknown parts and also the area, having given c = 27, B = M°4:'. A = 45° 56^ a = c sin A. f log a = log c + log sin A. log c = 1.43136 log sin A = 9.85645 log a = 1.28781 a = 19.40. 6 = c cos A. log 5 = log c + log cos A. log c = 1.43136 log cos A = 9.84229 log b = 1.27365 & = 18.778. F= J a6. log a = 1.28781 log b = 1.27365 colog 2 =9.69897-10 log F = 2.26043 -F = 182.15. 46. Compute the unknown parts and also the area, having given a = 47, ^ = 48° 49^. A = 41° IK b = a cot -4, log b = log a + log cot A. log a = 1.67210 log cot A = 10.05803 log& = 1.73013 b = 53.719. &iuA log c = log a + colog sin A. log a = 1.67210 colog sin ^ = 0.18146 logc = 1.85356 c = 71.377. F=lab. log a = 1.67210 . log 6 = 1.73013 colog 2 = 9.69897-10 logF = 3.10120 F = 1262.4. 47. Compute the unknown parts and also the area, having given 6 = 9, .5 = 34° 44^ A = 55° 16^ a = b tan A. log a = log b + log tan A. log 6 = 0.95424 log tan J. = 10.15908 log a = 1.11332 a = 12.98L TEACHERS EDITION. 33 c = mi A log c = log a 4- colog sin A. log a =1.11332 colog sin ^"==0.08523 log c = 1.19855 c = 15.7960 log a log 6 colog 2 logi^ i^- ^ ah. = 1.11332 = 0.95424 = 9.69897-10 = 1.76653 = 58.416. 48. Compute the unknown parts and also the area, having given c = 8.462, B = m°¥. ^ = 3° 56^ a = c sin A. log a = log c + log sin A. log c = 0.92747 log sin A = 8.83630 log a =9.76377-10 a = 0.58046. h = c cos A. log h = log c +log cos ^. log c = 0.92747 log cos A = 9.99898 log b = 0.92645 6 = 8.442. F = lah. log a = 9.76377- -10 log& = 0.92645 colog 2 = 9.69897 - -10 logi^ = 0.38919 F = 2.4501. 49. Find the value of F in terms of c and A. F=iab. Sin A =-' c a = c sin A. COS A = — c b = c COS A. Substitute, F=^ab = J (c^ sin A COS A). 50. Find the value of F in terms of a and A. F= J ab. cot A = — a b = a cot A. Substitute, F=^ab = ^ {a^ cot ^). 51. Find the value of F in terms of b and A. F=lab. tan ^ = -• a = b tan .4. Substitute, i^=Ja& = J (62 tan ^). 52. Find the value of F in terms of a and c. F= i ab. c^ = a^ + b^. b^ = c^ — a^, b = a/c^ — a^. Substitute, 34 TRIGONOMETRY. 53. Given i^= 58, a = 10 ; solve the triangle. F= I ah. a log 6 = log 2 i^ + colog a. log2P =2.06446 colog a = 9.00000 - 10 log h = 1.06446 h = 11.6. tan A = -■ log tan A = loga + colog b. log a = 1.00000 colog b = 8.93554 - 10 log tan^ = 9.93554 A = 40° 45^ 48^^ B =49°14M2^>'. sin^ log € = log a + colog sin A. log a = 1.00000 colog sin .4 = 0.18513 log c = 1.18513 c = 15.315. 54. Given i^= 18, 6 = 5; solve the triangle. F= I ah. 2F log a = log 2 i^ + colog 6. log2i?' =1.55630 colog b = 9.30103 - 10 log a = 0.85733 " a = 7.2. tan A = -' log tan A = log a + colog b. log a = 0.85733 colog 6 = 9.30103 -10 log tan J. = 10.15836 A 55° 13^ 20^^. B 34° 46^ 40^^ c = a sin^ logc = = log a + colog sin A. log a = 0.85733 colog sin A = 0.08546 log c = 0.94279 c 8.7658. 55. Given jP= 12, ^ = 29° solve the triangle. B = = 61°. ■ F= -lab = 12. ah = = 24. a = 24 ■ b' tan^ = a tan 29°- 24 " b^' fc2 = 24 tan 29° log 6 = i (log 24 + colog tan 29°). log 24 = 1.38021 colog tan 29° = 0.25625 log& b 2) 1.63646 = 0.81823 = 6.58. TEACHERS EDITION. 35 tan 29° = ^. a = & tan 29°. log a = log b 4- log tan 29°. log 6 = 0.81823 log tan 29° = 9.74375 log a = 0.56198 a = 3.6474. • A « c = sin 29° log c = log a + colog sin 29°. log a = 0.56198 colog sin 29° = 0.31443 log c = 0.87641 c = 7.5233. 56. Given i^= 100, c = 22; solve the triangle. F= 1 ab = 100. ab = 200. 2_00 b ' 40000 62 a = o3 + 52 = c2 = 484. Substitute, 40000 62 + 62 = 484. 40000 + 6* = 484 62. 6* -484 62 = -40000. ¥-{ ) +(242)2= 18564. log\/l8564 = 2 )4.26867 = 2.13434 ; but 2.13434 = log 136.25. .-. 62 - 242 = 136.25. 62 = 378.25. log 6 = Hlog 378.25). = 1.28889. 6 = 19.449. cos A = -' c log COS A = log 6 + colog c. log 6 = 1.28889 colog c = 8.65758 log cos A = 9.94647 A = 27° 52^ B =62° 8^ sin A a — ^. c = c sin A. log 5« = log c + log sin A log c = 1.34242 log sin A = 9.66970 log a = 1.01212 a = 10.283. 57. Find the angles of a right triangle if the hypotenuse is equal to three times one of the legs. Let c = hypotenuse, and let c = three times a, one of the legs. • A O, sm A = -' c log sin A = log a + colog c. TRIGONOMETRY. log a = 0.00000 colog c = 9.52288 -10 log sin A = 9.52288 A = 19° 28^ 17^^. B = 70° 3F 43^^ 58. Find the legs of a right tri- angle if the hypotenuse = 6, and one angle is twice the other. Let c = hypotenuse = 6, and let B = twice A ; then B = 60°, A = 30°. sin A a = c sin A. log a = log c + log sin A. log c = 0.77815 log sin A = 9.69897 log a a = 0.47712 = 3. sin B h h ~ c = c sin B. log h = log c + log sin B log c = 0.77815 log sin B = 9.93753 log h = 0.71568 h = 5.1961. 59. In a right triangle given c, and A = nB ; find a and h. B=^90°-A = 90° - nB. B{n + 1) = 90°. 90° B = n + l cos B a ~ g' 90° a cos 71 + 1 c 90° a = c cos h n + l sin 5 c . 90° _h sm n + l c b 90° = c sir L -. n + l 60. In a right triangle the differ- ence between the hypotenuse and the greater leg is equal to the jdif- ference between tt e two legs ; find the angles. c — a = a — 6. 2a-b = c. (1) a2 + 62 = c2. (2) Squaring (1). 4o2- -4a& + 62 = c2 + 62 = c2 3a2_4a6 = 3a2 = 4a&. 3a = 46. n 4& tan^ = -=i h 3 log tan ^ = log 4 + colog 3. log 4 = 0.60206 colog 3 = 9.52288 - 10 log tan ^ = 10.12494 A =53° 7M8/^. B = 36° 52^ 12^^ TEACHERS EDITION. 37 61. At a horizontal distance of 120 feet from the foot of a steeple, the angle of elevation of the top was found to be 60° 30^ ; find the height of the steeple. tan A = — b a = b tan A. log a = log b + log tan A. log 5 = 2.07918 log tan A = 10.24736 log a = 2.32654 a = 212.1. 62. From the top of a rock that rises vertically 326 feet out of the water, the angle of depression of a boat was found to be 24° ; find the distance of the boat from the foot of the rock. cot A = — a b = a + cot A. log b = log a + log cot a. log a = 2.51322 log cot A = 10.35142 log 6 = 2.86464 6 = 732.22. 63. How far is a monument, in a level plain, from the eye, if the height of the monument is 200 feet and the angle of elevation of the top 3° 30^ ? cot J. = — a b = a cot A. log 6 = log a + log cot A. log a = 2.30103 log cot J. = 1.21351 log b = 3.51454 b = 3270. 64. In order to find the breadth of a river a distance AB was meas- ured along the bank, the point A being directly opposite a tree C on the other side. The angle ABC was also measured. If AB = 96 feet, and ABO= 21° 14^ find the breadth of the river. If ABC= 45°, what would be the breadth of the river ? tsin B = AC ^AB. AC=ABxt&nB. log AC = log AB + log tan B. log AB = 1.98227 log tan B = 9.58944 log^C =1.57171 AC = 37.3 feet. log AC = log AB + log tan B. log AB = 1.98227 log tan B = 10.00000 log^C = 1.98227 AC =96 feet. 65. Find the angle of elevation of the sun when a tower a feet high casts a horizontal shadow b feet long. Find the angle when a = 120, 5 = 70. tan^ = -. b tan A 120 70 ' log tan ^ = log 120 + colog 70. 38 TRIGONOMETRY. log 120 = 2.07918 colog 70 = 8.15490 - 10 log tan J. = 10.23408 A = 59° 44^ 35^^ 66. How high is a tree that casts a horizontal shadow b feet in length when the angle of elevation of the sun is A° ? Find the height of the tree when b = 80°, A = 50°. tan A = -• a = h tan A. log a = log b + log tan A. log 6 = 1.90309 log tan A = 10.07619 log a = 1.97928 a = 95.34. 67. What is the angle of eleva- tion of an inclined plane if it rises 1 foot in a horizontal distance of 40 feet? tan A = -' log tan A = loga + colog b. log a = 0.00000 colog b = 8.39794-10 log tan A = 8.39794 A =1° 25' 56'^ 68. A ship is sailing due north- east with a velocity of 10 miles an hour. Find the rate at which she is moving due north and also due east. Let AB be the direction of the vessel, and equal one hour's progress = 10 miles. A C= distance due east passed over in one hour. As the direction of the ship is north-east, A = 45°. b = c cos A. log b = log c + log cos A. log 10 = 1.00000 log cos A = 9.84949 log b = 0.84949 b = 7.0712 miles due east, and also due north, since AP= AO. 69. In front of a window 20 feet high is a flower-bed 6 feet wide. How long must a ladder be to reach from the edge of the bed to the window ? tan J. = -• log tan A = log 20 + colog 6. log 20 = 1.30103 colog 6 = 9.22185-10 log tan A = 10.52288 A = 73° 18^ a c = sin -4 log c = log 20 -f- colog sin A. log a = 1.30103 colog sin A = 0.01871 log c = 1.31974 c = 20.88. TEACHERS EDITION. 39 70. A ladder 40 feet long may be so placed that it will reach a win- dow 33 feet high on one side of the street, and by turning it over with- out moving its foot it will reach a window 21 feet high on the other side. Find the breadth of the street. p 33 cos B =^ — 40 log 33 = 1.51851 colog 40 = 8.39794 - 10 log cos B = 9.91645 B = 34° 2V 45^^ tan 5 = —. 33 • b = = 33 tan B log 33 = 1.51851 log tan^- = 9.83571 log b - 1.35422 b = = 22.605. cos B'= = 21. 40 log 21 ^ = 1.32222 colog 40 * =8.39794-10 log cos B^= 9.72016 B^ = 58° 19^ 54^^ tan B'= — . 21 6^= 21 tan B^. log 21 = 1.32222 log tan ^^=0.20982 log ¥ = 1.53204 b^ = 34.044 h = 22.605 b + y 56.649 71. From the top of a hill the angles of depression of two succes- sive milestones, on a straight level road leading to the hill, are observed to be 5° and 15°. Find the height of the hill. 5280 sin 5° = 5280 a = 5280 sin 5°. log 5280 =3.72263 log sin 5° = 8.94030 log a = 2.66293 sin 10°= -. c a = c sin 10°. c = -^L_. sin 10° log a = 2.66293 colog sin 10°= 0.76033 logc = 3.42326 cos 75°= ^. c J = c cos 75°. log c = 3.42326 log cos 75°= 9.41300 log b = 2.83626 b = 685.9 feet - 228.63 yards 72. A fort stands on a horizontal plane. The angle of elevation at a certain point on the plane is 30°, 40 TRIGONOMETRY. and at a point 100 feet nearer the fort it is 45°. How high is the fort ? A 100 ft. A' C Let B represent the fort, AC the horizontal plane, £C a ± from fort to plane, BAC= angle made by line from eye of observer = 30°. BA^Q= 45° = angle of elevation 100 feet nearer. From A^ draw A^Nl. to AB. In rt. A AA^N, Z. NAA^=^ 30°, and ^ iV^.4^^ = 60°. .-. NA^= 50 feet. /. AN= \/(100)2- (50)2 = \/7500 = 50\/3 = 86.602. In rt. A BNA' ^^ = cot NBA'= cot 15°, NA' and log NA' log cot 15 log^iV BN AN AB B]Sr= NA' cot 15°. = 1.69897 = 0.57195 = 2.27092 = 186.60 = 86.60 = 273.20 In rt. A ABC, ZBAC= 30°, and Z ABC= 60°. .-. BC=^AB = ^x212>.20 = 136.60 feet. 73. From a certain point on the ground the angles of elevation of the belfry of a church and of the top of a steeple were found to be 40° and 51° respectively. From a point 300 feet farther off, on a horizontal line, the angle of elevation of the top of the steeple is found to be 33° 45''. Find the distance from the belfry to the top of the steeple. Draw DE ± to AB from D. In A BED ^ = sin 33° 45^. BD ^Z) = 300xsin33°45^ log 300 = 2.47712 log sin 33° 45^= 9.74474 log ED = 2.22186 Z EAD = 180°- 33° 45^ - (ISO" 51°) = 17° 15^ In A ADE ED AD = sin 17° 15^. TEACHERS EDITION. 41 AD = ED sin 17° 15^ log ED = 2.22186 colog sin 17° 15^= 0.52791 log AD = 2.74977 = cos 51°. In A ADC DC AD DC= AD cos 51°. log AD = 2.74977 log cos 51° = 9.79887 log DO = 2.54864 In A ADO ^= tan 51°. DC AC=DOiQ.uf>l°. log DC = 2.54864 log tan 51° = 10.09163 log^C = 2.64027 AO = 436.79 In A EDO EC DC = tan 40°. EC = DC tan 40 log DC = 2.54864 log tan 40° = 9.92381 log EC = 2.47245 EC = 296.79. AC- EC = :4o. 74. The angle of elevation of the top of an inaccessible fort C, ob- served from a point A, is 12°. At a point £, 219 feet from A and on a line AB perpendicular to AC, the angle ABC is 61° 45^. Find the height of the fort. In rt. A CAB ^ = cot ABC AC AC = AB cot ABC log AC = log AB + co\ogcot ABC. log AB = 2.34044 colog cot ABC = 0.26977 log AC = 2.61021 In rt. ^ADC ^ = sin CAD. AC CD = AC sin CAD. log CD = log AC+ log sin CAD. log AC =2.61021 log sin CAD = 9.31788 log CD = 1.92809 CD = 84.74 feet. 42 TRIGONOMETRY. Exercise VII. Page 25. 1. In an isosceles triangle, given a and A ; find C, c, h. C= 180° -2 A = 2 (90° - A). i^ = cos A. a c = 2 a cos J.. ^ • A - = sm -4. a h = a sin A. 2. In an isosceles triangle, given a and C; find J., c, A. C + 2^ = 180°. ^ = 90°- i (7. a cos A. c = 2 a cos A. h_ a sin A. h = a sin J.. 3. In an isosceles triangle, c and A ; find C, a, h. C^ 180° -2 A = -2{dO°-A). Jc ^r.c. A 4. In an isosceles triangle, given c and C; find A, a, h. ^ = 90° - J a ^^ A *- = cos J.. a r. a = 2 cos J. given 2a = a = cos J. ^ 2 COS J. — = sm A. a h = asm A. 5. In an isosceles triangle, given h and A ; find C, a, c. (7=2(90°-^). sm A = — a .'. a = h-T- sin A. a 2a .'. c = 2a COS A, 6. In an isosceles triangle, given Ih and C\ find J., a, c. J. = 90° - J G sm -d = — - = sin A. a A = a sin J.. a a = h -T- sin A. cos ^ = 2-2 =_^. a 2a c = 2 a cos J.. 7. In a and ^ an isosceles triangle, ; find A, C, c. given sin J.= Ah- a. C = 180° -2 A. TEACHERS EDITION. 43 = 2(90°-^). a 2a c = 2a cos A. 8. In an isosceles triangle, given c and h ; find A, C, a. tan A = (7= -180°-2^ = = 2(90°-^). sin A = h a = = A -f- sin A. 9. In an isosceles triangle, given a = 14.3 , c = 11 ; find A, C, U. cos A = a log COS A = -- log ^c + colog a. log ^c = - 0.74036 colog a = = 8.84466 - 10 log cos A = = 9.58502 A = 67° 22^ 50^^ C = 2 (90° - A) = 45° 14^ 20^^ sin A = a h = a sin A. log h = log a + log sin A. log a = 1.15534 log sin A = 9.96524 log h = 1.12058 h ^ 13.2. 10. In an isosceles triangle, given a = 0.295, A = 68° 10' ; find c, A, F. sin A = h a h = = a sin A. logA = = log a + Ic )g sin A log a = 9.46982 - -10 log sin A = = 9.96767 log A = 9.43749 - 10 h = 0.27384. cos A = a ic = -- a cos A. log \c = = log a + Ic >g cos A log a = 9.46982 - 10 log cos A = = 9.57044 logic = 9.04026 - 10 ic = 0.109713. c = 0.21943. F= ^ich. 2F^ ■- ch. \q^2F= ■■ log c + lo g^. logc -9.34130- 10 log A = 9.43749 - 10 Iog2i^ r 8.77879 - 10 2F = 0.060089. F 0.03004. 11. In an isosceles triangle, given c = 2.352, (7=69°49^ find a, h, F. I C= 34° 54' 30''. siii|C=l^. a a = + c sm I C loga = log|c+colog sin |C. 44 TRIGONOMETKY. log Jc = - 0.07041 cologsin J(7= = 0.24240 log a = 0.31281 a = = 2.0555. cos 1 C^ h a h = = a cos ^ 0. log A = log a + log cos io. log a = 0.31281 log cos ^ C= = 9.91385 log h = 0.22666 h = 1.6852. 2 F= ch. log 2 i^= log c + log A. log c = 0.37144 log h = 0.22666 log2i?' =0.59810 2F = 3.9637. jP = 1.9819. 12. In an isosceles triangle, given A =7.4847, ^=76° 14^; find a, c, F. sin A = -' sin -4 log a = log h + colog sin A. log h = 0.87417 colog sin A = 0.01266 log a = 0.88683 a = 7.706. tan A=—' Jc = tan .4 log J c = log A + colog tan A logh =0.87417 colog tan A = 9.38918 - 10 log^c =0.26335 ^c =1.8338. c = 3.6676. F=^ch. log i^=log|c + log/i. logjc =0.26335 log h = 0.87417 log i^ = 1.13752 F = 13.725. 13. In an isosceles triangle, given a = 6.71, A = 6.60; find ^, C, c. 8inJ. = ^. a log sin A = log h + colog a. log /i = 0.81954 colog a =9.17328-10 log sin A = 9.99282 A = 79° 36^ 30^^ = 20° 47^ C0S7l = 4^. a ^c = acosA. log J c = log a + log cos A log a = 0.82672 log cos A = 9.25617 log^c =0.08289 |c =1.2103. c = 2.4206. TEACHERS EDITION. 45 14. In an isosceles triangle, given c = 9, ^ = 20 ; find A, c, a. tan ^C=i^. log tan ^ c = log J c + colog A. log^c =0.65321 colog h = 8.69897 - 10 log tan ^(7= -- 9.35218 hG = 12° 40^ 49^^ C = 25° 2V 38^^ 2A = = 180° - a A^ - 77° 19^ 11^^ sin ^ = h a a = h sin JL log a = log A + colog sin A. log h = 1.30103 colog sin A = 0.01072 log a = 1.31175 a = 20.5. 15. In an isosceles triangle, given c = 147, i^= 2572.5; find^.C^^- c log h = log 2 F+ colog c. log2i?' =3.71139 colog c = 7.83268-10 log^ = 1.54407 h =35. tan J. = — log tan A = log h + colog ^ c. log colog log A G h = 1.54407 |c = 8.13371 - 10 tan A = 9.67778 = 25° 28^ = 2 {90° -A) = 129° 4^ a = h log log colog log a sin. A = log ^ + colog sin A. = 1.54407 sin A = 0.36655 a = 1.91062 = 81.41. 16. In an isosceles triangle, given h = 16.8, F= 43.68 ; find A, C, a, c. ic F h log^c = log F -\- colog h logi^ = 1.64028 colog h = 8.77469-10 log^c = 0.41497 ^G = 2.60. G = 5.2. tan A = A log tan A = log h + colog ^ c. log^ = 1.22531 colog I c = 9.58503 - 10 log tan A = 10.81034 A = 81° 12^ 9^^ 1 C =8° 47^ 51^^ C = 17° 35^ 42^^ 46 TRIGONOMETRY. COS A = ^c log a = log I c + colog cos A. logic =0.41497 colog cos A = 0.81547 log a = 1.23044 a =17. 17. In an isosceles triangle, find the value of F in terms of a and c. F= J ch. =v 4 a2 - c2 = *V4a2-c2. ■Z^=2c(jV4a2-c2) = |c\/4a2-c2. 18. In an isosceles triangle, find the value of F in terms of a and C. F= I ch. J c = a sin f C. h = a cos J 0. F= a sin ^ Cx a cos J C. = a^ sin J C cos J C. 19. In an isosceles triangle, find the value of F in terms of a and A. F=lch. ic= a cos A. h = a sin A. F= a cos Ax a sin A = a^ sin J. cos A. 20. In an isosceles triangle, find the value of F in terms of h and C. F= i cA. J c = A tan J C. J'=A(/itaniC) = h^ tan J C. 21. A barn is 40 X 80 feet, the pitch of the roof is 45°; find the length of the rafters and the area of both sides of the roof. 40-2 = 20 = ^c. cos^ = fc^a = 20 -f- a. 20 = a cos A. 20 a = cos -4 log a = log 20 + colog cos A. log 20 = 1.30103 colog cos A = 0.15051 log a = 1.45154 a = 28.284. 28.284 X 80 = 2262.72. 2262.72 X 2 = 4525.44. 22. In a unit circle what is the length of the chord corresponding to the angle 45° at the centre ? sin J (7= i^. a log J c = log a -j- log sin J C. log a = 0.00000 logsin ^(7= 9.58284 logjc =9.58284-10 Jc =0.382683. G = 0.76537. 23. If the radius of a circle = 30, and the length of a chord = 44, find the angle at the centre. TEACHERS EDITION. 47 sinK'=— • a logsin JC= log 2^ c + colog a. logic =1.34242 colog a = 8.52288 - 10 log sin ^(7= 9.86530 IC =47° 10^ C = 94° 20^ 24. Find the radius of a circle if a chord whose length is 5 subtends at the centre an angle of 133°. sin i (7= i^- a log a = log J c + colog sin J C. logic =0.39794 colog sin -^(7= 0.03760 log a =0.43554 a = 2.7261. 25. What is the angle at the cen- tre of a circle if the corresponding chord is equal to | of the radius ? Let a = 3, then c = 2, and J c = 1. sin h C= — ^ 3 log sin i C= log 1 + colog 3. log 1 = 0.00000 colog 3 = 9.52288 - 10 log sin J (7= 9.52288 i C =19° 28^ C = 38° 56^ 33^^^ 19° 28^ 16|^^. 26. Find the area of a circular sector if the radius of the circle = 12 and the angle of the sector = 30°. Area O = irE^. Area sector = 360 log area sector = log 30 -f colog 360 + log TT + 2 log R. log 30 = 1.47712 colog 360 = 7.44370 - -10. logTT = 0.49715 2 log B = 2.15836 log area = 1.57633 Area = 37.699. ExEECisE VIII. Page 26. 1. In a regular polygon given n = 10, c = 1 ; find r, A, F. i(7= 180= = 18°. 10 i e = 0.5. A = 72°. A = 2 c tan A. log h = log i c + log tan A. logic = 9.69897-10 log tan A = 10.48822 log h = 0.18719 h = 1.5388. log r = log i c + colog cos A. logic =9.69897-10 colog cos A = 0.51002 log r = 0.20899 r = 1.618. 48 TE-IGONOMETRY. F = hhp. log h = 0.18719 \ogp = 1.00000 \og2F = 1.18719 2F = 15.388 F = 7.694. 2. In a regular polygon given n=l2,p r 10 ■ find r, h, F. '^G = 15°. A = 75°. c = 70 H- 12 = 5.833. \c = 2.917. h = J c tan A. logic = 0.46494 log tan A = 10.57195 log A = 1.03689 h = 10.886. r = 1 c cos A. log |c = 0.46494 Colog COS A = 0.58700 log r = 1.05194 r = 11.27. F = lhp. log A = 1.03689 logp = 1.84510 A = r sin A. log r = 0.00000 log sin A = 9.99335 log h = 9.99335 h = 0.9848. ^ c = r cos A. log r = 0.00000 log cos A = 9.23967 log i c = 9.23967 - -10 Jc =0.17365. p = 6.2514. F=^hp. log h = 9.99335 - -10 logp =0.79598 log 2 i?' = 0.78933 2F =6.1564. F = 3.0782. log 2 i?' = 2.88199 2F = 762.07. F = 381.04. 3. In a regular polygon given n = 18, r = 1 ; find h, p, F. ^C= 10°. A = 80°. 4. In a regular polygon given n = 20, r = 20 ; find h, c, F. iC= 9°. A = 81°. 7i = r sin A. log r = 1.30103 log sin A = 9.99462 log h = 1.29565 h = 19.754. ^ c = r cos -4. log r = 1.30103 log cos A = 9.19433 log J c = 0.49536 ^c =3.1286. c = 6.257. p = 125.14. TEACHERS EDITION. 49 F^lhp. log A = 1.29565 logp == 2.09740 \og2F = 3.39305 2F = 2472. F = 1236. tan 1 /nr_ 2 <^ 2 O — — - 5. In a regular polygon, given n = 8, h = l] find r, c, F. ^C=22°30^ tan^a=t^. log J c = log ^+log tan \C. log h = 0.00000 log tan ^ C = 9.61722 log^c =9.61722-10 ^c = 0.41421. c = 0.82842. cos I (7=-. r log r = log ^. + colog cos \ C. log h = 0.00000 colog cos ^ C = 0.03438 log r = 0.03438 r = 1.0824. F=ihp. = 3.3137. 6. In a regular polygon, given n = 11, F= 20 ; find r, /i, c. ^0=ph. ^ = S' A = 40. |a=16°22^ h Substituting values of h and c, tania=^^^ = ^. ^ 22 p 880 logi> = i (log 880 + log tan J C). log 880 = 2.94448 log tan ^ a = 9.46788 2) 2.41236 logp = 1.20618 p = 16.076. c = 1.4615. sin 2 0= r log r = log \c-\- colog sir iC log |c : 9.86376 - 10 colog sin 1^ C = = 0.55008 log r = - 0.41384 r = = 2.592. cos 1 (7= r log A = log r + log cos \ 0. log r = 0.41384 log cos ^ C = 9.98204 log h = 0.39588 h = 2.4882. 7. In a regular polygon, given n = 7, i^= 7 ; find r, ^, p. 14 =pA. P |C=25°43^ 50 TRIGONOMETRY. tan J (7= t^. h 14 tanJ(7=^-^-L^=^. 14 p 196 logp = I (log 196 + log tan ^ C). log 196 = 2.29226 log tan ^C =9.68271 2) 1.97497 \ogp = 0.98749 p = 9.716, ^c = 0.694. tan J C=i-'. log h = log I c + colog tan | C. log f c = 9.84136 - 10 colog tan ^C =0.31729 \ogh = 0.15865 h = 1.441. siniC=i^- r logr = log J c + colog sin 1 C. log Jc = 9.84136 - 10 colog sin |(7 =0.36259 log r = 0.20395 r = 1.5994. 8. Find the side of a regular decagon inscribed in a nnit circle. sin J a= i^. r log c = log 2+log sin J C. log 2 = 0.30103 log sin J (7 = 9.48998 log c = 9.79101 - 10 c . =0.6181. 9. decag circle log log log ic c Find the side of a regular ;on circumscribed about a unit ^C=18°. taniC=i^. h log J c = log h + log tan | C. h =0.00000 tan J C = 9.51178 ^c =9.51178-10 == 0.32492. = 0.64984. ^ 10. If the sides of an inscribed regular hexagon is equal to 1, find the side of an inscribed regular do- decagon. Let be the centre of the circle, BC a side of the hexagon, and BA a side of the dodecagon. Also let OD be ± to BA. Then OB = 5C= 1.'' Z BOD = 15°. : In rt. A ODB sin BOD = ^AB^ OB AB ^ 2 OB X sin BOD. log AB = log 2 0B + log sin BOD. log 2 05 = 0.30103 log sin 15° = 9.41300 log AB = 9.71403 - 10 AB =0.51764.- - 11. Given n and c, and let b denote the side of the inscribed reg- ular polygon having. 2 w sides ; find b in terms of n and c. Let O be the centre of the cir- cle, BG ih^ j3i4e of the polygon TEACHERS EDITION. 51 having n sides, BA the side of the polygon having 2n sides. Then OA is ± to 5 C at its middle point D. 360° 180° ZBOA^ ZOBC=dO°- 2n n 180° n The A BOA is isosceles. .\/.OBA=-(l^Q)' Q0° 180^ n A ABC= Z OB A- Z OBC 180 /9Qo_90^\ ^90o_ n ^90° n ic 90° i— = cos " b n Whence, 1 J. 90° ^c = cos n h = he cos 90° n o 90° 2 cos — n 12. Compute the difference be- tween the areas of a regular octa- gon and a regular nonagon if the perimeter of each is 16. ^ 2n 16 1 QAO ^ = i^=22°30^ n , log A = log I c + log cot A. logjc = 0.00000 log cot ^ =10.38278 log A = 0.38278 log F= log h + log Ijp. log h logi^ = 0.38278 = 0.90309 Ic = 1.28587 = 19.3139. • ^=-^ = — = 0.8889. 2n^ 18 A'= 180° 20°. n' log A^= log ^ c'-f log cot A\ logic^ = 9.94885-10 log cot A' ' =10.43893 logA^ = 0.38778 log F^= log h^+ log Ip. log A^ = 0.38778 log^i5 = 0.90309 logi^-' = 1.29087 F' = 19.5377. F'-F = 19.5377 - 19.3139 = 0.2238. 13. Compute the difference be- tween, the perimeters of a regular pentagon and a regular hexagon if the area of each is 12. F= = 12, n = 5. ic== 180° 5 = 36 >°. F= --^hp. h = .24 P ic = -.2L- 2n 10 tan iC= h P 10 24" P' 240 p 52 TKIGONOMETRY. |j2 = 240 tan J C. log 240 = 2.38021 log tan ^C =9.86126 2)2.24147 logp = 1.12074 p = 13.205. n = 6, I C'= 30° tan ^^ ~ 24" 288 V' p^2= 288 tan \ C. log 288 = 2.45939 log tan J C''= 9.76144 2)2.22083 logp^ =1.11042 f =12.895. p-jt)/ =0.310. 14. From a square whose side is equal to 1 the corners are cut away so that a regular octagon is left. Find the area of this octagon. 1 /360' \C^ 22° 30^ A = 90°- 22° 30' = 67° 30^ tan A = \c \c^ h tan A log I c = log ^ + colog tan A. log n = 9.69897 - 10 colog tan ^ =9.61722-10 logic =9.31619-10 log F= log I c + log ^n. logic =9.31619-10 logjn =0.60206 log F = 9.91825 - 10 F = 0.82842. 15. Find the area of a regular pentagon if its diagonals are each equal to 12. C ZAOD AAOC- AACB: ZOAD. ZDAC= 180= = 36« : 180°- 36° = 144°. I (180° -144°) 18° =zaAo. 90°-ZAOD=5i°. 54° +18° = 72°. cos i)^C=^^ = AD^ic AO 12 logic log cos 72° ■■ log 12 logic log 12 + log cos 72°. 9.48998 1.07918 0.56916 TEACHERS EDITION. tani)^0=— • he log h = log J c +logtan 54°. logjc = 0.56916 log tan 54° = 10.13874 log A = 0.70790 p = i c X 2 n. log F= log jc+log n+log h. logjc =0.56916 log n = 0.69897 log h = 0.70790 logi?' =1.97603 F = 94.63. 16. The area of an inscribed reg- ular pentagon is 331.8; find the area of a regular polygon of 11 sides inscribed in the same circle. Let AB be a side of a regular in- scribed pentagon, and AD the side of a regular inscribed polygon of 11 sides. Let a be the radius of the circle whose centre is 0, and h and h^ the apothems of the 2 polygons, respec- tively. Given F the area of pentagon = 331.8. Find F\ the area of 11- sided polygon. Let p and p^ and c and c' repre- sent the perimeters and sides of the pentagon and 11-sided polygon, re- spectively. F== Iph. 331.8 = hph, ph = 663.6. 663.6 h P c = -2. 5 ic = 10 ZAOF= = 36°. tan 36° = ^- = -2. X -2~ h 10 663.6 = Pi-. 6636 log p2 = log tan 36° -h log 6636. log 6636 = 3.82191 log tan 36° = 9.86126 log ^2 =3.68317 logp =1.84159 Since ^ c = y^^ of p, log ^c = 0.84159. sinZ^O^=i^- R log R = log J c -f colog sin 36°. log^c =0.84159 colog sin 36° =0.23078 log R = 1.07237 Z A00= ^^ = 16°21/49^A 22 sin Z AOC= -^c'-^R. logR = 1.07237 log sin u4 0(7-= = 9.44985 log^c^ = 0.52222 tan^OC= Jc^ log h^= log I c^+ colog tan AOC. logjc^ =0.52222 colog tan ^0C= 0.53220 log^i^ =1.05442 54 TEIGONOMETRY. F= Ip'h' = lc'XllXh' log Jc^ = 0.52222 log 11 = 1.04139 log y = 1.05442 logF = 2.61803 F = 414.99. 17. The perimeter of an equilat- eral triangle is 20 ; find the area of the inscribed circle. Perimeter ■■ AB ZOAB tan 30° = 20. ix20 = -i/. i of 180° = 30°. r AB log AB log tan 30° log r Area log IT log r^ log area Area = 0.52288 = 9.76144 = 0.28432 = irr^, = 0.49715 = 0.56864 = 1.06579 = 11.636. 18. The area of a regular poly- gon of 16 sides, inscribed in a circle, is 100 ; find the area of a regular polygon of 15 sides, inscribed in the same circle. 1(7=360^^110 15/. 32 360° ia = ii^ = 12°. ^ 30 Let AC= h, AB = r, BC^ic - _ - F=}hp. 100 = lhp. A = 20p. P P. 32 200 l>2 = 6400 tan J 0. log 6400 = 3.80618 log tan I (7 = 9.29866 tan J C= rrrx- logp P 2 )3.10484 = 1.55242 = 35.68. ^0 = 35.68-32 = 1.115. sinJC=^ = 1.115 log 1.115 = 0.04727 cologsiniC = 0.70976 log r = 0.75703 — = cos J (7'' (12°). h^ = r X cos ^ C. TEACHERS EDITION. 55 log log log log log log r = 0.75703 cos J C^ - 9.99040 ic^ = 0.74743 log log log log F — - = sm 2 r J c^ = r X sin J C\ r =0.75703 sin ia = 9.31788 ^c^ =0.07491 F= log J c^+ log n + log h^. = 0.07491 = 1.17609 = 0.74743 15 1.99843 = 99.640. 19. A regular dodecagon is cir- cumscribed about a circle, the cir- cumference of which is equal to 1 ; find the perimeter of the dodecagon. Given circumference of inscribed = 1, n=12; find p. 2 Trr = circumference. r = circ. 27r \G - 3B0° _ 15 24 o tan 15° r tan 15° 3.1416 log tan 15° = 9.42805 colog 3.1416 = 9.50284 - 10 log c = 8.93089 - 10 log 12 = 1.07918 logp = 0.01007 p = 1.0235. 20. The area of a regular poly- gon of 25 sides is equal to 40 ; find the area of the ring comprised be- tween the circumferences of the in- scribed and the circumscribed circles. ic/i = — = 1.6. ^ 25 ^ C= 7° 12^ 360 A== 2n = 7° 12^. or 2^ = tan I C, h i^ = tan i a A^ = -^^. tan 2^ C log 1.6 colog tan ^ C log h? log h h = 0.20412 = 0.89850 = 1.10262 = 0.55131. r r = log A colog COS ^ C log r log r^ = COS ^ 07 h cos I = 0.55131 = 0.00344 = 0.55475 = 1.10950. irr^ = area of circumscribed O. log TT = 0.49715 log r2 = 1.10950 log F = 1.60665 F = 40.425. log TT = 0.49715 log A2 = 1.10262 logirh^ =1.59977 Area = 39.790,(inscrib'dO) 40.425-39.790 = 0.635. 56 TRIGONOMETRY. Exercise IX. Page 36. 1. Construct the functions of an angle in Quadrant II. What are their signs ? Sines and tangents extending up- wards from horizontal diameter are positive ; downwards, negative. Co- sines and cotangents extending from vertical diameter towards the right are positive ; towards the left, nega- tive. Signs of secant and cosecant are made to agree with cosine and sine, respectively. Hence, sin and esc are -f cos and sec are — tan and cot are — 2. Construct the functions of an angle in Quadrant III. What are their signs ? sin and esc are — cos and sec are — tan and cot are -h 3. Construct the functions of an angle in Quadrant IV. What are their signs ? sin and esc are — cos and sec are + tan and cot are — 4. What are the signs of the functions of the following angles : 340°, 239°, 145°, 400°, 700°, 1200°, 3800° ? 340° is in Quadrant IV. sin = — tan = — sec = + cos = + cot = — esc = — 239° is in Quadrant III. sin = — tan = + sec = — cos = — cot = -f esc = — 145° is in Quadrant II. sin = -t- tan = — sec = — cos = — cot = — CSC = -f- 400 = 360° -f 40° = signs of func- tions of 40°. 40° is in Quadrant I. sin = + tan = -f sec = + cos = + cot = + CSC = -t- 700° = 360° -f 340° = signs of the functions of 340°. 340° is in Quadrant IV. sin = — tan = — sec = + cos = + cot = — esc = — 1200° = 3 X 360° + 120° = signs of the functions of 120°. 120^ is in Quadrant II. sin = -h tan = — sec = — cos --= — cot = — CSC = + 3800° = 10 X 360° + 200° of the functions of 200°. signs 200° is in Quadrant III. sin = — tan = -f sec = — cos = — cot = + CSC = — 5. How many angles less than 360° have the value of the sine equal to -I- f, and in what quad- rants 'do they lie? . TEACHERS EDITION. 57 Since the sine is +, by § 24, the angles can lie in but two quadrants, the first and second. In the first quadrant, by | 3, the sine increases from to 1, and in the second, decreases from 1 to 0. This is a continually increasing and decreasing quantity. Therefore there can be but one angle whose sine is equal to + f in each quadrant, the first and second. 6. How many values less than 720° can the angle x have if cos x = +f, and in what quadrants do they lie ? 720° is twice 360°; hence the moving radius will make exactly 2 complete revolutions. The cosine has the + sign in the first and fourth quadrants, hence it will have four values : two in Quad- rant I. and two in Quadrant IV. 7. If we take into account only angles less than 180°, how many values can x have if sin a; = f ? if cos a; = ^ ? if cos x = — | ? if tan x = f? ifcotaj = -7? (i.) Sign being +, the angle can be in Quadrant I. or II. .-. two values, one in Quadrant I. and one in Quadrant II. (ii.) Sign being -f, the angle is in Quadrant I. or IV. .•. two values, one in Quadrant I. and one in Quadrant IV. (iii.) Sign being — , the angle can be in Quadrant II. or III. .•. two values, one in Quadrant II. and one in Quadrant III. (iv.) Sign being +, the angle can be in Quadrant I. or III. .•. two values, one in Quadrant I. and one in Quadrant III. (v.) Sign being — , the angle can be in Quadrant II. or IV. .•. two values, one in Quadrant II. and one in Quadrant IV. 8. "Within what limits must the angle x lie if cosx = — |? if cot a; = 4? if sec a; = 80? ifcsca; = -3? {x to be less than 360°.) If cos X = — f , X must lie in the second or third quadrant, or between 90° and 270°. If cot a; = 4, X is between 0° and 90°, or 180° and 270°. If sec a; = 80, a; is between 0° and 90° or 270° and 360°. If CSC a; = — 3, a; is between 180° and 360°. 9. In what quadrant does an angle lie if sine and cosine are both negative ? if cosine and tangent are both negative ? if the cotangent is positive and the sine negative? (i.) Sine is negative in Quadrants II, and III. ; cosine is negative in Quadrants III. and IV. .•. angles having both sine and cosine negative are in Quadrant III. (ii.) Cosine is negative in Quad- rants II. and III. ; tangent is nega- tive in quadrants II. and IV. .•. angles having both cosine and tangent negative are in Quadrant II. (iii.) Cotangent is positive in Quadrants I. and III. ; sine is neg- ative in Quadrants III. and IV. .'. angles having cotangent posi- tive and sine negative are in Quad- rant III. 58 TRIGONOMETRY. 10. Between 0° and 3600° how many angles are there whose sines have the absolute value f ? Of these sines how many are positive and how many negative ? Between 0° and 3600° there are 10 revolutions, and in each there are 4 angles whose sines have the absolute value f. .*. there are 40 angles. The sine is positive in Quadrants I. and II., and negative ing Quadrants III. and IV. .•. there are 20 angles with the sine positive, and 20 with the sine negative. 11. In finding cos x by means of the equation cos x = ± Vl — sin^ x, when must we choose the positive sign and when the negative sign ? Since cosines only of angles in Quadrants I. or IV. are positive, we use the sign + only when angle x lies within these limits. Also, since cosines of angles in Quadrants II. and III. are nega- tive, we use the sign — , when x is known to lie in either of these. 12. Given cos a; = - V| ; find the other functions when x is an angle in Quadrant II. sin^x + cos^a; = 1. sin X = Vl — cos^o; = Vl - {-VlY = VJ. 11/9 esc X = = — r: = "v^- sm x V^ sec x = cos 03 tan 05 = cos a; —VI = -V2. sin^^_Vj_ ^ _ 1 cot X = = — = — 1. tan X — 1 13. Given tana;= V3; find the other functions when x is an angle in Quadrant III. tana3= VS. cot X = - — = ^y/S. tancc V3 sin X cos a; tan X X cos x = sin x. Vs cos a; = sin x. 3 cos% — sin^a; = cos^x + sin^a; = 1 4 cos^a; cos'^a; cos X = 1 ±i The angle being in Quadrant III. the cosine is negative. .". cos X = -i- sin X = Vi-(- -W = V| = ± JV3. Sine is negative. .*. sin X = -jV3. sec X 1 1 -2. esc X 1 = -# -*V3 14. Given sec a; = + 7, and tan x negative ; find the other functions of x. X must be in Quadrant IV. .•. sine, cosine, tangent, and co- tangent will be negative, and cosine positive. TEACHEES EDITION. 59 cos X = sec a; sma; ^ 49 >'49 CSC a; = tana; = cot a; = - f V3, 1 1 sin a; _ |V3 = -tW3. sin X - -|V3 cos X 1 = - 4 V3. 1 tan X 1 4V3 = i^^V3. 15. Given cot a; = — 3; find all the possible values of the other functions. By [3] tan x = — ^, and may be in Quadrant II. or IV. By [1], sin^a; = 1 — cos'^x. sin X = Vi — cos^a;. By [2], 1 Vi — coa^a; 3 ( 308 a; 1_ 1- cos^a; y cos-'a; cos^a; = 9 — 9 cos^a;. cos^a; = 9 lo' cos X = Vio and is — in Quadrant II., + in IV By [1], sia x = \l = \ — \ in \in 10 ^10 = i^7VT0, and is + in Quadrant II., — in IV. Vio sec a; = iVio. CSC a; = VTO, 16. What functions of an angle of a triangle may be negative ? In what case are they negative ? When an angle of a triangle is acute, its functions are all positive. When an angle is obtuse, its func- tions are those of an angle in Quad- rant II. .•. sine and cosecant are positive, and cosine, tangent, cotangent, and secant are negative. 17. What functions of an angle of a triangle determine the angle, and what functions fail to do so ? The sine and cosecant being posi- tive in the first and second quad- rant, leave it doubtful whether the angle is obtuse or acute ; but the other functions, if positive, deter- mine an angle in the first quadrant, that is to say, an acute angle ; if negative, an angle in the second quadrant, an obtuse angle. 18. Why may cot 360° be con- sidered equal either to -I- oo or to -oo? The nearer an acute angle is to 0°, the greater the positive value of its cotangent ; and the nearer an angle is to 360°, the greater the 60 TRIGONOMETRY. negative value of its cotangent. When the angle is 0° or 360°, co- tangent is parallel to the horizontal diameter and cannot meet it. But cotangent 360° may be regarded as extending either in the positive or in the negative direction ; and hence either + oo or — oo . 19. Obtain by means of Formu- las [l]-[3] the other functions of the angles given : (i.) tan 90° = oo. (ii.) cos 180° = - 1. (iii.) cot 270° = 0. (iv.) esc 360° = - 00. (i-) tan90° = oo=l. cot 90° = ~ CO sin 90° cos 90° cos 90° -0 sin 90° = 0. cos2 90° + sin^ 90° = 1. sin2 90° = 1. sin 90° = 1. (ii.) cos 180° = -1. sin2 180° + cos2 180° = 1. sin2 180° + 1 = 1. sin 180° = 0. sin 180° tan 180° cot 180° = cos 180° - 1 = -0. cos 180° _ - 1 sin 180° (ill-) cot 270° = 0. tan 270° = - = oo. cos 270° 0. sm 270° cos 270° = sin 270° = 0. sin2 270°-hcos2 270° = l. sin2 270° + = l. sin2 270° = l. sin 270° = - 1. (IV.) CSC 360° = - oo. sin 360° = — = - 0. — 00 sin2 360° + cos2 360° = l. cos2 360° = l. cos 360° = 1. tan 360° = ^ = - 0. cot 360° = -!- = - 00. -0 20. Find the values of sin 450°, tan 540°, cos 630°, cot 720°, sin 810°, CSC 900°. sin 450° = sin (360° + 90°) = sin 90° = 1. tan 540° = tan (360° + 180°) = tan 180° = 0. cos 630° = cos (360° + 270°) = cos 270° = 0. TEACHERS EDITION. 61 cot 720° = cot (360° + 360°) = cot 360° = 00. sin 810° = sin (2 x 360° + 90°) = sin 90° = 1. esc 900° = esc (2 x 360° +180°) = CSC 180° = 00. 21. For what angle in each quadrant are the absolute values of the sine and cosine alike ? The sine and cosine of 45° are equal in absolute value. Corre- sponding to the angle of 45° in the first quadrant are the angles (90° + 45°), (180° + 45°), (270° + 45°) in the second, third, and fourth quad- rants. Hence the sines and cosines of 45°, 135°, 225°, 315°, etc., are all equal in absolute value. 22. Compute the value of a sin 0° + 5 cos 90° - c tan 180°. sin 0° = 0. cos 90° = 0. tan 180° = 0. Substituting, axO + 6xO-cxO=0. 23. Compute the value of a cos 90° - h tan 180° + c cot 90°. cos 90° = 0. tan 180° = 0. cot 90° =0. Substituting, ax0-6x0 + cx0 = 0. 24. Compute the value of a sin 90° - 5 cos 360° + (a - 6) cos 180°. sin 90° = 1. cos 360° = 1. cos 180° = - 1. ■ Substituting, axl-6xl+(a-5)x-l = 0. 25. Compute the value of (a2-52)cos 360°- 4 ah sin 270°. cos 360° = 1. sin 270° = - 1. Substituting, (a2-52)xl-4a&X-l = a2-62 + 4a5. Exercise X. Page 41. 2. Express sin 172° in terms of the functions of angles less than 45°. sin 172° = sin (180° - 8°) = sin 8°. 3. Express cos 100° in terms of the functions of angles less than 45°. cos 100° = cos (90° + 10°) = - sin 10°. 4. Express tan 125° in terms of the functions of angles less than 45°. tan 125° = tan (90° - 35°) = - cot 35°. 62 TRIGONOMETRY. 5. Express cot 91° in terms of the functions of angles less than 45°. cot 91° = cot (90° + 1°) = - tan 1°. 6. Express sec 110° in terms of the functions of angles less than 45°. sec 110° = sec (90° + 20°) = - CSC 20°. 7. Express esc 157° in terms of the functions of angles less than 45°. CSC 157° = esc (180° -23°) = CSC 23°. 8. Express sin 204° in terms of the functions of angles less than 45°. sin 204° = sin (180° + 24°) = - sin 24°. 9. Express cos 359° in terms of the functions of angles less than 45°. cos 359° = cos (360° - 1°) = cos 1°. 10. Express tan 300° in terms of the functions of angles less than 45°. tan 300° = tan (270° + 30°) = - cot 30°. 11. Express cot 264° in terms of the functions of angles less than 45°. cot 264° = cot (270° -6°) = tan 6°. 12. Express sec 244° in terms of the functions of angles less than 45°. sec 244° = sec (270° -26°) = -csc26°. 13. Express esc 271° in terms of the functions of angles less than 45°. CSC 271° = CSC (270° - 1°) = -secl°. 14. Express sin 163° 49'' in terms of the functions of angles less than 45°. sin 163° 49^= sin (180°- 16° 110 = sin 16° IV. 15. Express cos 195° 33^ in terms of the functions of angles less than 45°. cos 195° 33^= cos (180°+ 15° 33^ = - cos 15° 33^ 16. Express tan 269° 15^ in terms of the functions of angles less than 45°. tan 269° 15^= tan (270°- 450 = cot45^ 17. Express cot 139° 17^ in terms of the functions of angles less than 45°. cot 139° 17^= cot (180°- 40° 430 = - cot 40° 43^ 18. Express sec 299° 45'' in terms of the functions of angles less than 45°. sec 299° 45^= sec (270°+ 29° 450 = CSC 29° 45^. 19. Express esc 92° 25'' in terms of the functions of angles less than 45°. esc 92° 25^= CSC (90°+ 2° 250 = sec 2° 25^ 20. Express all the functions of — 75° in terms of those of positive angles less than 45°. TEACHERS EDITION, 63 sin (- 75°) = sin (270° + 15°) = - cos 15°. cos (- 75°) = cos (270° + 15°) = sin 15°. tan(- 75°) = tan (270° + 15°) = - cot 15°. cot (- 75°) = cot (270° + 15°) = - tan 15°. 21. Express all the functions of — 127° in terms of those of positive angles less than 45°. sin (- 127°) = sin (270° - 37°) = - cos 37°. cos (- 127°) = cos (270° - 37°) = - sin 37°. tan (- 127°) = tan (270° - 37°) = cot 37°. cot (- 127°) = cot (270° - 37°) = tan 37°. 22. Express all the functions of — 200° in terms of those of positive angles less than 45°. sin (- 200°) = sin (180° - 20°) = sin 20°. cos (- 200°) = cos (180° - 20°) = - cos 20°. tan (- 200°) = tan (180° - 20°) = -tan20°. cot (- 200°) = cot (180° - 20°) = - cot 20°. 23. Express all the functions of — 345° in terms of those of positive angles less than 45°. sin (- 345°) = sin 15°, etc. 24. Express all the functions of — 52° 37'' in terms of those of posi- tive angles less than 45°. sin (- 52° 370 = sin (270°- 37° 23^ = - cos 37° 23^ cos (- 52° 370 = cos (270°+ 37° 23^ = sin 37° 23^ tan (- 52° 370 = tan (270°+ 37° 230 = - cot 37° 23^. cot (- 52° 370 = cot (270°+ 37° 230 = -tan37°23^ 25. Express all the functions of — 196° 54-^ in terms of those of posi- tive angles less than 45°. sin (- 196° 540 = sin (180°-16°540 = sin 16° 54^ cos (- 196° 540 = cos (180°-16° 540 = - cos 16° 54^, tan (- 196° 540 = tan (180°-16° 540 = -tanl6°54^ cot (- 196° 540 = cot (180°-16° 540 = -cotl6°54^ 26. Find the functions of 120°. sin 120° = sin (90° + 30°) = cos 30° = |V3. cos 120° = cos (90° + 30°) = - sin 30° = - J. tan 120° = tan (90° + 30°) = - tan 30° = -Vs. cot 120° = cot (90° + 30°) = -cot30° = -^V3. sec 120° = - 2. CSC 120° = f Vs. 64 TRIGONOMETRY. 27. Find the functions of 135°. sin 135° = sin (90° + 45°) cos 45° = I V2. cos 135° = cos (90° + 45°) = -sin45° = - jVi sin 135° tan 135° = cos 135° cotl35° = ^"^^^^° = -l. sin 135° 1 sec 135° = ^— = - \/2. CSC 135° = cos 135° 1 V2. sin 135° 28. Find the functions of 150°. sin 150° = sin (180° - 30°) = sin 30° = J. cos 150° = COS (180° - 30°) = -cos30° = -jV3. tanl50° = tan(180°-30°) sin 30° = -tan30° = - cos 30° = -iV3. cot 150° = cot (180° - 30°) - cos 30° = - cot 30° = sin 30° = -V3. sec 150° = sec (180° - 30°) = - sec 30° = - cos 30° = -fV3. CSC 150° = CSC (180° - 30°) = CSC 30° = sin 30° = 2. 29. Find the functions of 210°. sin 210° = sin (180° + 30°) = - sin 30° = - |. cos 210° = cos (180° + 30°) = -cos30° = -A\/3. tan 210° = tan (180° + 30°) = tan30° = J\/3. cot 210° = cot (180° + 30°) = cot 30° = V3. 30. Find the functions of 225°. sin 225° = sin (180° + 45°) = -sin45° = -|\/2. cos 225° = cos (180° + 45°) = -cos45° = - J\/2. tan 225° = tan (180° + 45°) = tan 45° = 1. cot 225° = cot (180° + 45°) = cot45° = l. 31. Find the functions of 240°. sin 240°= sin (270° - 30°) = -cos30° = -jV3. cos 240° = cos (270° - 30°) = - sin 30° = -h tan 240° = tan (270° -30°) = cot 30° = Vs. cot 240° = cot (270° - 30°) = tan 30° = ^ Vs. 32. Find the functions of 300°. sin 300° = sin (270° + 30°) = -cos30° = -jV3. cos 300° = cos (270° + 30°) 1 2- TEACHERS EDITION. 65 tan 300°= tan (270° + 30°) = - cot 30° = - V3. cot 300° = cot (270° + 30°) = -tan30°=-jV3. 33. Find the functions of - 30°. sin - 30° = - sin 30° = - i. cos - 30° = tan - 30° = cot - 30° = sec - 30° = esc - 30° = cos 30° = J Vs. -tan30° = -^V'3. - cot 30° = - Va. sec 30° = |V3. - CSC 30° = - 2. 34, Find the functions of -225°. - 225° = 90° + 45°. sin - 225° = sin (90° + 45°) = cos 45° = \V2. cos - 225° = cos (90° + 45°) = - sin 45° = ~ I V2. tan - 225° = tan (90° + 45°) = - cot 45° = - 1. cot - 225° = cot (90° + 45°) = - tan 45° = - 1. 1 sec - 225° = cos (90° + 45°) = -V2. 1 =V2. CSC - 225° = — sin (90° + 45°) 35. Given sin a; = —Vl, and cos a? negative ; find the other functions of X, and the value of x. Since sin 45° == V|, and the signs of both the sine and cosine are neg- ative, the angle must be in Quadrant III., and must be, therefore, 180° + 45° = 225°. Then cos45°=\/J. Hence cos (180° + 45°) = - VJ. sin 225^ -T ^^ ) — cos 225° = -x4 (180° + 45°) = 1 tan 225° 225° 1 1 COS 225 ° -vi = - V2. CSC 225° 1 1 sm 225° _VJ = -V2. 36. Given cot x = — \/3, and x in Quadrant II. ; find the other func- tions of X, and the value of x. Since cot30°=\/3, and the sign is negative, the angle is in Quad- rant II. tan X = ■ = ■ = — jVS. cot a; — V3 !ilLiE = -|V3. cos X sin x = — I VS cos X. sin^a; = ^ cos^a:. But sin'^aj + cos^a? = 1, .'. J cos^a; + cos^aj = 1 ; and and cos^a; ,*. cos x= ^v3 ; • 2 1 4 .'. sin x = — 2 sec X = = fV3. cos a? esc X = - — = 2. sm X 66 TKIGONOMETBY. 37. Find the functions of 3540°. 3540« = 9 X 360° + 300°. sin 300° = sin {S&)P - m^} = sin 60° = - 1 VS. cos aOO^' = cos (360° - 60°) tan 300^ = cot 300° cos 60^ = i. 2 smg00^ ^ -|V3 cos SOO'* i 1 1 secSCKT esemF = tan 300° _v^ -|V3. 1 €Os3(X>'^ i = - = 2L sin 300*^ ^V3 Sa What angles less ^an 360°' have a sine equal to — J ? a tangent equal to — V3 ? (i.) ^aee sin 30^= J and the sign is negative, tlie angle nrast be in Qnadraat III. or IV., and mnst be therefore 180° + 30° = 210°, or 360° -30° = 330°. (ii.) Since tan 60° =V3 and the sign is negative, the angle must be in Quadrant II. or IV., and must be therefore 180°-60° = 12CP, or 360° -60° = 300°. 39. Which of the angles men- tioned in Examples 27-34 have a cosine equal to — VJ? a cotangent equal to — V3 ? (i.) Since cos45°=V^ and the sign is negative, the angle must be in Quadrant II. or III., and must be therefore 180° - 45° = 135°, or 180° + 45° = 225°. Also, the func- tions of — 225° are the same as the functions of 360° - 225° = 135°. Hence the angles are 135°, 225°, or -225°. (ii.) Since cot 30° = V3 and the sign is negative, the angle must be in Quadrant II. or IV., and must be therefore 180° - 30° = 150°, or 360° - 30° = 330°, or - 30°. Hence the angles are 150° or — 30°. 40. What values of »■ between O*' and 720° will satisfy the equation sin X = 4r 2? Since sin 30° = -| and the sign is positive,, the angle must be in Quad- rant I. or II., and must be therefore 30°, or 180°- 30° = 150°, the first revolution. In the second revolu- tion, these angles must be increased by 360°. Hence the angles are 30°, 150°, 390°. and 510°. 41. In each of the following eases find the other angle between 0° and 360° for which the corresponding function (sign included) has the same value : sin 12°, cos 26°, tan 45°, cot 72° ; sin 191°, cos 120°, tan 244°, cot 357°. In order that the sign shall be the same, sin 12° must be in Quadrant II. = sin(lSO^- 12°) = sin 168°. cos 26° miist be in Quadrant IV. = cos (360° - 26°) = cos 334°. tan 45° must be in Quadrant III. = tan (180° + 45°) = taa 225^ TEACHERS EDITION. 67 cot 72° must be in Quadrant III. = cot (180° + 72°) = cot 252°. sin 191° must be in Quadrant IV. = sin (360° - 11°) = sin 349°. cos 120° must be in Quadrant III. = cos (180° + 60°) = cos 240°. tan 244° must be in Quadrant I. = tan (244° - 180°) = tan 64°. cot 357° must be in Quadrant II. = cot (357° - 180°) = cot 177°. 42. Given tan 238° = 1.6; find sin 122°. tan 238° = tan (180° + 58°) -tan 58°. sin 122° = sin (180° - 58°) , = sin 58°. But tan 238° = 1.6. .-. tan 58° = 1.6. sin 58° tan 58° = 1 cos 58° sin 58° Vi-sin258° 2.56 - 2.56 sin2 58° = sin2 58°. 3.56 sin2 58° = 2.56. sm 58° = -\/ ■ ^'3.56 = 0.848. 43. Given cos 333° = 0.89 ; find tan 117°. cos 333° = 0.89. = cos (270° + 63°) = sin 63° = tan (180° -63°) = - tan 63°. Bin2 63° + cos2 63° = 1. (0.89)2 + cos2 63° = 1. cos2 63° = 0.2079. cos 63° = 0.456. tan 63°= '^^ ^^° cos 63° 0.89 - 1.952 0.456 44. Simplify the expression a cos (90° -x)+h cos (90° + x) = a sin X + b{— sin x) = sin x{a — h). 45. Simplify the expression m (cos 90° - a;) sin (90° -x). cos (90° — x) = sin x. sin (90° — x) = cos x. .'. the expression =msmx cos x. 46. Simplify the expression (a -J) tan (90° -k) + (a + &)cot(90°+4 tan (90° - x) = cot x. cot (90° + x) = — tan x. .'. the expression equals (a — b) cot X — {a + b) tan x. 47. Simplify the expression a2 + 62_2a6cos(180°-a;) = a^ -\- b^ — 2ab {— cos x) = a^ + b^ + 2ab cos x. 48. Simplify the expression sin (90° + x) sin (180° + x) + cos (90° + x) cos (180° - x) = (cos x) (— sin a?)+(— sin x)(— cos x) = — sin cos X + sin cos x = 0. 49. Simplify the expression cos (180° + x) cos (270° - y) - sin (180° + x) sin (270° - y). 68 TRIGONOMETRY. COS (180° + x) = — cos X. cos (270° —y) = — sin y. sin (180° + x)^ — sin x. sin (270° — y) = — cos y. Hence the expression = cos X sin y — sin x cos 3/. 50. Simplify the expression tan x + tan (— y) — tan (180° — y). tan (— 2/) = — tan 1/. -tan (180° -3/) = tan 2/. Hence the expression = tan x. 51. For what values of x is the expression sin x + cos x positive, and for what values negative ? Rep- resent the result by a drawing in which the sectors corresponding to the negative values are shaded. If X be any angle in Quadrant I., sin X + cos X must be positive since both the sine and cosine are posi- tive. In Quadrant II. the sine is positive and cosine negative ; hence, so long as the sine is greater than, or equal to, the cosine, the expres- sion sin a? -f cos re is positive; but after passing the middle of Quad- rant II., viz., 135°, the cosine of Zx is greater than sine, and the ex- pression is negative. In Quadrant III. both sine and cosine are nega- tive, and hence their sum must be negative. In Quadrant IV. the sine is negative and cosine posi- tive. The sine and cosine are equal at 315°, after which the cosine is greater than sine. Hence the ex- pression sin X + cos X is negative from 135° to 315°, and positive be- tween 0° and 135°, and 315° and 360°. 52. Answer the question of last example for sin x — cos x. As X increases from 0° to 45°, the sine increases in value, and cosine decreases, until at 45° sine = cosine. Hence up to this point sin .r — cos it is negative. For the remainder of Quadrant I. sine is greater than co- sine, and consequently the expres- sion sin X — cos X is positive. In Quadrant II. sine is positive and cosine negative, so the expression sin X — cos X is uniformly positive. In Quadrant III. sine is negative and cosine negative ; hence, so long as sine is less than cosine, the ex- pression is positive, viz., to 225°; after that point, sine is greater than cosine, and sin x — cos x is negative. In Quadrant IV. sine is negative and cosine positive : therefore sin x — cos a; is uniformly negative. The expression is, then, negative be- tween 0° and 45°, and 225° and 360°; positive between 45° and 225°. 53. Find the functions of (x — 90°) in terms of the functions ^^^- X- 90° = 360° - (90° - x) = 270° -f X. sin {x - 90°) - sin (270° + x) = — cos X. cos {x - 90°) - cos (270° -1- x) = sin X. tan {x - 90°) - tan (270° -f x) = — cot X. cot {x - 90°) = cot (270° + x) = — tan X. 54. Find the functions of {x — 180°) in terms of the functions of cc. TEACHERS EDITION. 69 X - 180° = 360° - (180° - X) = 180° + X. sin {x - 180°) = sin (180° + x) = — sin X. cos {x - 180°) = cos (180° + x) = — cos X. tan (a; - 180°) = tan (180° - x) = tan X. cot {x - 180°) = cot (180° + x) = cot X. Exercise XI. Page 48. 1. Find the value of sin {x + y) and cos {x + y) when sin a? = |, cos x = f, sin 2/ -tV. cosy 12 13- sin (x + y) = sin x cos y + cos x sin y + '5^13 36 20 _ 56^ 65 65 65* 5^13 cos (a? + 3/) = cos X cos :« — sin x sin 1/ "'5^13) (5^13 48 65 15 65 33 65" 2. Find sin (90° — y) and cos (90° -y) by making a; = 90° in Formulas [8] and [9]. sin (90° - y) = sin 90° cos y — cos 90° sin y. sin 90° = 1. cos 90° = 0. .-. sin (90° - y) = (1 X cosy) — (0 X sin y) = cos y. cos (90° - y) = cos 90° cos y + sin 90° sin y = (0 X cos y) + (1 X sin y) = sin y. 3. Find, by Formulas [4H11], the first four functions of 90° + y. sin (90° + y) = sin 90° cos y + cos 90° siny = (1 Xcosy) + (Oxsiny) = cos y. cos (90° + y) = cos 90° cos y — sin 90° sin y = (0 X cos y) — (1 X sin y) = — sin y. (tan 90° + y) cos y , = -, — ^ = — cot y. sin y cot (90° + y) sin V , = ^ = — tan y. cosy 4. Find, by Formulas [4]-[ll], the first four functions of 180° — y. sin (180° - y) = sin 180° cosy — cos 180° siny = (0 X cos y) — (— 1 X sin y) = sin y, cos (180° - y) = cos 180° cos y + sin 180° sin y = (— 1 X cos y) + (0 X sin y) = — cosy. tan (180° - y) = ^ = — tan y. cosy cot (180° - y) cos V , = ; — ^ = — cot y. sin y 70 TRIGONOMETEY. 5. Find, by Formulas [4]-[ll], the first four functions of 180° + y. sin (180° + y) = sin 180° cos y + cos 180° sin y = (0 X cos y) + (- 1 X sin y) = — sin y. cos (180° + y) = cos 180° cos y — sin 180° sin y = (— 1 X cos 2/) — (0 X sin y) = — cos y. tan (180° + y) — sin v , = ^ = tan y. — cosy cot (180° + y) — cos y , = : — ^ = cot y. — sin y 6. Find, by Formulas [4]-[ll], the first four functions of 270° — y. sin (270° - y) = sin 270° cos y — cos 270° sin y = (— 1 X cos 2/) — (0 X sin y) = — cos y. cos (270° - y) = cos 270° cos y + sin 270° sin y = (0 X cos 2/) + (- 1 X sin y) = — sin y. tan (270° -2/) — cos v . = — ; — ^ = cot y. — sin y cot (270° - y) = :i^iM = tan2/. — cos 2/ 7. Find, by Formulas [4]-[ll], the first four functions of 270° + y. sin (270° + y) = sin 270° cos y + cos 270° sin y = (- 1 X cos 2/) + (0 X sin y) = — cos y. cos (270° + y) = cos 270° cos y — sin 270° sin y = (0 X cos 2/) — (— 1 X sin y) = sin y. tan (270° + y) — cosy , = — : ^ = — cot y. sm y cot (270° + y) sin V i. = "^ = — tan y. — cos y 8. Find, by Formulas [4]-[ll], the first four functions of 360° — y. sin (360° - y) = sin 360° cos y - cos 360° sin y = (0 X cos 2/) — (1 X sin y) = — sin y. cos (360° - y) = cos 360° cos y + sin 360° sin y = (1 X cos 2/) + (0 X sin y) = cos 2/. tan (360° - y) — sin v , = -^ = — tan y. cosy cot (360° -2/) cos y , = — r-"^ = — cot y. — sin 2/ 9. Find, by Formulas [4]-[ll], the first four functions of 360° + y. sin (360° + y) = sin 360° cos y + cos 360° sin y = (0 X cos 2/) + (1 X sin y) = sin 2/. cos (360° + y) = cos 360° cos y — sin 360° sin y = (1 X cos y) — {Ox sin y) = cos y. tan (360° +2/) _ sm_y ^ ^^^ ^^ cos 2/ TEACHERS EDITION. 1 cot (360° + y) cos y , = — — ^ = cot y. siny 10. Find, by Formulas [4]-[ll], the first four functions of x — 90°. sin {x - 90°) = sin X cos 90° — cos x sin 90° = (0 X sin a;) — (1 X cos x) = — cos X. cos (x - 90°) = cos X cos 90° + sin x sin 90° = (0 X cos x) +{lx sin x) = sin X. tan (a; - 90°) — cos a; = — cot X. sin X cot {x - 90°) _ sin X — cosx = — tan X. 11. Find, by Formulas [4]-[ll], the first four functions of a; — 180°. sin {x - 180°) = sin x cos 180° — cos x sin 180° = sin x (— 1) — cos a; X = — sin X. cos (a; - 180°) = cos X cos 180° + sin x sin 180° = cos X (— 1) + sin a; X = — cos X. tan (a; - 180°) — sin X = tan X. — cos X cot (a; - 180°) — cos X cot a;. sin X 12. Find, by Formulas [4]-[ll], the first four functions of a;— 270°. sin (a; - 270°) = sin X cos 270° — cos x sin 270° = sin a; X — cos a; X (— 1) = cos X. cos (a; - 270°) = cos X cos 270° + sin x sin 270° = cos a; X + sin x (— 1) = — sin X. tan (a; -270°) cos X = — cot X. — sm X cot (a; - 270°) _ — sin X _ cos X — tan a;. 13. Find, by Formulas [4]-[ll], the first four functions of — y. sin (0° - y) = sin 0° cos y — cos 0° sin y = (0 X cos y) — (1 X sin y) = — sin y. cos (0° - y) = cos 0° cos y + sin 0° sin y = (1 X cos 2/) + (0 X sin y) = cos y. tan(0°-3/) — sin V . = ^ = — tan y. cosy cot(0°-y) cos y , = — -"^ = — cot v. — siny 14. Find, by Formulas [4]-[ll], the first four functions of 45° — y. sin (45° — y) = sin 45° cos y — cos 45° sin y = 2^V2 cos y — ^\/2 siny = ^ V2 (cos y — sin y). 72 TRIGONOMETEY. COS (45° — y) = cos 45° cos y + sin 45° siny = |V2 cos 2/ + ^ y/2 sin y = \yf2 (cos 2/ + sin y). tan (45° - y) _ cos y — sin y _ 1 — tan y cos y + sin 3/ 1 + tan y cot (45° - y) _ cos y + sin y _ cot y + 1 cos y — sin y cot y — 1 15. Find, by Formulas [4]-[ll], the first four functions of 45° + y. sin (45° + y) = sin 45° cos y + cos 45° sin y = iV2 cosy + jVlsiny = J V2 (cos y + sin y). cos (45° + y) = cos 45° cos y — sin 45° sin y 5= J V2 cos y — ^\/2 sin y = J V2 (cos y — sin y). tan (45° + y) _ cos y + sin y _ 1 + tan y cos y — sin y 1 — tan y cot (45° + y) _ cos y — sin y _ cot y — 1 cos y 4- sin y coty + 1 16. Find, by Formulas [4] -[11], the first four functions of 30° + y. sin (30° + y) = sin 30° cos y + cos 30° sin y = \ (cos y + \/3 sin y). cos (30° + y) = cos 30° cos y — sin 30° sin y = J ( V3 cos y — sin y). tan (30° + y) _ cos y + \/3 sin y ^ Vo cos y — sin y divide each term by VS cos y, _ ^ V3 + tan y 1 — jV3 tany cot (30° + y) _ V3 cos y — sin y . cosy + VSsiny divide each term by sin y, = "V^ cot y — 1 coty + VS 17. Find, by Formulas [4]-[ll]. the first four functions of 60° — y. sin (60° - y) = sin 60° cos y — cos 60° sin y — K"^ cos y — sin y). cos (60° - y) = cos 60° cos y + sin 60° sin y = 2 (cos y + \/3 sin y). tan (60° - y) _ Vs cos y — sin y cos y + V3 sin y _ V3 — tan y 1 + VS tan y cot (60° -y) _ cos y + VB sin y VS cos y — sin y _ ^V3coty + 1 cot y — \ Vs 18. Find sin 3 a; in terms of sin x. sin 3x = sin (2a; + x) = sin 2 a; cos X + cos 2 a; since. TEACHERS EDITION. 73 sin 2x = 2 sin x cos x. cos 2 a; = cos^o; — sin^rc. Substituting, sin 3x = 2 sin re cos^a; + sin X cos^a; — sin^x = 3 sin X cos^aj — sin^a;. But cos^a; = 1 — sin^a;. Substituting, sin 3 a; = 3 sin x — 3 sin^x — sin'a; = 3 sin x — 4: sin^a;. 19. Find cos 3 a; in terms of cos a;, cos 3x = cos {2x + x) = cos 2 X cos X — sin 2 x sin x. sin 2 a; = 2 sin x cos x. cos 2 a; = cos^a; — sin^a;. Substituting, cos 3 a; == cos^a; — sin^a; cos x — 2 sin^x cos x = cos^a; — 3 sin'^a; cos x. But sin% = 1 — cos^a;. Substituting, cos 3 a; = cos^a; — 3 cos a; + 3 cos'a; = 4 cos% — 3 cos x. 20. Given tan | a; = 1 ; find cos x. 1 tan cos a; + cosa; =V} cos X 1 = + cosa; 1 —cos a; • 1 + cos X 1 + cos a; = 1 — cos a;. 2 cos X = 0. cos x = 0. 21. Given cot | a; = V3 ; find sin a;. cot Ja;='Y- +cosa; cos a; V3 i + cos X. ^l- 3 = cos a; 1 + cos X 1 — cos X 3 — 3 cos x = 1 + cos X. — 4 cos X = — 2. cosx = -• 2 sin^x = 1 — cos^x = 1- 1^3 4 4' sm X -4-iVs. 22. Given sin x = 0.2 ; find sin ^ x and cos^x. sin X = 0.2. cos^x = 1 — sin'^x = 1 - 0.04. cos X = VOM. sin ix = ^- — cosx 1-Va96 1-oaVq = 0.10051 cos J X = ^ - + cos X =^ I'l + 0.4 VS = 0.99494. 74 TRIGONOMETRY. 23. Given cos a; = 0.5 ; find cos 2 x and tan 2 x. cos 2 .<; = cos'^a; — sin^a;. sin a; = W'-(iy-' Vs. cos 2a; = 0.25 -0.75 = _ 0.50 = - 1. 2 tan a; = tan 2 a; = sm X COS a; ^ 2 tan a; 2\/3 1 — tan^a; 1 — 3 = -V3. 24. Given tan 45° = 1 ; find the functions of 22° 30^ tana; = sin X cos a; .'. sin X = cos X. %va?-x + cos'^a; = 1. sin^a; + sin'^a; = 1. 2sin'^a; = 1. sin^a; = \. sin a; = ^ \/2 = sin J a; or sin 22° 30^ cos a;. _ ll-^\/2 = aV2-V2 2 = 0.3827. cos A a; or cos 22° 30' ^ l + iV2 2 = aV2+V2 2 = 0.9239. tan \ x sm * a; A a; AV2-V2 2i''_ f cosja; 1V2+V2 2 multiply by ^%+V2 2-V2 2-V2' cot 2 ^ = _ | (2-V2f \ 4-2 = ^ V(2 - V2)2 X V2 = (1-JV2)XV2 = \/2- 1 = 0.4142. cos I a; sin ^x ^V2+V2 iV2-\/2 I2+V2 2-V2 = \/2 + 1 = 2.4142. 25. Given sin 30° = 0.5 ; find the functions of 15°. sin 30° = 0.5 = -• 2 cos 30° =V-^V! = iV3. sin ^ a; = ^- cosa; sin 15° = l-iV^ = jV2-V3 = 0.25885. cos 15° = -v — —^ \ 2 = iV2 + V3 = 0.96592. TEACHERS EDITION. 75 tan 15° = ^1 + ^ 2V3 -4 2- V3 2 + V3 2-V3 ^ 2-V3 2 + V3 2 + V3 (2-V3)2 4-3 = 2- \/3 = 0.26799. cot 15° -V; + ^V3 iV3 = 2 + V3 = 3.7321. 26. Prove that ^^^13o^sjn33^+sin3_°. cos 33° + cos 3° Let X = 18°, y = 15°. Then (1) 2 sin X cosy = sin (a;+y) + sin {x—y). (2) 2 cos a; cosy = cos (x+y) + cos {x—y). Divide (1) by (2). tan a; = sm (a;+y) + sin (a;-y) cos {x+y) + cos {x—y) Substitute values of x and y, ^^^^go_sjn33^J:_sin3f_ cos 33° + cos 3° 27. Prove the formula • o 2 tan x sin 2 a; = 1 + tan^a; sin 2 a; = 2 sin x cos x. 2 tan a; = z sma ; cos a; 1 + tan^j; = 1 + cos^j; _ cos^a; + sin^a; cos^a; But cos^a; + sin^a; = 1. .-. 1 + tan^a; = -1-. cos^a; 2sina^cosaj = 2siT^^cos^^ cos a? 1 •. 2 sin X cos a; = 2 sin a: cos x. 28. Prove the formula 1 — tan^a; cos 2 a; = 1 + tannic 1 cos 2 a; = cos^a? 1 + sin-'a; cos'^aj cos 2 a; = cos^a; — sin^a;. 1- cos^a; — sin^a; = ein^a? cos'-^a; 1 + sin^B cos^a; _ cos% — sin^a; cos^a; + sin^a; _ cos^a; — sin^a; ~ i = cos^a; — sin^aj. 29. Prove the formula sin a; tan i a; = tan 2 a; = 1 + cos X vT^ cos a? vr + cos a; 76 TRIGONOMETRY. sin X Vi- - CO'&^X 1 + cos X 1 + COS X Vl — COS.T _ Vi- - cos^aj 1 + cos X Vi + cos X 1 — COS X 1 — cos^a; 1 ± 2 sin \ X cos ^x = l ± sin x. ± 2 sin J a; cos J a; = ± sin a;. 2 sin ^ X cos J x 1 4- cos X (1 + cos «)^ _ 1 — cos a; 1 + cos X 30. Prove the formula cot J a; = sm X 1 — cos X sin a; = Vl — cos'^a;. ^ 1 — cos X By substituting, V 1 + cos X Vl cos^'a; 1 — cos X 1 — cos X 1 + cos X _ 1 — cos^a; 1 — cosa; (1 — cosa;)^ _ 1 + cos X 1 — cos X 31. Prove the formula sin I a; db cos J a; = Vl db sin x. By squaring, sin^ J a? ± 2 sin J a; cos J a; + cos^ | x = 1 ± sin X. But sin h X =V^ COS a; and COS lx = -V /I + COS a; — cos'^.r ± sm X = ±A' /L cos zx •.±2-^. 1 — cos'-^a; = .^| 1 — coa 2 a; 1 — cos^x = 1 — cos 2 aj Substitute values of sin^a; and 1 cos f x. 1— cosx ±2sin^a;cos|a; + = 1 ± sin X. 1 -f cos a; /-^ ^ ^h —cos 2 a; Vl — cos''a; = •\l .'. sin a; = sin a;. 32. Prove the formula t^M±l^M = ±tana;tan2/. cot a; ± cot y tan a; ± tan y = ± tan X cot a; tan y + cot y tan y tan a; But tan a; cot a; = 1 , and tan3/coty = l. .-. tan X ± tan y == tan aj ± tan y. 33. Prove the formula 1 — tan X tan (45° - a;) = tan (45° - a;) = 1 + tan X sin (45° — x) cos (45° — x) sin (45° - a;) = sin 45° cos x — cos 45° sin x = J V2 cos X — A V2 si sm X = I V2 (cos X — sin a;), cos (45° — a;) = cos 45° cos X + sin 45° sin x TEACHERS EDITION. = ^ \/2 COS a; + J \/2 sin a; = 2"\/2 (cos tc + sin x). 4- / 1 Ko \ cos X — sin a? tan (45° — x) = -. — • cos X + sin X Dividing numerator and denomi- nator by cos X, 1 — tan X tan (45° - a;) = 1 + tan X 34. If A, B, C are the angles of a triangle, prove that sin A + B\n B -{■ sin C = 4 cos J J. cos ^ ^ cos I C. Bin ^ + sin j5 + sin C = sin^+sin ^+sin [180°-(^+5)] = sin ^ + sin B + sin {A + B) = 2 sin ^ (^ + B) cos 1{A--B) + 2sin H^ + B) cos ^- {A + 5) = 2sini(^ + B) [cos ^(^ - B) + cos 1(^ + 5)]. ButK^ + -S) = ^^ Then, by [22], cos A^-{- cos J5^ = 2 cos ^ -4 cos I 5. .-. == 4 sin I (J. + B) cos | J. cos J ^. But cos J C= cos [90° -1{A + B)] = ^mi{A + B). .'. sin A + smB + sin = 4 cos ^ -4 cos ^ ^ cos J C. 35. If ^, 5, C are the angles of a triangle, prove that cos A + cos B + cos C = 1 + 4 sin -J ^ sin J ^ sin | C. cos (7= cos [180°- (^ + ^)] = — cos (A + B). .'. cos A + cos B + cos C = cos J. + COS B — cos {A + 5). By [22], = 2 cos J (^ + B) cos J (^ - B) — cos (J. + B). By [17], = 2 cosK^ + B) cos H^ - -5) - 2 cos2 ^ (^ + ^) + 1 = [2cos^(^ + ^)] X [cos i Ia-B)-co& ^{A+B)]-{-h By [23], = [2cosH^+^)] X [2 sin ^ J. sin ^ 5] + 1 = (2 sin J C) (2 sin ^ ^ sin ^ 5) + 1 = 1 + 4 sin ^ ^ sin J .B sin ^ C. 36. If A, B, C are the angles of a triangle, prove that tan A + tan B + tan C = tan ^ X tan B X tan C. Since ^ + 5 + (7= 180°, C=1S0°-{A+ B). .-. tan C= tan [180° -(^1 + B)] = - tan {A + ^). Again, tan A + tan 5 = tan {A + B){1 — tan A tan P) = tan {A + B) — tan {A + ^) tan ^ tan B, .•. tan J. + tan B + tan Q = tan (^ + ^) - tan {A + ^) — tan {A + 5) tan J. tan B = - tan {A + 5) tan ^ tan B = tan j4 tan B tan C. 78 TRIGONOMETRY. 37. If A, B, C are the angles of a triangle, prove tliat cot ^ J. + cot J ^ + cot J C = cot J J. X cot J i? X cot f C. Since^^ + J5 + iC=90°, .■.cotiC=ta,n^{A + B), and cot J ^ = tan H^ + Q^ and cot J J. = tan J (^ + C). .'. cot J J. + cot f ^ + cot J C = tan J ( J. + 5) + tan i{A+ C) + t&n}{B+C) = tan J ( A + -g) X tan J (^ + C) X tan K^ + Q- By substitution, cot 2" -4. + cot J ^ + cot J C = cot^ilX cot ^ 5 X cot J C 38. Change to a form more con- venient for logarithmic computation cot X + tan X. cot X + tan aj _ cos X sin a; sin X cos a; _ cos^a; + sin^g; sin X cos X _ 2 (cos^a; + sin^o;) 2 sin a; cos x 2 sin 2 a; 39. Change to a form more con- venient for logarithmic computation cot X — tan X. cot X — tan X _ cos K _ sin X sin a; cos x _ cos% — sin% sin X cos a; cos 2x sin a; cos a; 2 cos 2 a; 2 sin X cos a; 2 cos 2 a; sin 2 a; 2 cot 2 a;, [13] [12] 40. Change to a form more con- venient for logarithmic computation cot X + tan y. cos a; J. sin y cosy cot a; = ^^^^^^, tan?/=^^^^-^- [2] sin X Adding, _ cos X cos y + sin x sin y sin a; cosy Substitute for cos x cos ?/-l-sin x sin y its equal cos {x — y), [9] _ cos (x — y) sin X cos y 41. Change to a form more con- venient for logarithmic computation cot X — tan y. , sm v tany = ^• cosy , cos a; cot X = sm X cot a; — tan y cos a; sin y sin a; cosy cos X cos y — sin a; sin y sin a; cos y _ cos (a; + y) sm a; cos y 42. Change to a form more con- venient for logarithmic computation 1 — cos 2 a; 1 -h cos 2 a; TEACHERS EDITION. 79 1 — cos 2x 1 + cos 2x 1 — cos 2 X 2 1 + cos 2 X 2 sin^a; cos-'a; = tan^a;. 43. Change to a form more con- venient for logarithmic computation 1 + tan X tan y. 1 + tan X tan y ^ sin X sin v = 1 + X cos X cos 2/ cos X cos y + sin x sin 3/ cos X cos 3/ _cos(a; — y) cos a; cosy 44. Change to a form more con- venient for logarithmic computation 1 — tan X tan y. 1 — tan X tan y _ ^ sin X sin 3/ cos X COS y _ cos X COS 2/ — sin x sin 3/ COS X cos 2/ cos (a; -f y) COS a; cos y 45. Change to a form more con- venient for logarithmic computation cot a; cot 2/ + 1. cot X cot 2/ + 1 cos a; cos y sm X sin 2/ + 1 By [9] _ COS X cos 2/ + sin a; sin 2/ sin X sin 3/ _ cos {x — y) sin X sin 2/ 46. Change to a form more con- venient for logarithmic computation cot xcoiy — 1. cot X cot 3/ — 1 cos a; cos y -1 sm a; sm y _ cos a; cos 3/ — sin a; sin 3/ sin a; sin 3/ ^ _ cos [x + 3/) sin a; sm y 47. Change to a form more con- venient for logarithmic computation tan X + tan y cot a; + cot y tana; -f tang/ cot a; -h cot y sin a; sin 3/ cos X cos 2/ cos a; cos 2/ sin a; sin 2/ sin a; cos y + co&x sin 3/ cos X cos 2/ sm a; cos y + cos a; sm y sin a; sin y sin a; sin 2/ cos X cos 2/ tan X tan y. 80 TBIGONOMETEY. Exercise XII. Page 53. 1. What do the formulas of § 36 become when one of the angles is a right angle ? B If angle C is a right angle, a sin A ■ . - = = sm A c sin C c _ sin C _ 1 b sin B sin B a _ sin A b sin B a c = tan A ; sin A sin C b ^ c sin B sin C = c; = c. 2. Prove by means of the Law of _Sines that the bisector of an angle of a triangle divides the opposite side into parts proportional to the adjacent sides. Let CD bisect angle C. Then and AD _ sm^O CD sin A ' DB ^ sin ^ (7 CD sin B ' By division, AD _ sin B DB sin A But ?iB4 = ^. sin A a " BD a 3. What does Formula [26] be- come when ^ = 90° ? when ^ = 0° ? when ^-180°? What does the triangle become in each of these cases ? Formula [26] is a^ = b^ + c^ — 2bc cos A. When A = 90°, cos A = 0°. .-. a^ = b'^ + c\ When A = 0°, cos A = l. .-. a2 = 62 + c2-26c. When A = 180°, cos ^ = - 1. .•.a2 = 62 + c2 + 2Jc. B A a = BC. b = AC B a = Ba b^AC c=AB. a = b — c. c=BA. a = b + c. 4. Prove that whether the angle B is acute or obtuse c = a cos B + b cos A. What are the two symmetrical formulas obtained by TEACHERS EDITION. 81 changing the letters? What does the formula become when B = 90° ? A Fig. 2. D Case I (Fig. 1). (1) . When angle B a is acute h ,'. DB = a cos B, and AD = h cos A. Add, Z)5+^i) = acos5 + 6cos^. But Z)5+^i) = c. .". c = a cos 5 + 6 cos A. Case II. When angle B is obtuse (Fig. 2), ^D = cos ^. :?:^ = cos(180°-.B) = — cos ^. .-. AD = 6cos^, and 5Z) = — a cos B. Subtract, observing that the sign of cos B is minus. AD — BD = 6 cos J. + a cos B. But AD-BD = c. ,'.c = acosB + b cos A The symmetrical formulas are b = a cos C+ ccos^, a = 6 cos C + c cos -S. When B = 90°. (3) cos ^ = -• 6 .'. c = 6 cos -4. 5. From the three following equa- tions (found in the last exercise) prove the theorem of ^ 37 : c = a cos 5 + 6 cos J., 6 = a cos C + c cos A^ a = 6 cos (7 + c cos B. & = ac cos B -{■ be cos A. (1) 52 = a6 cos C+bc cos J.. (2) a^ = ab cos C + ac cos .8. (3) Add (2) and (3), a' + b^ = 2ab cos C+bc cos .4 + ac cos j5. (4) Subtract (4) from (1), c^ — a^ — b'^ = — 2ab cos 0. .•.c2 = a2 + 62_2a6cosa ^7 6. In Formula [27] what is the maximum value of ^{A — B)l of a — b _ tan ^(A — B) a+b tanJ(^ + 5) 82 THIGONOMETRY. The limit of ^ - 5 is 180°. .'. the limit of the maximum value ofH^-^) 180° 2 = 90°. The limit of ^ + 5 is 180°. /. the limit of the maximum value of^{A + B) = 1^0! =90° 2 7. Find the form to which Form- ula [27] reduces, and describe the nature of the triangle when (i.) C=90°; (ii.) A- B = 90°, and B=0. a — b ^ tan ^{A — B) a + b isin^{A + B) (i.) When C= 90°. A + B = 90°. B=90°-A. Q-5^ tann^-(Q0°-^)1 a + b tan 45° _^ tan (^-45°) 1 = tan (^-45°). Since C is a right angle, the tri- angle is a right triangle. (ii.) When ^-5 = 90°, and 5=a a — b _ tan ^{A — B) a + b tan ^ ( J. + 5) A + B + 0^180°, or A + 2B = 180° A-B =90° ,'. 3 5= 90° B= 30°, C= 30°, and A = 120°. a — b tan 45° a + b tan 75° tan 45° cot 15° 1 2+VS •'• a + b = (a-6)(2 + >/3). Since A ■- celes. = B, the triangle is isos 1. Given a = 500, A = 10° 12^ B = 46° 36' ; Exercise XIII. find C= 123° 12^ b = 2051.48, c = 2362.61. a = 500. A= 10° 12' B = 46° 36' A + B= 56° 48' J. C= 123° la^ Page 55. log a = 2.69897 colog sin A = 0.75182 log sia B = 9.86128 log b = 3.31207 6 = 2051.48. log a = 2.69897 colog sin A = 0.75182 log sin C = 9.92260 logc =3.37339 c = 2362.61. TEACHERS EDITION. 83 2. Given find a = 795, C= 55° 20^ J. = 79°59^ 6=567.688, B = 44° 4F; c == 663.986. a = 795. ^= 79° 59^ B = 44° 4F A-\-B= 124° 40^ .-. C= 55° 20'. log rt = 2.90037 colog sin ^ = 0.00667 log sin B = 9.84707 log b = 2.75411 b = 567.688. log a = 2.90037 colog sin A = 0.00667 log sin C = 9.91512 log c = 2.82216 c = 663.986. 3. Given find a = 804, C=35°4^ ^ = 99°55^ 6 = 577.31, ^ = 45° 1' ; c = 468.93. a = 804. A= 99° 55' ^= 45° V A + B = 144° 56' .-. C= 35° 4'. log a = 2.90526 colog sin A = 0.00654 log sin B = 9.84961 log 6 = 2.76141 6 = 577.31. log a. = 2.90526 colog sin A = 0.00654 log sin Q = 9.75931 log c = 2.67111 c = 468.93. 4. Given find a = 820, C= 25° 12', A = 12° 49', 6 = 2276.63, 5 =141° 59'; c = 1573.89. a = 820. A= 12° 49' ^ = 141° 59' ^ + .5 =154° 48' .-, C= 25° 12'. log a = 2.91381 colog sin A = 0.65398 log sin B = 9.78950 log 6 =3.36729 b = 2276.63. log a = 2.91381 colog sin A = 0.65398 log sm = 9.62918 log c = 3.19697 c = 1573.89. 5. Giv^n find c = 1005, C= 47° 14', A = 78° 19', a = 1340.6, ^ = 54° 27'; ,6 = 1113.8. c = 1005. ^= 78° 19' B = 54° 27' ^ + ^ = 132° 46' .-. C= 47° 14'. 84 TRIGONOMETKY. logc = 3.00217 colog sin C = 0.13423 log sin A = 9.99091 log a = 3.12731 a = 1340.6. log c = 3.00217 colog sin C = 0.13423 log sin B = 9.91042 log b = 3.04682 b =1113.8. 6. Given b = 13.57, 5=13°57^ (7=57° 13^ find ^ = 108°50^ a = 53.276, c = 47.324. 6 = 13.57. £^ 13° 57^ C= 57° 13' £ + 0= 71° 10' .-. A = 108° 50'. log b = 1.13258 colog sin B = 0.61785 log sin A = 9.97610 log a = 1.72653 a = 53.276. log a = 1.72653 colog sin A = 0.02390 log sin C = 9.92465 log c = 1.67508 c = 47.324. 7. Given a = 6412, A = 70° 55', C = 52° 9' ; find B = 56° 56', 6 = 5685.9, c = 5357.5. a = 6412. A= 70° 55' C= 52° 9' A+ 0=123° 4' .-. B = 56° 56'. log a = 3.80699 log sin B = 9.92326 colog sin A = 0.02455 log 6 = 3.75480 b = 5685.9. log a = 3.80699 log sin C = 9.89742 colog sin A = 0.02455 log c = 3.72896 c = 5357.5. 8. Given b = 999, A = 37° 58', C= 65° 2' ; find B = 77°, a = 630.77, c = 929.48. 6 = 999. A= 37° 58' C= 65° 2' .4 + C=103° .•.^= 77°. log b = 2.99957 colog sin 5=0.01128 log sin A = 9.78902 log a = 2.79987 a = 630.77. log 6 = 2.99957 colog sin B = 0.01128 log sin C = 9.95739 logc = 2.96824 c = 929.48. TEACHERS EDITION. 85 9. In order to determine the dis- tance of a hostile fort A from a place B, a line BC and the angles ABC and BCA were measured, and found to be 322.55 yards, 60° 3V, and 56° 10^, respectively. Find the dis- tance AB. a = 322.55. B = 60° 34^ C= 56° 10^ B+C= 116° 44^ /. A = 63° 16^ log a 2.50860 colog sin A = 0.04910 log sin C = 9.91942 logc 2.47712 c 300. 10. In making a survey by tri- angulation, the angles B and Cof a triangle ABC were found to be 50° 30^ and 122° 9^ respectively, and the length BC is known to be 9 miles. Find AB and AC. (7=122° 9^ B= 50° 30^ B+C= -. 172° 39^ .'. A = = 7° 21^ log BC = 0.95424 colog sin A = 0.89303 log sin B = 9.88741 log b 1.73468 b = AC = 54.285. log BC = 0.95424 colog sin A = 0.89303 log sin C = 9.92771 logc c^AB 1.77498 59.564. 11. Two observers 5 miles apart on a plain, and facing each other, find that the angles of elevation of a balloon in the same vertical plane with themselves are 55° and 58°, respectively. Find the distance from the balloon to each observer, and also the height of the balloon above the plain. B= 58° A= 55° A + B = 113° C = 67°. logc = 0.69897 colog sin c = 0.03597 log sin A = 9.91336 log a = 0.64830 a = BC = 4.4494. logc = 0.69897 colog sin C = 0.03597 log sin B = 9.92842 log 6 = 0.66336 h^AC = 4.6064. ) find h. h a = sin B. .• .h = a sin B. log a = 0.64830 log sin B = 9.92842 log A = 0.57672 h = 3.7733. 12. In a parallelogram, given a diagonal d and the angles x and y which this diagonal makes with the sides. Find the sides. Compute 86 TRIGONOMETRY. the results when cZ = 11.237, x Id" V, and y = 42° 5V. d= 11.237. x= 19° V 2/= 42° 54^ x + y= 61° 55^ .-. z = 118° 5^ \ogd = 1.05065 colog sin z = 0.05440 log sin X = 9.51301 log a = 0.61806 a = 4.1501. logd = 1.05065 colog sin 2 = 0.05440 log sin y - 9.84297 logc = 0.93802 c = 8.67. 13. A lighthouse was observed from a ship to bear N. 34° E. ; after sailing due south 3 miles, it bore N. 23° E. Find the distance from the lighthouse to the ship in both posi- tions. c = 3. 5 = (180^ A= 23° '-34°) = 146° A + B = 169° .\C= 11°. log c = 0.47712 colog sin C = 0.71940 log sin A = 9.59188 log a = 0.78840 a = 6.1433. log c = 0.47712 colog sin Q = 0.71940 log sin B = 9.74756 log h = 0.94408 h = 8.7918. 14, In a trapezoid, given the parallel sides a and 6, and the an- gles X and y at the ends of one of the parallel sides. Find the non- parallel sides. Compute the results when a = 15, 6 = 7, a; = 70°, y = 40°. Given parallel sides, ^5 =7 and DC=15; also, ADG= 40° and BCD =10°; required AD and BC. Draw AE II BC; then AB = EC (lis comp. bet. lis), and DE=DC-AB = 15 - 7 = 8. Also AED = BCD=10°{ext. int. A). Now i)JL.£;=180°-(40° + 70°) = 70°. But since AED = DAE =70°, the A is isosceles, and side DA=DE=8. 'Now AE= BC, and we are to find BC. AE_ sin ADE DE sin DAE log DE = 0.90309 log sin ADE =9.80807 colog sin DAE = 0.02701 log AE - = 0.73817 AE=BC =5.4723. TEACHERS EDITION. 87 15. Given h = 7.07107, A = 30°, C= 105° ; find a and c without using logarithms. Let p and q denote the segments of c made by the ± dropped from C. 5 = 45°. sin A= — 2 sin 5 = iV2. a_ b~ a = i iV2 _b_ V2 7.07107 = 5. b P 1.41421 cos ^ = |\/3 = 0.86602. = bx 0.86602 = 7.07107 X 0.86602 = 6.12369. ? = sin 5 = JV2 = 0.70711. q = ax 0.70711 = 5 X 0.70711 = 3.53555. c=p + q = 6.12369 + 3.53555 = 9.65924. 16. Given c = 9.562, A = 45°, B = 60° ; find a and b without using logarithms. C= 75°. e sin A a = = ^V2xiV3 + ^V2xi 9.562 x*V2 b = (v/6+V2) _ 19.124 xV2 V6+V2 _ (19.124xV2)(\/6-V2) 6-2 = 9.562 (VS-l) = 6.999 = 7. a sin 5 7 X iVS sin J. 7V3 7\/6 f V.4J a = sin C sin a= sin (45° + 30°) = sin 45° cos 30° + cos45°sin30°. V2 2 = 3.5 V6 = 8.573. 17. The base of a triangle is 600 feet, and the angles at the base are 30° and 120°. Find the other sides and the altitude without using log- arithms. AB = 600. A = 30°. 5=120°. .-. C= 30°. A=C. a = c = 600 feet. 7 _ a sin 5 sin A _ 600 X sin (180°- 60°) sin 30° ^ 600x|a/3 1 = 600x1.732051 = 1039.2. h = b sin A == 1039.2 X J = 519.6 feet. TEIGONOMETRY. 18. Two angles of a triangle are, the one 20°, the other 40°. Find the ratio of the opposite sides with- out using logarithms. Let X = 20°, 37 = 40'', and a and h be opposite sides. sin X _a Then sin y nat sin a; = 0.3420. nat sin y = 0.6428. .-. a : 6 : : 3420 : 6428. : : 855 : 1607. 19. The angles of a triangle are as 5 : 10 : 21, and the side opposite the smallest angle is equal to 3. Find the other sides without using logarithms. Since the angles A, B, C are as 5:10:21, ^ = ^6 of 180° = 25°. ^=10 of 180° = 50°. C=|iof 180° = 105°. a sin 5 3 X 0.766 sin J. = 5.43775. a sin C sin A = 6.857. c = 0.4226 3 X 0.9659 0.4226 20. Given one side of a triangle equal to 27, the adjacent angles equal each to 30°. Find the radius of the circumscribed circle without using logarithms. 2E sin^ sin A = sin 120° = sin (180° - 60°) = sin 60°. sin 60° = |\/3. 27 _54 _ 54xV3 V3 3 •.2i2 = ^V3 = 18 Vs. i2=9V3 = 15.588. Exercise XIV. Page 59. 1. Determine the number of solu- tions in each of the following cases : (i.) a = 80, 6 = 100, A = 30°. a = 80 < 6 = 100, but > 5 sin ^ = 100 X |, and A < 90°. .*. two solutions. (ii.) a = 50, ft = 100, ^ = 30°. a = 50 = 6 sin il = 100 X h .'. one solution. (iii.) a = 40, 6 = 100, ^ = 30°. a = 40< bsmA^lOOxh and A < 90°. .'. no solution. (iv.) a = 13.4, 6 = 11.46, ^=77° 20'. a = 13.4 > 5=11.46. .*. one solution. (v.) a = 70, 5 = 75, A = 60°. a = 70 < & = 75, but >6sin^ = 75xiV3, TEACHERS EDITION. 89 and ^ = 60° < 90°. .*. two solutions. (vi.) a = 134.16, b = 84.54, B = 52° 9^ 11^^. b» a sin J5, and ^ < 90°. .•. two solutions log a = 1.53148 log sin B = 9.70332 colog b = 8.65758 - 10 log sin A = 9.89238 A = 51° 18' 27". A^= 128° 41' 33". .-. C= 98° 21° 33". .-. C'= 20° 58' 27". log a =1.53148 log sin (7=9.99536 colog sin A = 0.10762 log c = 1.63446 c = 43.098. log a = 1.53148 log sin C"= = 9.55382 colog sin J.= = 0.10762 log c' = 1.19292 c' = 15.593. 10. Given b = = 19, c = = 18, C = = 15° 49' ; nd B = = 16° 43' 13", ^'= = 163° 16' 47", A = = 147° 27' 47", A'= = 0° 54' 13", a = = 35.52, a'= = 1.0415. log b : 1.27875 log sin (7 = 9.43546 colog c 8.74473 - -10 log sin 5 = 9.45894 .5 16° 43' 13". ^' - = 163° 16' 47". A = 147° 2 7' 47". ^'= 0^ '54' 13". log 6 1.27875 colog sin B = 0.54106 log sin A = 9.73065 log a 1.55046 a = 35.519. log b = 1.27875 colog sin 5'= 0.54106 log sin J.'= 8.19784 log a' = 0.01765 a' = 1.0415. 11. Given V^. log& log sin colog sin B = 1.51970 = 9.70418 = 0.27348 logc c = 1.49736 = 31.431. 10. Given find a = 47.99, ^ = 2° 46^ 8^', b = 33.14, .5=1° 54^ 42^^, C= 175° 19^ 10^^ ; c = 81.066. a + 5 = 81.13 a-b = 14.85 ^ + 5 = 4° 40^ 50^^ |(^ + ^) = 2° 20^ 25^^ I {A-B) = 0° 25^ 43^^ A = 2°W 8^\ ^ = 1° 54^ 42^^ log (a -J) =1.17173 colog (a + b) = 8.09082 - 10 log tan J [a + B) = 8.61138 log tan H^ - ^) = 7.87393 m-B) = 0° 25' 43^^ log 6 = 1.52035 log sin = 8.91169 colog sin B = 1.47680 logc = 1.90884 c = 81.066. 11. If two sides of a triangle are each equal to 6, and the included angle is 60°, find the third side. 96 TRIGONOMETRY. Since e a = h, A = B, A + B = 120°. ■.A = 5 = 60° log a = 0.77815 log sin C = 9.93753 colog sin B = 0.06247 logc = 0.77815 c = 6. 12. If two sides of a triangle are each equal to 6, and the included angle is 120°, find the third side. A + B = 60°, .\A = B = 30°, a = 6 = 5. log a =0.77815 log sin (7=9.93753 colog sin A = 0.30103 log c = 1.01671 c = 10.392. 13. Apply Solution I. to the case in which a = b or the triangle is isosceles. If a = 6 and the A is isosceles, the angles A and B are equal, be- ing opposite the equal sides. Now as a = 6 and A= B, the formula tan J (A-B) = ^^ X tan J (A+B) a+b will become tan J(0) = Oxtan J(2^). = 0. 14. If two sides of a triangle are 10 and 11, and the included angle is 50°, find the third side. a + b ^21 a-b = l A + B =130° UA + B)= 65° ^{A-B)= 5° 49^ 51^^ A= 70°49^5F^ B = 59° 10^ 9^\ log (a -5) = 0.00000 colog (a + 6) = 8.67778-10 log tan 1(^ + 5) = 10.33133 log tan J (J. -^)= 9.00911 ^{A-B) = 5° 4:9' 5V^. logb log sin colog sin B logc c 1.00000 9.88425 0.06617 0.95042 8.9212. 15. If two sides of a triangle are 43.301 and 25, and the included angle is 30°, find the third side. a + 5 = 68.301. a-5 = 18.301. ^ + ^ = 150° UA + B)= 75° ^{A-B)= 45° A = 120°. B= 30°. .'. in isos. triangle ABQ c = 6 = 25. log(a-&) = 1.26247 colog (a + 6) = 8.16557-10 log tan^(^ + ^) = 10.57195 log tan J (^-^)=^ 9.99999 ^{A-B) = 45°. TEACHERS EDITION. 97 16. In order to find the distance between two objects A and B sep- arated by a swamp, a station C was chosen, and the distances CA = 3825 yards, CB = 3475.6 yards, together with the angle ACB =^ 62° SV, were measured. Find the distance from ^ to B. b + a = 7300.6 b-a = Sid A B+ A = 117° 29^ i{B + A)= 58° 44^ 30^^ B = A = 4° 30^ 30^^ 63° 15^ 54° 14^ log(&-a) = 2.54332 colog(6 + a) = 6.13664-10 log tan ^(5 + ^) = 10.21680 log tan J (5 - ^) = 8.89676 H^ - ^) = 4° 30^ 30^^ log 6 log sin C colog sin B logc c 3.58263 9.94799 0.04916 3.57978 3800. 17. Two inaccessible objects A and B are each viewed from two stations C and D 562 yards apart. The angle ACB is 62° 12^ BCD 41° 8^ ABB 60° 49^ and ADC 34° 5V ; required the distance AB. e In triangle ACD A = 1S0°-{C+D) = 41° 49^ b ^ sin 34° 51^ 562 sin 41° 49^' . ^ _ 562 sin 34° 5V sin 41° 49^ log 562 = 2.74974 log sin 34° 51^ = 9.75696 colog sin 41° 49^ = 0.17604 log b = 2.68274 b = 481.65. In triangle CBD B = 1S0°-{C+D) = 43° 12^. a ^ sin 95° 40^ 562 sin 43° 12^' _ 562 cos 5° 40^ sin 43° 12' log 562 = 2.74974 log cos 5° 40' = 9.99787 colog sin 43° 12^ = 0.16460 log a = 2.91221 = 816.98. In triangle ACB tan }iA-B)= "*— ^ x tan | (A+B) a + b H^ + ^) = Hi80°-C) = 58° 54'. a- 6 = 816.98 -481.65 = 335.33. a + 6 = 816.98 +481.65 = 1298.63. 98 TRIGONOMETRY. log (a— b) = 2.52547 colog {a + b) = 6.88651 -10 log tan i(^ + ^)= 10.21951 logtani(.4-5)= 9.63149 i(^-^)= 23° 10^26'^. ^ = 82° 4^26^A log a log sin C colog sin A logc c 2.91221 9.94674 0.00418 2.86313 729.68. 18. Two trains start at the same time from the same station, and move along straight tracks that form an angle of 30°, one train at the rate of 30 miles an hoar, the other at the rate of 40 miles an hour. How far apart are the trains at the end of half an hour ? < A i{A I + 6 = 35 1-6 = 5 + B= 150° + 5)= 75° H^ -B)= 28° ¥ B= 46° 56^ ^ = 103° 4^ log (a - b) colog (a + b) log tan J {A = 0.69897 = 8.45593-10 + B) = 10.57197 log tan i {A ■1 K^ ~B)= 9.72687 -B) = 28° V. log 6 log sin C colog sin B = 1.17609 = 9.69897 = 0.13634 logc c = 1.01140 = 10.266. 19. In a parallelogram given the a - 6 = 0.5 two diagonals 5 and 6, and the an- a + 6 = 5.5 gle that they form 49° 18^. Find ^ + ^=130° 42^ the sides. ^{A+B)= 65° 21^ In the parallelogram ABDE ^{A-B)= 11° 12^ 20^^ let EB = 6, A= 76°33^20^^ AD^b, ^=54° 8^40^^. and Z BCA = 49° 18^. log («- -6) = 9.69897-10 In triangle ACB colog {a + b) = 9.25964-10 let .5C=o = 3, log tan i(^ + ^) = 10.33829 ^C=6=2.5. log tan 1{A-B)= 9.29690 Find AB = c. i (^ - J5) = 11° 12^ 20^^ teachers' edition. 99 log a 0.47712 log a = 0.47712 colog sin J. = 0.01207 colog sin A = 0.34238 log sin Q = 9.87975 log sin c = 9.87975 - 10 logc c =AB = 0.36894 log G = 0.69925 2.3385. c =EA = 5.0032. 20. In a triangle one angle equals In triangle AEQ 139° 54^, and the sides forming the angle have the ratio 5 : 9. Find the other two angles. EC= - a = S, AC = -. h = 2.5, a = 9 ZACE = = 130° 42^ 6 = 5 A + E = = 49° 18^ a + 6= 14 K^ + ^) = - 24° 39^ a-& = 4 H^-^) = = 2° 23^ 20^^ ^ + 5 = 40° 6^ A = = 27° 2^20^'. ^{A + B) = 20° y 1{A-B)^ 5° 57^ 10^^ ^ = 26° OMO^^ log (a - 6) = 9.69897-10 ^ = 14° 5^50^^ colog (a + b) = 9.25964 - 10 log {a-h) = 0.60206 log tan ^{A + E) = = 9.66171 colog (a + 6) =8.85387-10 log tan n^ - ^) = = 8.62032 log tan ^{A + B)= 9.56224 i(^-^) = = 2° 23^ 20^^ log tan H^-^) = 9.01817 ^ = . 27° 2^ 20^^ *(^-^)=5°57M0^^ Exercise XVI. Page 67. 1. Given a = 51, b = 65, C = 20; colog s = 8.16749 - 10 find the angles. a= 51 b= 65 colog {s-a) = 8.76955 - 10 log {s-b) = 0.47712 c= 20 log {s-c) = 1.68124 2s=136 2)19.09540-20 s= 68. s-a= 17. s-b= 3. log tan ^^= 9.54770 J^ = 19°26^24^^ 8-c= 48. A = 38° 52^ 48^^ 100 TRIGONOMETRY. colog s = 8.16749-10 colog (s-b) = 9.52288 - 10 log (s- a) = 1.23045 log (s-c) = 1.68124 2 )20.60206-20 log tan 1 5 = 10.30103 J ^ = 63° 26^ 6'\ B = 126° 52^ 12^^ A^- B = 165° 45^. .-. (7= 14° 15^ 2. Given a = 78, 6 = 101, c = 29 ; find the angles. a= 78 & = 101 c = _29 2s = 208 s = 104. s — a= 26. s-h= 3. s — c= 75. colog s = 7.98297-10 colog (s - a) = 8.58503 - 10 log(s-5) = 0.47712 log \s-c) = 1.87506 2)18.92018-20 logtanJ^= 9.46009 J^ = 16° 5^27^^ A = 32° 10^ 54^^ colog s = 7.98297-10 colog (s - 5) = 9.52288 - 10 log(s-a) = 1.41497 log (s - c) = 1.87506 2 )20.79588-20 logtan 1^ = 10.39794 ^B= 68° 11^ 55^^ ^=]36°23^50^^ ^ + 5 = 168° 34^ 44^^. .-. C= 11° 25^ 16^^ 3. Given a = 111, 5 = 145, c = 40; find the angles. a = 111 5 = 145 c= 40 2 s = 296 s = 148. s — a= 37. s - 5 = 3. s-c = 108. colog s = 7.82974 - 10 colog (s- a) = 8.43180-10 log(s-5) = 0.47712 log {s-c) = 2.03342 2)18.77208-20 logtanJ^= 9.38604 ^J. = 13°40^16^^ A = 27° 20^ 32^^. colog s = 7.82974-10 log(s-a) = 1.56820 colog (s - 5) = 9.52288 - 10 log {s-c) = 2.03342 2 )20.95424-20 log tan|5= 10.47712 ^B= 71° 33^ 54^^. ^ = 143° 7M8^^. 5 + ^ = 170° 28^ 20^^. .-. C= 9° 3F 40^^. 4. Given a = 21, 6 = 26, c = 31 ; find the angles. TEACHERS EDITIOX 101 a = 21 6=26 c = 31 2s = 78 s = 39. s-a=18. s-6 = 13. s-c= 8. colog s = 8.40894-10 colog (s-a) = 8.74473 - 10 log (s-h) = 1.11394 log (s - c) = 0.90309 2)19.17070-20 logtani^= 9.58535 .-. A = 42° 6^ 13^^ colog s = 8.40894-10 log {s-a)= 1.25527 colog (s - 6) = 8.88606 - 10 log {s-c) = 0.90309 2 )19.45836-20 logtaiiJ^= 9.72668 .-. B = 56° 6^ 36^^ ^ + ^ = 98° 12M9^^ .-. C= 81° 47^ 11^^ 5. Given a = 19, 5 = 34, find the angles. a= 19 &= 34 c = _49 2s=102 s= 51. s-a'= 32. s-h= 17. 8-C= 2. 49 colog s = 8.29243 - 10 colog (s-a) = 8.49485 - 10 log (s - 5) = 1.23045 log (s-c) = 0.30103 2)18.31876-20 logtanJ^= 9.15938 M = 8° 12^ 48^^. A = 16° 25^ 36 v^ colog s 8.29243 - 10 colog (s - -h) = 8.76955 - 10 log (s - -c) = 0.30103 log (s - -a) = 1.50515 lo^ 2 )18.86816-20 tan ^5= 9.43408 J5= 15° 12^. B = 30° 24^ .-. C= 133° 10^ 24^^ 6. Given a = 43, 5 = 50, c- find the angles. = 57 a= 43 6= 50 c= 57 2s =150 s= 75. s-a= 32. s-b= 25. s-c= 18. colog s = 8.12494- -10 colog {s~a) = 8.49485 - -10 log(s-6) = 1.39794 log (s-c) = 1.25527 2)19.27300- -20 logtanJ^= 9.63650 J^ = 23° 24^47^^. A = 46° 49^ 35^^ 102 TRIGONOMETRY. cologs = 8.12494-10 log(s-a)= 1.50515 colog (s - 6) = 8.60206 - 10 log {s-c)= 1.25527 2) 19.48742 - 20 logtanJ^= 9.74371 J5 = 28°59^52^^ B = 57° 59^ 44^^. .-. (7= 75° 10^ 41^^ 7. Given a = 31, 5 = 58, c = 79 ; find the angles. a= 37 6= 58 c=_79 2s = 174 s= 87. s — a= 50. s-b= 29. s-e= 8. colog s = 8.06048 - 10 colog {s-a) = 8.30103 - 10 log(s-&) = 1.46240 log (s-c) = 0.90309 2 )18.72700-20 logtan^^= 9.36350 J^ = 13°0M4^^. A = 26° 0^ 29^^. colog s = 8.06048 - 10 log(s-a)= 1.69897 colog (s - 6) = 8.53760 - 10 log (s - c) = 0.90309 2)19.20014-20 logtanJ5= 9.60007 ^B= 21°42M0^^ B = 43° 25^ 20^^. .-. C= 110° 34^ 11^/. 8. Given a =73, 6 = 82, c = 91; find the angles. a= 73 &= 82 c= 91 2s = 246 s = 123. s — a = 50. s-b= 41. s — c= 32. colog s = 7.91009 - 10 colog (s-a) = 8.30103 - 10 log(s-6) = 1.61278 log (s - c) = 1.50515 2) 19.32905 - -20 log tan i^ 9.66453 i^ = 24° 47^ 29^^. A = 49° 34^ 58 // colog s = 7.91009 - -10 log (s - -a) = 1.69897 colog (s - -b) = 8.38722- -10 log (s - -c) = 1.50515 2) 19.50143 - ■20 log tan|^= 9.75072 ^B= 29°23^29^^ B = 58° 46^ 58^^. .-. C= 71° 38^ 4^^. 9. Given a = 14.493, b = 55.4363, e = 66.9129 ; find the angles. a= 14.493 b= 55.4363 c= 66.9129 2s =136.8422 s= 68.4211. s-a= 53.9281. s-&= 12.9848. s-c= 1.5082. TEACHERS EDITION. 103 cologs = 8.16481-10 colog(s-a) = 8.26819-10 log(s-6) = 1.11344 log (s-c) = 0.17846 2)17.72490-20 log tan J A 8.86245 A = 8° 20^ cologs = 8.16481-10 log(s-a) = 1.73181 colog (.s -b) = 8.88656 - 10 log (s-c) - 0.17846 2)18.96164-20 logtaiiJ5= 9.48082 ^B= 16° 50^ B = 33° 40^ .-.(7=138°. 10. Given a=y/5,b = V6, c = V7] find the angles. a =V5 = 2.2361 6 =V6 = 2.4495 c=V7 = 2.6458 2s =7.3314 s = 3.6657. s-a = 1.4296. 8-6 = 1.2162. s-c = 1.0199. log (s-b) = 0.08500 log (s-c) = 0.00856 colog s = 9.43585 - 10 colog (s-a) = 9.84478 - 10 2 )19.37419-20 logtanJ^= 9.68709 |^ = 25°56^36^^ A = 51° 53^ 12^^ colog (s - -b) = 9.91500- 10 log (s - -c) = 0.00856 colog s = 9.43585 - 10 log (s - -a) = 2) 0.15522 19.51463 - 20 log tan 15 = 9.75732 i5 = 29° 45^ 54 // B = 59° 31^ 4S // .:C= 68° 35^ 11. Given a = 6, 6 = 8, c = = 10 find the angles. a = 6. 6 = 8. c = 10. s = 12. s — a = 6. s -6 = 4. s — c = 2. colog s = 8.92082 - 10 colog (s-a) = 9.22185-10 log (s - 6) = 0.60206 log (s-c) = 0.30103 2 ) 19.04576 - 20 log tan I ^= 9.52288 i^ = 18°26^ 6^^ A = 36° 52^ 12^^. Since this is a right triangle, C= 90°. B = 90° - A = 53 ;c T/ 48^^. 12. Given a = 6, 6 = 6, c = 10; find the angles. a= 6 6= 6 c = 10 2s=22 104 TRIGONOMETRY. s = ll. s — a= 5. s - 6 = 5. s — c = 1. colog s = 8.95861 - 10 colog {s-c) = 0.00000 log (s - 6) = 0.69897 log (s - a) = 0.69897 2 ) 20.35655 - 20 log tan^ (7= 10.17828 ^C= 56° 26^ 33^^. C= 112° 53^ 6'\ Since this is an isosceles triangle, ^ = ^=1(180°- C) = 33° 33^ 27^^. 13. Given a = 6, & = 6, c = 6 ; find the angles. The triangle is equilateral and also equiangular. .-. A = B=C=^of 180° = 60°. 14. Given a = 6, 6 = 5, c = 12 ; find the angles. The sum of the two sides a and b is less than the side c. .•. the triangle is impossible. 15. Given a = 2, h =V6, c = VS — 1 ; find the angles. a = 2 6 = V6 = 2.4495 c= VS- 1 = 0.732 2 6 = 5.1815 5 = 2.5908. s-a = 0.5908. s- 6 = 0.1413. s-c= 1.8588. log(s-a) = 9.77144-10 log(5-&) = 9.15014-10 log {s-c) = 0.26923 colog s = 9.58656 - 10 log r2 = 18.77737 - 20 logr = 9.38869-10 log tan ^ J. = 9.61725 logtan|5 = 10.23855 logtan^(7= 9.11946 iA= 22° 30^ ^B= 60°. i C = 7° 30^. J.= 45°. B = 120°. C= 15°. 16. Given a = 2, 6 = Ve, c \/3 + 1 ; find the angles. a = 2 6 = V6 = 2.4495 c = V3 + 1 = 2.732 2s= 7.1815 s = 3.5908 s-a = 1.5908 s-& = 1.1412 s-c = 0.8588 log {s-a) = 0.20162 log (s - 5) = 0.05740 log (s-c) = 9.93385 - 10 colog s = 9.44481 - 10 logr2 =19.63768- 20 logr = 9.81884- 10 logtanJJ. = 9.61721. log tan J5 = 9.76146. logtan J (7=9.88494. TEACHEES EDITION. 105 hA = 22° 30^. J5 = 30°. iC = 37° 30^ A = 45° B = 60° C = 75° 17. The distances between three cities A, B, and C are as follows : AB = 165 miles, AC= 72 miles, and BC = 185 miles. B is due east from A. In what direction is Cfrom A ? What two answers are admissible ? a =185 6= 72 c = 165 2s = 422 s = 211. s-a= 26. s-6 = 139. s-c= 46. cologs = 7.67572-10 colog (s-a) = 8.58503 - 10 log-(s-6) = 2.14301 log [s-c) = 1.66276 2 ) 20.06652 - 20 logtan^^ = 10.03326 iA = i1° IF 30^^. A = 94° 23^ Angle BAC= 94° 23^ Subtract 90° of the quadrant U to N, and we obtain 4° 23^ W. of N. But C may be to the southward of A. Hence two answers are ad- missible: W. of N. or W. of S. 18. Under what visual angle is an object 7 feet long seen by an observer whose eye is 5 feet from one end of the object and 8 feet from the other end ? a= 5 &= 8 c = _7 2s = 20 s = 10. s — a= 5. s-h= 2. s — c= 3. cologs = 9.00000-10 colog (s - a) = 9.30103 - 10 log(s-&) = 0.30103 log(s-e) = 0.47712 2)19.07918-20 logtan|^= 9.53959 J^ = 19° 6^24^^. A = 38° 12^ 48^^. colog s = 9.00000 - -10 log (s - -a) = 0.69897 colog (s - -6) = 9.69897- -10 log (s - -c) = 0.47712 2)_ 19.87506 - -20 log tan hB = 9.93753 iB = 40° 53^ 36^^ B = 81° 47^ 12^^. .\C= 60°. 19. When Formula [28] is used for finding the value of an angle, why does the ambiguity that occurs in Case II. not exist? When Formula [28] is used for finding the value of an angle, the ambiguity that occurs in Case II. does not exist because the sides are all known and the angle can 106 TEIGONOMETRY. have but one value ; while in Case II. the side opposite the angle is not known, and may have two val- ues ; therefore the angle also may have two values. 20, If the sides of a triangle are 3, 4, and 6, find the sine of the largest angle. a = 3 h= 4 c = _6 2s = 13 5 = 6.5. s — a = 3.5. s -h = 2.5. s{s- -c) = 3.25. log {s- -a) = 0.54407 log (s- -h) = 0.39794 colog S{S' -c) = 9.48812- 2)20.43013- 10 20 log tan \G = J 0.21507 \G = 58°38^2£ f^. C=117°16^50^^ log sin C = 9.94879. sin C " = 0.88877. 21. Of three towns A, B, and C, A is 200 miles from B and 184 miles from (7, B is 150 miles due north from C; how far is A north of (7 ? a = 150 s = 267. 6 = 184 s-a = 117. c = 200 s-h= 83. 2s = 534 s-c= 67. colog s = 7.57349 - 10 colog (s-c) = 8.17393-10 log (s - a) = 2.06819 log (s-b) = 1.91908 2 )19.73469-20 Iogtan^^= 9.86735 itl a' G J^ = 36°22^58^^ A - 72° 45^ 56^^ Draw ± from A to BC. To find a^ (part cut off by ± on BC from c). a^= b cos C. log b = 2.26482 log cos C = 9.47171 log a^ = 1.73653 a' = 54.516. TEACHERS EDITION. 107 Exercise XVII. Page 69. 1. Given a = 4474.5, b = 2164.5, C= 116° 30' 20^' ; find the area. F= Ja&sin C. log a = 3.65075 log b = 3.33536 colog 2 = 9.69897 - 10 log sin C = 9.95177 log F = 6.63685 F =4333600. 2. Given b = 21.66, c = 36.94, A = 66° 4' 19^' ; find the area. F=^bcsmA. log 6 = 1.33566 logc = 1.56750 log sin A = 9.96097 \og2F = 2.86413 2F = 731.36. F = 365.68. 3. Given a = 510, c = 173, B = 162° 30' 28'' ; find the area. log a = 2.70757 log c = 2.23805 log sin B = 9.47795 colog 2 = 9.69897 - 10 log F = 4,12254 F =13260. 4. Given a = 408, 6 = 41, c = 401; find the area. a = 408 6= 41 c=401 2s = 850 s = 425. s — a = 17. s-b = 384. s-c= 24. log s = 2.62839 log(s-a) =1.23045 log {s-b) = 2.58433 log(s-c) =1.38021 2)7.82338 log F = 3.91169 F = 8160. 5. Given a = 40, 6 = 13, c = id the area. a = 40 = 37; 6 = 13 c = 37 2s = 90 s = 45. s — a = 5. s-b = 32. s-c= 8. log s = 1.65321 log (s - a) = 0.69897 log (s - 6) = 1.50515 log (s-c) = 0.90309 2 )4.76042 log F = 2.38021 F = 240. 6. Given a = 624, 5 = 205, c = 445; find the area. a= 624 6= 205 c= 445 2s = 1274 108 TRIGONOMETRY. s= 637. s — a= 13. s-b= 432. s-c= 192. log s = 2.80414 log {s-a) = 1.11394 log Is-b) = 2.63548 log {s-c) = 2.28330 2 log i^ = 8.83686 log F = 4.41843. F = 26208. 7. Given h = 149, ^ = 70° 42^ 30^^ - 39° 18^ 28^^ ; find the area. A = 70° 42^ 30^'' B = 39° 18^ 28^^ .'.C = 69° 59^ 2^^ log 6 = 2.17319 colog sin B = 0.19827 log sin A = 9.97490 log a = 2.34636 log a = 2.34636 colog sin yl = 0.02510 log sin C = 9.97294 . logc = 2.34440 colog 2 = 9.69897 - 10 log a = 2.34636 log& = 2.17319 log sin C = 9.97294 logF = 4.19146 F = 15540. log c = 2.48813 log sin A = 9.62852 colog a = 7.66575 — 10 log sin C =9.78240 C= 37° 17^ 38^^. -•. 5 = 117° 32^ 51^^ Or, C = 142° 42^ 22^^ .'.B^= 12° 8^ 7^^. colog 2 = 9.69897 log a = 2.33425 log c =■ 2.48813 loR sin B = 9.94774 10 8. Given a = 215.9, c- 307.7,^ 25° 9^ 31^^ ; find the area, n < c and > c sin A. A < 90°. .'. two solutions. logi^ = 4.46909 F = 29450. colog 2 = 9.69897 - -10 log a = 2.33425 logc = 2.48813 log sin B' = 9.32269 logjP'' = 3.84404 F' = 6983. 9. Given 6 = 8, c = 5, ^ = 60°; find the area. F=\'bc sin J. = 1(8x5) (0.86602) = 20 X 0.86602 = 17.3204. 10. Given a =7, c = 3, ^ = 60°; find the area. colog a = 9.15490 - 10 log c = 0.47712 log sin A = 9.93753 log sin C = 9.56955 C ^ 21° 47^ 12^^. .-. B = 98° 12^ 48^^ TEACHEES EDITION. 109 log a = 0.84510 log sin B = 9.99552 log sin A = 0.06247 log b = 0.90309 b = 8. F= ■■ J be sin A = J X 24 X W3 = 6V3, or 10.3923 11. Given a = 60, ^ = 40° 35^ 12^^ area = 12 ; find the radius of the inscribed circle. ^ ac sin B = 12. 24 log 24 colog a colog sin B log c c a sin B = 1.38021 = 8.22185 = 0.18665 10 = 9.78875-10 = 0.61483. tanH^-C) a a + c Xtan^(J. + C) = ^^-^^^^"^ X tan (69° 42^ 24^0- 60.61483 ^ ^ log {a-c)= 1.77368 colog (a + c) = 8.21742 - 10 logtani(^+g)= 0.43206 log tan H^-C') = 0-42316 i{A-C)= 69° IV 19^^ i{A+C)= 69° 42^ 24^'' A =139° 1M3^^ B a sinB a sin A sm-d log a = 1.77815 log sin B =9.81331 colog sin A =0.18331 log 5 = 1.77477 b = 59.534. a= 60 6= 59.534 c= 0.61483 2s =120.14883 s= 60.07442. s - a = 0.07442. s -b= 0.54042. s -c= 59.45959. log {&-a)= 8.87169 - 10 log (s - 5) = 9.73274 - 10 log {&-c)= 1.77422 colog c = 8.22131-10 2)18.59996-20 logr = 9.29998-10 r = 0.19952. 12. Obtain a formula for the area of a parallelogram in terms of two adjacent sides and the included angle. By Geometry, area of parallelo- gram = base X height. In this case = bli. But A = a sin J.. .•. area oi LJ ^ ab sin A. 110 TRIGONOMETRY. a 13. Obtain a formula for the area of an isosceles trapezoid in terms of the two parallel sides and an acute angle. Let AB = a. F=^{a + h) c. - = tan A P c =p tan A. .-. F= l{a^h)x^{a—h) tan A = \{a?- ¥) tan A. 14. Two sides and included angle of a triangle are 2416, 1712, and 30° ; and two sides and included angle of another triangle are 1948, 2848, and 150°; find the sum of their areas. Let a = 2416, c = 1712, B = 30°. F= h ac sin B. log a =3.38310 log c = 3.23350 colog 2 = 9.69897 - -10 log sin B = 9.69897 log F = 6.01454 F =1034000.. Let a^= 1948, c'= 2848, 5^=150= jr/=iaVsin5^. P log a^ = 3.28959 log c^ = 3.45454 colog 2 = 9.69897 - 10 log sin B^ = 9.69897 log F^ = 6.14207 F^ = 1387000. F+F^ =2421000. 15. The base of an isosceles tri- angle is 20, and its area is lOO-r-VS ; find its angles. a = b. c = = 20. F= = 100^ Vs. ^ch = 100 V3 10 h = 100 V3 h^ 10 V3 h = tan A. log h = - 0.76144 log Jc = = 9.00000 - 10 log tan A = 9.76144 A = 30°. B = 30°. C = 120°. TEACHERS EDITION. Ill Exercise XVIII. Page 70. 1. From a ship sailing down the English Channel the Eddy stone was observed to bear N. 33° 45^ W. ; and after the ship had sailed 18 miles S. 67° 30^ W. it bore N. 11° 15^ E. Find its distance from each position of the ship. a = 18 miles. ACE^ 33° 45^ DCB= 67° 30°. ABF= 11° 15^ ACB = 1S0°~{ACE + DCB) = 78° 45^ CBD = 90° -DCB = 22° 30^ ABC= 90° - {CBD + ABF) = 56° 15^ .^^C=45°. 6_sin_S c _ sin C a sin J. sin A log a = 1.25527 log sin B = 9.91985 colog sin A = 0.15051 logb = 1.32563 b = 21.166. log a = 1.25527 log sin C = 9.99157 colog sin A = 0.15051 logc = 1.39735 c = 24.966. 2. Two objects, A and B, were observed from a ship to be at the same instant in a line bearing N. 15° E. The ship then sailed north- west 5 miles, when it was found that A bore due east and B bore north-east. Find the distance from Ato B. N W- E „'^ 45° 105° /. ^\45° 75°; \ d^ V / / ^n60°/ N / NS 112 TRIGONOMETRY. S'A sin ASS' log 140 = 2.14613 SS' sin S'AS colog sm A = 0.00665 log SS' = 0.69897 log AC = 2.15278 colog sin SAS' = 0.01506 HCA = 10°, log sin ASS' = 9.93753 MCB = 40°, log S'A = 0.65156 .-. ^C5 = 40°; CAB = 10°, AB sin BS'A .-. ^5(7=130°. S'A sm S'BA 4P_ -ACsin C log /S^^ = 0.65156 sin -B colog sm S'BA = 0.30103 log.4C =2.15278 log sin BS'A = 9.84949 log sin C =9.80807 log AB = 0.80208 colog sin B = 0.11575 AB = 6.3399. log AB = 2.07660 • AB --= 119.29. 3. A castle and a monument stand on the same horizontal plane. The angles of depression of the top and the bottom of the monument viewed from the top of the castle are 40° and 80° ; the height of the castle is 140 feet. Find the height of the monument. 110= height of castle. AB = height of monument. MCB = ^0°. BAC =S0°. HCA = 10°. HC =140 ft. 140 AC= sin A 4. If the sun's altitude is 60°, what angle must a stick make with the horizon in order that its shadow in a horizontal plane may be the longest possible ? The shadow of the stick will be the longest when the stick is per- pendicular to the rays of the sun. Let BC represent the stick, and ^Cthe horizontal plane. B = 90°. A = 60°. .-. C = 30°. 5. If the sun's altitude is 30°, find the length of the longest shadow cast on a horizontal plane by a stick 10 feet in length. Let a be a stick A. to rays of sun, and c be the longest shadow. - = sm A. c TEACHERS EDITION. 113 log a colog sin A logc c 1.00000 0.30103 1.30103 20. 6. In a circle with the radius 3 find the area of the part comprised between parallel chords whose lengths are 4 and 5. (Two solu- tions.) log(s-6) = 0.30103 log (s-c) = 0.30103 cologs(s-a) = 9.30103-10 2 )19.90309-20 logtanJ-BOC= 9.95155-10 J50C= 41° 48^ 38^^ BOQ= 83° 37^ 16^^ By Table VI. B = Z. :. area O = 28.274. 301036 BOQ- 1296000 75259 X360° 324000 area sector BOC 75259 X360^ X 28.274. log 75259 log 28.274 colog 324000 log area Area sector 50C= 6.5675. 324000 = 4.87656 = 1.45139 = 4.48945 - 10 = 0.81740 In triangle BOO F=Vs{s- - cos a + COS c 1 + cos = cos a . 1 — cos 6 _ c os a — cos c 1 + cos 6 cos a + cos c But by [18], page 47, 1 — cos 6 1 + cos 6 = tan^ ^ 6. And if in [23] and [22], page 48, we write a and c in place of A and B and divide [23] by [22], we get cos a — cos c cos a + cos c = — tan ^{a -j- e) tan i{a — c) = tan J (c + a) tan J (c — a). .'. tan^ J 6 = tan ^ (c + a) tan f (c — a). 5. Deduce from [37]-[42] the for- mula tan2 (45° -^A) = tan J (c — a) cot J (c + a). 120 TRIGONOMETRY. From [38], sin A = sm a sm c when, operating as in Example 4, we have 1 — sin A _ sin c — sin a 1 + sin A sin c + sin a If in [19], page 47, we substitute 90° -r A for z, and remember that cos (90° + A) = — sm A, [19] reduces to the form ^^^^ = cot2(45°+J^) 1 + sm ^ = tan2(45°-J^), (since 45° + J J. and 45° — ^A are complementary angles). And by dividmg [21] by [20], page 48, and writing c for A and a for B, we have sin c — sin ^4 sin c + sin J. = tan ^{c — a) cot J (c + a). .•.tan2(45°-J.4) = tan }{c — a) cot J (c + a). 6. Deduce from [37]-[42] the for- mula tan^ ^ B = sin (c — a) esc (c + a). From [39], by operating as be- ' 1 — cos B _ tan c — tan a 1 + cos B tan c + tan a By [18], page 47, l^^^^ = tanH5. 1 + cos i? And we write on the second side sin G ■ -, c i. A sin a • ■ m place oi tan c, and m cos c cos a place of tan a, and reducing, we obtain tan c — tan a _ sin (c — a) tan c + tan a sin (c + a) 1 p _ sin (c — g) sin (c + a) sin (c — a) esc (c + a). t&n^B 7. Deduce from [37]-[42] the for- mula tan^ J c = — cos {A+B) sec {A—B). By [42], cos c = cot ^ cot B _ cot J. . tan B' whence, as before, 1 — cos c _ tan B — cot A 1 + cos c tan B + cot A By [18], page 47, 1 — cos c and tan^ I c, 1 + cos c tan B — cot A tan B + cot A ■ sin ^ cos A cos ^ sin J. sm B cos ^ cos B sin J. _ sin Asm B — cos J. cos ^ sin Asm B + cos J. cos B _ — cos (^ + B) ~ cos{A-B) ' .-. tan^ J c = - cos {A+B) sec {A—B). 8. Deduce from [37]-[42] the for- mula tan2 J J. = tan [J ( J. + ^) - 45°] tan [J (^-5) + 45°]. From [40] cos A cos a = cos A CSC 5 = sin B TEACHERS EDITION. 121 whence, by proceeding as before, 1 — cos a _ sin B — cos A 1 + cos a sin B + cos A By [18], page 47, 1 cos a 1 + cos a tan^ h a. If in [6], p. 44, we make x = 45°, we have tan (45°+ 2/) = ^^ cosy _ cos y + sin y cos y — sin y And, in like manner, making y = 45° in [10], page 46, we have tan (x -45°) = sm X — cos X sin X + cos a; .-. tan (a;-45°) tan (45° + y) (sin X — cos x) (sin y + cos y) (sin a; + cos x) (cos y — sin y) sin {x — y) — cos (a^ + y) sin (a? — y) + cos (cc + y) If in this equation, in which x and y may have any values, we make x = \{A^-Bl y = l{A-B\ whence x-y = B, x + y = A,' the equation becomes tan [^(^ + ^)- 45°] tan [^ (^ - 5) + 45°] _ sin B — cos A sin J? + cos -4 .-. tan2 ^ a = tan [} {A + B) - 45°] tan [i {A-B) + 45°]. Exercise XXI. Page 109. 1. Show that Napier's Rules lead to the equations contained in For- mulas [38], [39], [40], and [41]. sin a = cos co. c cos co. A = sin c sin A. sin h = cos co. c cos co. B = sin c sin B. sin CO. B = tan a tan co. c. cos 5 = tan a cot c. sin CO. A = tan 6 tan co. c. cos J. = tan b cot c. sin CO. A = cos a cos co. ^. cos A = cos a sin B. sin CO. 5 = cos b cos co. J., cos B = cos 6 sin A. sin a = tan co. B tan 5 = cot B tan 6. sin b = tan a tan co. A = tan a cot -4, 2. What will Napier's Rules be- come, if we take as the five parts of the triangle, the hypotenuse, the two oblique angles, and the comple- ments of the two legs. (i.) Cosine of middle part equals product of cotangents of adjacent parts. (ii.) Cosine of middle part equals product of sines of opposite parts. 122 TRIGONOMETRY. Exercise XXII. Page 114. 1. Solve the right triangle, given a =36° 27^ 5 = 43° 32^ ?>V\ Taking a as the middle part, we have, hy Eule I., sin a = tan h cot B, whence tan h — sin a tan B, ■3 ^ -r, tan h and sin a log tan h = 9.97789 colog sin a = 0.22613 log tan B = 10.20402 B = 57° 59^ 19.3^^ Taking h as the middle part, hy Rule I., sin h = tan a cot A. tan a = sin & tan A. tan a tan A sin h log tan a = 9.86842 colog sin h = 0.16185 log tan J. =10.03027 A = 46° 59^ 43.2^^. Taking co. c as the middle part, by Rule II., cos c = COS a cos h. log cos a = 9.90546 log cos h = 9.86026 log cos c = 9.76572 c = 54° 20^ 2= Solve the right triangle, given a = ^Q° 40^ h = 32° 40^ cos c = COS a cos h. log cos a = 8.76451 log cos h = 9.92522 log cos c = 8.68973 c = 87° 11^ 39.8^^. tan A = tan a esc 5. log tan a =11.23475 log CSC h = 0.26781 log tan ^ =11.50256 A = 88° 11^ 57.8^^ tan B = tan h esc a. log tan h = 9.80697 log CSC a = 0.00074 log tan 5 =9.80771 B = 32° 42^ 38.7^^ 3. Solve the right triangle, given = 50°, b = 36° 54^ 49^^ cos c = COS a cos b. tan A = tan a esc b. tan B = tan 5 esc a. log cos a = 9.80807 log cos b = 9.90284 log cose =9.71091 c = 59° 4^ 25. 7^^ log tan a =10.07619 log CSC b = 0.22141 log tan ^ =10.29760 A =63° 15M3.13^^ log tan b = 9.87575 log CSC a = 0.11575 log tan 5 =9.99150 B = 44° 26^ 21 .6^^ TEACHEES EDITION. 123 4. Solve the right triangle, given a = 120° 10^ b = 150° 59^ U^\ cos c = COS a cos h. t&nA = tana esc b. tan B = tan b esc a. log cos a =9.70115 log cos & = 9.91180 log cos c = 9.64295 c = 63° 55' 43.3'^ log tan a = 10.23565 log esc 6 = 0.31437 log tan A = 10.55002 J..= 105°44'21.25''. log tan 5 =9.74383 log esc a = 0.06320 log tan B = 9.80703 5 = 147°19'47.14'^ 5. Solve the right triangle, given c = 55° 9' 32'^ a = 22° 15' 7''. cos B = tan a cot c. log tana =9.61188 log cote = 9.84266 log cos B = 9.45454 B = 73° 27' 11.16". tan h = sin a tan B. log sin a = 9.57828 log tan B = 10.52709 log tan 6 =10.10537 b =51° 53'. cos A = tan b cot c. log tan b = 10.10537 log cote = 9.84266 log cos A = 9.94803 A - 27° 28' 25.71". 6. Solve the right triangle, given c = 23° 49' 51", a = 14° 16' 35". cos b = cos c sec a. log cose =9.96130 log sec a = 0.01362 log cos & =9.97492 b = 19° 17'. sin A = sin a esc c. log sin a =9.39199 log CSC e = 0.39358 ^ — — « log sin A =9.78557 A = 37° 36' 49.4'^ cos B = tan a cot c. log tan a = 9.40562 log cote = 10.33488 log cos B = 9.76050 B = 54° 49' 23.3". 7. Solve the right triangle, given c = 44° 33' 17", a = 32° 9' 17". cos b = cos c sec a. log cose =9.85283 log sec a = 0.07231 log cos b = 9.92514 b =32° 41'. sin A = sin a esc c. log sin a =9.72608 log CSC e = 0.15391 log sin A = 9.87999 A = 49° 20' 16.3". COS B = tan a cot c. log tan a = 9.79840 log cot c = 10.00675 log cos B = 9.80515 B = 50° 19' 16'^ 124 TRIGONOMETEY. . 8. Solve the riglit triangle, given c = 97° 13^ 4^^ a = 132° 1¥ 12'\ cos b = cos c sec a. log cose =9.09914 log sec a = 0.17250 log cos 6 =9.27164 h = 79° 13^ 38.18^^ sin A = sin a esc c. log sin a = 9.86945 log esc c = 0.00345 log sin A = 9.87290 A = 48° 16^ 10^^ But A and a must be of the same kind, .-. A = 131° 43^ 50^^ cos B = tan a cot c. log tan a = 10.04196 log cot c 9.10259 log cos B = 9.14455 B = 81° 58^ 53.3^^. 9. Solve the right triangle, given a = 77° 21^ 50^^ A = 83° 56^ 40^^ sin c = sin a esc A. log sin a = 9.98935 log CSC ^ =0.00243 log sin c = 9.99178 c = 78° 53^ 20^^. Since c is found from its sine, it may have two values which are supplements of each other. Hence also c = 101° 6^ 40^^ sin h = tan a cot A. log tana =10.64939 log cot A = 9.02565 log sin b = 9.67504 or b = 28° 14^ 31.3^^. = 151° 45^ 28.7^^ sin B = sec a cos A. log sec a =0.66004 log cos A = 9.02323 log sin B = 9.68327 B = 28° 49^ 57.4^^, or = 151° 10^ 2.6^^ 10. Solve the right triangle, given a = 77° 2V 50^^ A = 40° 40^ 40^^ sin c = sin a esc A. log sin a = 9.98935 log CSC J. = 0.18588 log sin c = 10.17523 .". sin c > 1, which is impossible. 11. Solve the right triangle, given a = 92° 47^ 32^^ B = 50° V V\ tan e = tan a sec B. log tan a =11.31183 log sec B = 0.19223 log tan c = 11.50406 c =91°47M0^^ tan b = sin a tan B. log sin a = 9.99948 log tan B = 10.07671 log tan b = 10.07619 b =50°. cos A = cos a sin B. log cos a = 8.68765 log sin B = 9.88447 log cos .4 =8.57212 A = 92° 8^ 23^^. TEACHERS EDITION. 125 12. Solve the right triangle, given o = 2° 0^ 55^^ B = 12° 40\ tan h = sin a tan B. log sin a =8.54612 log tan 5 -9.35170 log tan & = 7.89782 h =0° 27^ 10.2^^ tan c = tan a sec B. log tan a = 8.54639 log sec ^ -0.01070 log tan c = 8.55709 c = 2° 3' 55.7^^ cos A = cos a sin ^. log cos a = 9.99973 log sin B = 9.34100 log cos ^ =9.34073 A = 77° 20^ 28.4^^ 13, Solve the right triangle, given a = 20° 20^ 20^^ B = 38° 10^ 10^^ tan b = sin a tan B. log sin a = 9.54104 log tan B = 9.89545 log tan b = 9.43649 b = 15° 16^ 50.4^^ tan c = tan a sec B. log tan a = 9.56900 log sec -S = 0.10448 log tan c = 9.67348 c = 25° 14^ 38.2^^ cos A = cos a sin B. log cos a = 9.97204 log sin B = 9.79098 log cos J. = 9.76302 A = 54° 35^ 16.7^^ 14. Solve the right triangle, given a = 54° 30^ B = 35° 30^ tan c = tan a sec B. log tan a = 10.14673 log sec B = 0.08931 log tan c =10.23604 c = 59° 51^ 20.7^^. tan b = sin a tan B. log sin a =9.91069 log tan B = 9.85327 log tan 6 =9.76396 b = 30° 8^ 39.3^^ cos A = cos a sin B. log cos a = 9.76395 log sin B = 9.76395 log cos A = 9.52790 A = 70° 17^ 35^^. 15. Solve the right triangle, given c = 69° 25^ 11^^, A = 54° 54^ 42^^. sin a = sin c sin A. log sin c = 9.97136 log sin A = 9.91289 log sin a = 9.88425 a =50°. tan b = tan c cos A. log tan c = 10.42541 log cos A = 9.75954 log tan 6 =10.18495 b = 56° 50^ 49.3^^ cot B = cos c tan A. log cos c = 9.54595 log tan A = 10.15335 log cot B = 9.69930 B = 63° 25^ 4^^. 126 TRIGONOMETRY. 16. Solve the right triangle, given c = 112° 48^ A = 56° 1 V 56^^ cot B = cos c tan A. log cose = 9.58829 log tan J. = 10.17427 log cot ^= 9.76256 £ = 120° 3^ 50^^ sin a = sin c sin A. log sin c = 9.96467 log sin ^ = 9.91958 log sin a =9.88425 a = 50°. tan b = cos A tan c. log cos A = 9.74532 log tan c =10.37638 log tan 5 =10.12170 b = 127° 4^ 30^^ 17. Solve the ri^ ^ht triangle, given c = 46° 40^ 12^^ A = 37° 46^ d''. sin a = sin A sin c. log sm A = 9.78709 log sin G = 9.86178 log sin a = 9.64887 a = 26° 27^ 24^^ tan b = tan c cos A. log tanc = 10.02533 log cos A = 9.89789 log tan b = 9.92322 b = 39° 57^ 41.5^^ cot^ = tan A cos c. log cos c = 9.83645 log tan J. = 9.88920 log cot 5 = 9.72565 £ = 62° 0^ 4'^ 18. Solve the right triangle, given c = 118° 40^ V^, A = 128° 0^ 4^^ sin a = sin c sin A. log sin G = 9.94321 log sin A = 9.89652 log sin a = 9.83973 a = 136° 15^ 32.3^^. tan b = tan c cos A. log tanc =10.26222 log cos J. = 9.78935 log tan 6 = 10.05157 b = 48° 23^ 38.4^^. cot B = cos c tan A. log cos c = 9.68098 log tan J. = 1010717 log cot 5= 9.78815 B = 58° 27^ 4.3^^. 19. Solve the right triangle, given ^ = 63°15^12^^ ^=135° 33^ 39^^. cos c = cot A cot B. log cot ^= 9.70241 log cot B = 10.00850 log cos c = 9.71091 c = 120° 55^ 34.3^^. cos a = cos A CSC B. log cos A = 9.65326 colog sm B = 0.15480 log cos a = 9.80806 a = 49° 59^ 56^^ cos b = cos B CSC A. log COS .5 = 9.85369 colog sin A = 0.04915 log COS b =-9.90284 b = 143° 5^ 12^^ TEACHERS EDITION. 127 20. Solve the right triangle, given A = 116° 43^ 12^^ B = 116° 3V 25'\ cos a = cos A CSC £. log cos A = 9.65286 log CSC B = 0.04830 log cos a = 9.70116 a =120° 10^ 3^^ cos b = cos -B CSC A. log cos 5 = 9.64988 log CSC J. = 0.04904 log cos h = 9.69892 b =119°59M6^^. log cos c = cot A cot B. log cot ^ = 9.70190 log cot 5 = 9.69818 log cos c = 9.40008 c = 75° 26^ 58^^ 21. Solve the right triangle, given A = 46° 59^ 42^^, B = 57° 59^ 17^^. cos a = COS A CSC B. log cos A = 9.83382 log esc B = 0.07164 log cos a = 9.90546 a = 36° 27^ cos b = cos B CSC A. log cos B = 9.72435 log CSC A = 0.13591 log cos b = 9.86026 b = 43° 32^ 37^^ cos c = cot A cot B. log cot ^ = 9.96973 log cot B = 9.79599 log cos c = 9.76572 c = 54° 20^ 22. Solve the right triangle, given A = 90°, B = 88° 24^ 35^^. cos c = cot A cot B. But cot J. = 0. .•. cos c = 0. c =90°. cos a = cos J. CSC B. But COS .A = 0. .•. cos a = 0. a =90°. cos b = cos B CSC J.. CSC A = l. b =B. b = 88° 24^ 35^^. 23. Define a quadrantal triangle, and show how its solution may be reduced to that of the right triangle. A quadrantal triangle is a tri- angle having one or more of its sides equal to a quadrant. Let A^B'Q^ be a quadrantal tri- angle with side A'B'= 90°, or a quadrant. Let ABQ\)^ its polar triangle. Then since A'B'^ (7=180°, C=90°. .-. ABQ\% a right triangle. .•. all parts of the polar triangle may be found by formulas for right triangle. The parts of A^B^C^ may then be found by subtracting proper parts of A^Cfrom 180°. 24. Solve the quadrantal triangle whose sides are : a = 174° 12^ 49.1^^ b= 94° 8^20'^, c = 90°. 128 TRIGONOMETRY. Let A^, B^, C, a\ ¥, c' repre- sent the corresponding angles and sides of the polar triangle. Then A'^ 5=> 47^ 10.9^^, B'= 85° bV W\ C^= 90°. tan^ \ c' = - cos {B'+A') sec {B^-A'). B'+ A^= 91° 38^ 50.9^^ B^- A^= 80° 4^ 29. F^. log cos (5^+ J-O =8.45863 log sec (^^- .40 = 0.76356 2 )9.22219 log tan Jc^ = 9.61110 Jc^= 22° 12^561^^. c^ = 44° 25^ 53^^. C==135°34^ 7^^ tan2 J b'= tan [J {B'+ A^ - 45°] tan[45°+H^^-^0]- J (^^+ .SO - 45° = 49^ 25.5^^ 45° + i {B'- A^) = 85° 2^ 14.6^^ log tan 0° 49^ 25.5^^= 8.15770 log tan 85° 2^4.6^^= 11.06133 2) 9.21903 log tan 1 6^= 9.60952 J&^= 22° 8^35^^ 6^= 44° 17^ 10^^. 5 =135° 42^ 50^^ tan2 J a'= tan [J (^^+ ^0 - 45°] tan[45°-H^'-^0]- H-S^+^0 - 45° = 0° 49^ 25.5^^. 45° - J {B'-A') = 4° 57^ 45.4^^ log tan 0° 49^ 25.5^^= 8.15770 log tan 4° 57^ 45.4^/= 8.93867 2 ) 7.0963 7 log tan ^a^= 8.54819 ia^= 2° 1^25^''. a^= 4° 2^50^^. ^ = 175°57^10^^ 25. Solve the quadrantal triangle in which c = 90°, A = 110° 47^ 50^^ 5=135°35^34.5^^ Let A\ B\ (7, a', ¥, c^ repre- sent the corresponding angles and sides of the polar triangle. Then a'= 69° 12^ 10^^. &/= 44° 24^ 25.5^^ (7= 90°. tan A^= tan a^ esc h^. log tan a' = 10.42043 log CSC y = 0.15505 log tan A^= 10.57548 A^= 75° 6^58^^ a = 104° 53^ 2^^. tan B^= tan h^ csc a^. log tan 5^= 9.99101 log CSC a^ = 0.02926 log tan B^= 10.02027 B^= 46°20M2^^ b = 133° 39^ 48^^ cos c' = cot J.'' cot B^. TEACHERS EDITION. 129 26. A, Q angle. log cot A'=^ 9.42452 log cot B^= 9.97973 log cos c^ = 9.40425 c/ = 75° 18^ 2V^. (7=104°4F39^^ Given in a spherical triangle and c = 90° ; solve the tri- sin a = sin c sin A. = 1x1. .'. a = 90°. tan b = tan c cos A = cox0. .-. b = 45°. cot 5 = cos c tan A. = 0xco. .-. B = 45°, 27. Given ^ = 60°, C= 90°, and c = 90° ; solve the triangle. sin a = sin c sin A. tan b = tan c cos yl. cot 5 = cos c tan J.. sin c = 1. .•. sin a = sin A. a = A = 60°. tan c = 00. .". tan 5 = 00. 6 = 90°. cos e = 0. .'•.cot^ = 0. B = 90°. 28. Given in a right spherical triangle, A = 42° 24^ 9^^ ^ = 9° 4^ W^ ; solve the triangle. cos c = cot J. cot B. log cot tI = 10.03943 log cot B = 10.79688 log cos c = 10.83631 which is impossible. .". triangle is impossible. 29. In a right triangle, given a = 119° IF, ^ = 126°54^ solve the triangle. tan c = tan a sec B. log tan a = 10.25298 log cos B = 0.22154 log tan c = 10.47452 c = 71° 27^ 43^^ tan b = sin a tan 5. log sin a = 9.94105 log tan B = 10.12446 log tan b = 10.06551 6 =130°4F42^^ , cos .A = cos a sin 5. log cos a =9.68807 log sin B = 9.90292 log cos A = 9.59099 ^ = 112° 57^ 2^\ 30. In a right triangle, given c = 50°, b = 44° 18^ 39^^ ; solve the triangle. cos a = cos c sec b. log cos e = 9.80807 colog cos b = 0.14535 log cos a = 9.95342 a = 26° 3^ 51^^. 130 TRIGONOMETRY. sin A = sin a CSC c. log sin a = 9.64284 log CSC c = 0.11575 log sin A = 9.75859 A = 35°. tan 5 = tan h CSC a. log tan b = 9.9S955 log CSC a = 0.35716 log tan B = 10.34671 £^ 65° 46^ 7'' 31. In a right triangle, given A = 156° 20' 30^^ a = 65° 15^45^^; solve the triangle. It is impossible, because a and A are unlike in kind. And, in Case III., they must be alike in kind; otherwise, impossible. 32. If the legs a and 5 of a right spherical triangle are equal, prove that cos a = cot J. =Vcosc. • COS c = COS a cos b. But cos a = cos b. •, COS c = cos^a. and •. cos a = Vcos c. sin a = cos a sin b tan J.. Since sin a = sin 5, 1 = cos a tan A. cos a = tan A . cos a = cot A. 33. In a right triangle prove that co&^A X sin^c = sin (c — a) sin (c + a). cos J. sin c = cos a sin b. .". cos^^ sin^c = cos^a sin^b. (1) sin^6 = sin^c sin^^ = sin^c (1-C0S2.5). (2) cos^B = tan^a cot^c sin^a cos^a cos^c -^ ■ 9 ■ sm-^c Substitute m (2), sin^ b = sin^? — sin^a cos^c cos'^^a sin^c cos^a— sin^a cos^c cos'^'a Substitute in (1), cos^a sin^Z) = sin^c coshes — sin^a cos^c = (sin c cos a + sin a cos c) (sin c cos a — sin a cos c) = sin (c + a) sin (c — a). Substitute in (1), cos^^ sin^c = sin (c + a) sin (c — a). 34. In a right triangle prove that tan a cos c = sin b cot ^. sin & = tan a cot -4. , > sin & cot -4= ■ ^• tan a cos c = cot -4 cot B. cot A cos c cotB cos c sin 6 cot B tan a tan a cos c == sin 6 cot ^. 35. In a right triangle prove that sin^-4 = cos^^ + sin% sin^B. sin o = sin ^ sin c. • o A sin^a sin^c cos B = tan a cot e. cos^5 = tan^a cot^c. TEACHERS EDITION. 131 sinM— cos'^B : — tan^a cot^'c sin^a sin^a cos^c sin^c cos^a sin^c sin% cos^a — sin^a cos^c cos'^a sm^c _ sin% (cos^a — cos^c) Now cos^'a sm'^c cos c = cos a cos 6. (1) cos^a = cos^c cos^S cos^c = cos^a cos^5. sin b sm c = sm^c = sin^ s in^Z) Substitute these values in (1), sin^a — co&'^B = sin'^a 'cos-'c cos^6 cos^a cos^S cot^^c sin-6 X = sin^a = sin^a _ cos'-^i sin^^ '" cos^c — cos^g cos^5 cos^6 cos^c sin^6 sin^i? cos^6 sin^5 cos'^c— cos^acos*& cos^c sin^6 • 2 • 9 D /cos^c — cos'^acos^SX sin^a sm^ij f : ) cos^c sin^6 J But cos^c — cos% cos*6 cos^c sin^5 -Q^2 _ ODx OE^ - 0~D' 0E\ DE' 0E\ OD^ - OE^ 0E\ ED'' on^~OE^ ED" = 1. .•. sin^^ — cos^^ = sin^a sin^^. .•. sinM = cos^^ + sin^a sin^^. 36. In a right triangle prove that sin (6 + c) = 2 cos^ J a cos 6 sin c. sin (b + g) = sin 5 cos c + cos h sin c sin & cos c , T \ 7 • ; — : — + 1 cos sm c cos sm c J = (tan b cot c+1) cos b sin c. (1) But tan b cot c = cos A, .•. tanJ cote +1 ^ cos J. + 1. (2) 132 TRIGONOMETRY. 4 cos ^ + 1 COS hA. Substitute in (2), (tan bc^e + 1) = 2 cos^ J A. Substitute in (1), sin (6 + c) = 2 cos^ ^A cos h sin c. 37. In a right triangle prove that sin {c — b) = 2 sin^ ^a cos 6 sin c. sin (c — b) = sin c cos b — cos c sin b 7 A-, cosf sin b\ ^^s = sm c cos 1 — : (1) \ sin c cos oy = sin c cos 6(1— cote tan 6). (2) cot c tan b = cos A. 1 — cos A 1— cote tan 5 = 2 (3) sinM = '\~ cos A Substitute in (3), 1 — cot c tan 5 = 2 sin^ | A. Substitute in (2), sin (c — 5) = 2 sin^ J a sin e cos b. 38. If, in a right triangle, p de- note the arc of the great circle pass- ing through the vertex of the right angle and perpendicular to the hypotenuse, w and n the segments of the hypotenuse made by this arc adjacent to the legs a and b respec- tively, prove that (i.) tan^a = tan c tan m, (ii.) sin^p = tan m tan n. In triangle BCA cos B = tan a cot c. , cos B :. tan a = cot c In triangle CBD tan 5(7= tan 5Z) sec 5. tan a = tan ni sec B _ tan m cos 5 Multiplying the two equations, , o tan m ^, cos 5 tan''a = X cos B cot c = tan m tan c. 2d. In triangle CBD sinp = tan m cot ili"; and in triangle CAD sinp = tan n cot iV! But cot Mx cot i\^= 1. .-. if+iV"=90°. sin^p = tan m tan w. TEACHERS EDITION. 133 Exercise XXIII. Page 116. 1. In an isosceles spherical tri- angle, given the base h and the side a ; find A the angle at the base, B the angle at the vertex, and h the altitude. Let ABA^ be an isosceles trian- gle, A and A^ being the equal angles, a and a^ the equal sides. Let h the arc of a great circle be drawn from B perpendicular to AA'. Let h and co. c be the given parts. 6 = J 6 in triangle ABA\ c = a in triangle ABA\ B = ^B in triangle ABA\ cos A = cot a tan | b. sin 2" & = sin a sin ^ B. sin^ B = esc a sin | b. cos A = cos a sec J &. 2. In an equilateral spherical triangle, given the side a ; find the angle A. In the equilat. triangle AA^A^^ draw &rc AC ± to A^A^^. Then in right triangle AA^C, given a. sin ^ J. = sin J a esc a =V^ cos a 1 sma / 1 — cos a ^'2 (1 - cos^a) '2(1 -(-COS a) 2 ^ 1 4- cos a ^ sec I a. Also, cos -4 = tan ^ a cot a V 1— cosa cos% l-hcosa 1— cos^a = cos a X 1 -I- cos a = cos a J sec^ J a. 3. Given the side a of a regular spherical polygon of n sides ; find the angle A of the polygon, the distance H .from the centre of the polygon to one of its vertices, and the distance r from the centre to the middle point of one of its sides. In the regular polygon ABDE draw arcs from the vertices A, B, etc., through the centre C, and from Cto M, the middle of one side. 134 TRIGONOMETRY. Then n also n and CAM = I A, AM=\a, AC=B, and MO=r. In equilateral right triangle CMA represent the parts by letters A^, B\ C\ etc. sin A^ = cos B' sec V. Substituting known values, sin ^ ^ = sec \ a cos n Or, sin c' = sin h^ esc B^, sm li = sm ^ a esc 180° Or, sin a' = tan b^ cot B^, 180^ sin r = tan i a cot 4. Compute the dihedral angles made by the faces of the five regu- lar polyhedrons. Let the figure ABCD represent a tetrahedron. It is required to find the dihedral angle ADCB or the corresponding plane angle DFO. DAF= 60° (since it is an angle of an equilat- eral triangle). DFA = rt. triangle (DF being JL to AC, since it is a side of the plane angle DFO corre- sponding to the dihedral angle ABCB). Let AD = 1, Then DF AD sin 60°. DF = AD sin ( 30°. log AD = 0.00000 log sin 60° = = 9.93753 -10 log DF = AF AD = 9.93753 = cos 60°. -10 AF=^ = AD cos 60°. log AD = = 0.00000 log cos 60° = = 9.69897 -10 log AF = 9.69897 - 10 Draw OA from centre of triangle ABC. OAF = I CAB =2,0° (since OA produced to the middle of the opposite side CB will pass through the centre and will also bisect the angle CAB, CA and AB being equal), OF A = rt. triangle (since Oi^ is ± to ^C). OF = tan 30°. AF Oi^=Ai^ tan 30°. log AF = 9.69897 log tan 30°= 9.76144 log OF 10 OF 9.46041 - 10 = cos DFO. DF log OF = 9.46041 - 10 colog DF = 0.06247 log cos DFO = 9.52288 - 10 DFO = 70° 31^ 43^^ TEACHEES EDITION. 135 Let AC^ be a cube. E,equired the dihedral angle AB^D^. The lines AA^ and BB^ deter- mine the plane ABB^A\ and the lines A^D^ and B^C^ determine the plane A^D^C^B^- and as AA^ and BB' are ± to A' B' and B'C\ re- spectively, the planes must be per- pendicular. .". the angle required is 90°. Let E-ABCD-F be an octahe- dron. Required the dihedral angle E-BA-F, EKO = 54° 44^ 9^^. EKF=^ 2 EKO = 109° 28^ 18^^. Draw ^^and FK 1. to AB, also 0.2" from intersection of axes. Then is EKF the plane angle required. Let AB=\, Then 0K= 0.5 (since 0^, OB, and 0^ are perpen- dicular and equal), and 0E{= OA) = sin 45° = 0.7071. cot EKO = ^• OE log OiT - 9.69897 colog OE =0.15051 log cot EKO = 9.84948 Let AE and DE be two faces of a dodecagon. It is required to find the dihedral angle AOD. B y '^ >^ \ / Take each side of the pentagons ^^andi)^=2. • 5 = 108°, and A = 36°. log AC = log 2 -h log sin 108° + colog sm 36° log 2 = 0.40103 log sin 108° =9.97821 colog sin 36° = 0.23078 log AC = 0.51002 ^C=3.286L ^C0 = 72°. log AC = 0.51002 log sin 72° = 9.97821 log AO = 0.48823 ^0=3.0777. CB = AC and AO = DO. ACD = 108°, being an angle of a regular pentagon. CD A = 36°. log AD = log AC + log sin 108° + colog sin 36°. 136 TRIGONOMETEY. log AC = 0.51002 log sin 108° =9.97821 colog sin 36° = 0.23078 log AD = 0.71901 ^Z) = 5.2361. J ^Z) = 2.61805. log sin J J. Oi) = log IAD + colog J. 0. log I AD =0.41798 colog AO = 9.51177 log sin |AOi)= 9.92975 IA0D= 58°16^52^^ AC>i)=116°33M4^^ Let AOB, BOC, COD, etc., be equilateral triangles forming five of the surfaces of a regular icosahe- dron, and let AB, BC, CD, etc. = 1. Kegarding ABCDE as a plane pentagon, each angle = 108°. .-. A5C=108°, BAC=2>Q°. In a triangle of which sides are AB, BC, and ADC (regarding ADC as a straight line joining centres of bases of triangles AOB and BOC), sin DCB :AB:: sin ABC: ADC log AB = 0.00000 log sin ABC = 9.93753 colog sin DCB = 0.30103 log ADC ' But = 0.23856 ADC = 1.73204:. AD=^iADC = 0.86602. AC is a diagonal bf plane penta- gon ABCDE. .-. sin FCB -.AB:: sin ABC: AC. AC^ AB^mABC .-.ADC^ AB sin ABC s'm DCB ' sin FCB log AB = 0.00000 log sin ADC= 9.97821 colog sin FBC = 0.23078 log AC- = 0.20899 AC= 1.61804. But AF=^AC = 0.80902. In right triangle AFD sin ADF= — , AD or log sin ADF= log Ai^+colog AD. log Ai^ = 9.90796 colog AD = 0.06247 log sin ADF= 9.97043 ADi^= 69° 5^ 48^^ But ADi^=JA-OD-C. .-. A-OB~C= 138° IF 36^^ 5. A spherical square is a spheri- cal quadrilateral which has equal sides and equal angles. Its two diagonals divide it into four equal right triangles. Find the angle A of the square, having given the side a. TEACHERS EDITION. 137 B '""fl ^ y / \ \ \^ z' / / \ \^ ^^V \ \ XcB X ~~^-~ 1 \ \ / \ /"-- --*: , \ T? "■■---A ^\\ H yh Let ABCX be a spherical square, and CA, BX the two diagonals; also, let the side BA = x, then XA will = X (being equal sides). By [37], cos a = cos x X cos x = cos'^o;. Also cos X = Vcos a. BE sm X = tana; = OD BD OD Dividing first equation by the second, DE , ^ cos X = -zr^ = cot X. But BD X^iA (diagonals of a square bisect the angles). Substitute cot J A for cos x ; cot i A= Vcos a. Exercise XXIV. Page 119. 1. What do Formulas [43] be- come if ^=90°? ifj5 = 90°? if (7=90°? if a = 90°? \fA = B = 90°? ifa = & = 90°? If ^ = 90°, sin a sin ^ = sin h, sin a sin C = sin c. If 5 = 90°, sin a = sin 5 sin A, sin 6 sin C = sin c. If C= 90°, sin a = sin c sin A, sin b = sin c sin B. If a = 90°, sin B = 8'mb sin A, sin C = sin c sin A. If A^B = 90°, sin a = sin b. If a = b =90°, sin B = sin A. 2. What does the first of [44] be- come if ^ = 0°? if A = 90°? if A = 180° ? If^ = 0°, cos a = cos 5 cos c + sin 5 sin c. If ^ = 90°, cos a = cos b cos c. If ^ = 180°, cos a = cos b cos c — sin b sin c. 3. From Formulas [44] deduce Formulas [45], by means of the relations between polar triangles (I 45). Substituting in Formulas [44] for a, b, and c, their equals, 180°— J.^, 180° - B', 180° - C', we obtain cos (180°-^^) = cos (180° - B^) cos (180° - (7) + sin (180° - B^) sin (180° - CO cos (180° - A^). 138 TRIGONOMETRY. Substituting for cos (180° -^0- etc., their equals — cos J.', etc., we obtain — cos A^= cos B^ cos C^ — sin B^ sin C^ cos a^. Multiply by - 1, cos A^= — cos B^ cos C^ 4- sin B^ sin C^ cos a^; and similarly, cos B^= ~ cos J.'' cos (7 + sin J.'' sin C^ cos 5''. cos (7= — cos A^ cos 5'' + sin A^ sin .5'' cos c^. Exercise XXV. Page 124. 1. Write formulas for finding, by Napier's Rules, the side a when h, c, and A are given, and for finding the side b when a, c, and B are given. (i.) By Rule I., cos A = cot 6 tan m, whence tan m = tan 5 cos A. By Rule II., cos a = cos 71 cosp, whence cosp = cos a sec n. cos 5 = cos m cosp, whence cosj9 = cos 6 seem. .". cos a seen = cos 6 seem. Or, since ?i ={c — m), cos a = cos b sec m cos (c— m). (li.) By Rule I., cos B = tan n cot a, whence tan n = tan a cos -S. By Rule II., cos b = cos m cosp, whence cosp = cos b sec m. cos a = cos n cosp, whence cos p = cos a sec w. .•. cos b sec m = cos a sec n. Or, since m = {c— n), cos 6 = cos a sec w cos (c — n). 2. Given find a= 88° 12^ 20^^, ^= Q3°15nV^, 5 = 124° 7a7^^ ^=132°17^59^^ C= 50° 2^ 1^^; c= 59° 4^8^^ H& - a) = 17° 57^ 28.5^^. ^{a + b)^10Q° 9M8.5^^ JC= 25° F 0.5^^ logcosJ(&-a) =9.97831 log sec J (a + 5) =0.55536 log cot J (7 =9.33100 log tan J (^ + ^)= 0.86467 log sec H^ + ^) = 0.86868 log cos J (a + 6) =9.44464 log sin J (7 = 9.62622 log cos ^ c = 9.93954 J c = 29° 32^ 9'^ logsinJ(&-a) =9.48900 log CSC J (a + 6) =0.01751 log cot iC =0.33100 log tanJ(^-^) = 9.83751 ^{B-A)= 34° 31^ 24^^ i{A + B)= 97° 46^ 35^^ A = 63° 15^ 11^^ .5=132° 17^59^^ c = 59° ¥ 18'^ TEACHERS EDITION, 139 3. Given find a =120° 55^ 35^^, J. = 129° 58^ 3^^ 5= 88°12^20^^ B= 63° 15^ 9^^ (7= 47° 42^ V^; c= 55°52^40^^ i(a-5)= 16°2F37.5^^ J (a + 5) = 104° 33^-57.5^^ J (7= 23° 51^ 0.5^^ log cos ^ (a -5) = 9.98205 log sec Ha + 5) = 0.59947 log cot J C = 0.35448 log tan 1{A + B)= 10.93600 J(^ + ^)=96°36^36^^. log sin Ka-^) =9.44976 logcsc-^a + J) =0.01419 log cot ^C =0.35448 log tan |(J.-^) = 9.81843 K^-^) = 33°2F27^^ HA+^)=96°36'36^^. A = 129° 58^ 3^^ B = 63° 15' 9^^. log sec J (J. + ,5) = 0.93890 logcosKa + &) =9.40053 log sin ^ C = 9.60675 log cos ^-c =9.94618 Jc = 27°56^20^^. c = 55° 52^ 40^^. 4. Given find 5 = 63°15a2^^ .5 = 88° 12^ 24^^ c = 47°42' V\ C = 55°52'42^^ ^ = 59° 4^25^^; a = 50° F40'^ l{h + c)^ 55° 28^ 36.5^^ Hb-c) = 7° 46^ 35.5^^ iA = 29° 32^ 12.5^^ logcosi(&-c) = 9.99599 colog cos i{b + c) = 0.24662 log cot J ^ =10.24671 log tan i{B+C)= 10.48932 ^5 + C) = 72° 2^ 33^^ logsin^S-c) = 9.13133 colog sin K& + c) = 0.08413 log cot ^ J. = 10.24671 log tan |(^- (7)= 9.46217 ^{B-C) = 1Q° 2' 5V'. iIb + C) = 72° 2^33^^ B = 88° 12^ 24^^ C= 55° 52^ 42^^. logcosH6 + c) =9.75338 colog cos J(5+ (7) = 0.51101 log sin ^^ =9.69284 log cos I a = 9.95723 f a = 25° 0^ 50^^. a = 50° V W. 5. Given find 6= 69°25ai^^ 5= 56°1F57^^ c = 109° 46^9^^, (7 = 123°2F12^^, A= 54° 54^ 42^^; a= 67° 13'. \{c-h) = 20° 10' 34''. ^c + 6) = 89° 35' 45". I A =27° 27' 21". log cos ^c- 6) = 9.97250 colog cos J (c + ^) = 2.15157 log cot J ^ =10.28434 log tan J (C+ 5) = 12.40841 \{C\B) = 89° 46' 34.5". 140 TRIGONOMETRY. log sin J (c- 6) = 9.53770 colog sin Kc + ^) = 0.00001 log cot J ^ = 10.28434 log tan J (C- .5)= 9.82205 ^{Q-B)= 33°34^37.8^^ C= 123° 21^ 12^^ B= 56°1F57^^ logcos J(c + 5) =7.84843 colog cos 1{C+ B) = 2.40837 log sin h A lo g cos ^ a = 9.66376 = 9.92056 J a = 33° 36^ 30^^ a = 67° 13^ Exercise XXVI. Page 126. 1. What are the formnlas for com- puting A when B, C, and a are given ; and for computing B when A, C, and b are given? By Rule I., cos a = cot y cot B, .'. cot y = tan B cos a. By Rule II., cos A = cosp sin x, .'. cos p = COS A CSC X. cos B = cos JO sin y, .'. cos p = COS B CSC y. .'. COS J. CSC a? = COS B esc y. Or, since a; = C— y, cos J. = cos^csc?/sin(C— 2/). (ii.) By Rule I., cos b = cot a; cot A, .'. cot X = tan A cos 6. By Rule II., cos A = C0SJ9 sin ar, .*. cos p = cos ^ CSC X. cos ^ = cos jD sin y, .'. cos p = COS ^ CSC y. .'. COS B cscy = COS .4 esc a;. Or, since y = (7— a;, COS 5 = cos ^ esc X sin (C— a;). 2. Given find ^= 26°58M6^^ a= 37°14a0^^ B= 39° 45^0^^, 6=121°28aO^^ c=154°46^48^^ C=161°22ai^^ i{B-A)= 6° 23^ 12^^. ^{B + A) = 33° 21^ 58^^. f c = 77° 23^ 24^^. logcosi(^--^)= 9.99730 log sec |(^ + ■^) = 0.07823 log tan I c =10.65032 log tan J (5 + a) =10.72585 log sin i(^ + ^y= 9.74035 logsecJ(&-a) =0.12972 log cos J c = 9.33908 log cos 2- = 9.20915 1 (7= 80° 41^5.4^^. logsin J(5-74)= 9.04625 log CSC H^ + -4) = 0.25965 log tan Jc' =10.65032 log tan J (J -a) = 9.95622 1(6 -a) = 42° 7^ i(b + a) = 79°21M0^^ & = 121° 28^ 10^^ ■ a = 37° 14^ 10^^. (7= 161° 22^ 11^^ TEACHERS EDITION. 141 3, Given find ^ = 128°4F49^/, a=125°4F44^^ 5= 107° 33^ 20^^ h= 82°47^34^^ c= 124° 12^ 31^^; (7-127°22^ 1{A-B)= 10° 34^ 14.5^^. 1{A + B) = 118° 7^ 34.5^^ Jc =62° 6M5.5^^. log cos J (J. - 5) = 9.99257 colog cos 1{A-^B)= 0.32660 log tan J c =10.27624 log tan J (a + 5) =10.59541 l{a+h) =104° 14^ 38.5^^. log sin i (J. -5)= 9.26351 colog sin ^{A + B)= 0.05457 log tan ^-c =10.27624 logtan|(a-6) = 9.59432 l{a-l) = 21° 27^ 5^^ a =125°41^44^^. h = 82°47^34^^ log sin ^(^ + 5) = 9.94543 colog cos J (a -h) = 0.03118 log cos ^ c = 9.67012 log cos J a = 9.64673 1(7= 63° 4P. C=127°22^. 4. Given find 5 = 153° 17^ &^, 5 = 152° 43^ 5K^ C= 78°43^36^^ c= 88° 12^ 21^^, a= 86° 15^5^^; A== 78°15M8^^. i(^+C') = 116° 0^21^^. i(5-C)= 37°16M5^^ J a =43° 7^37.5^^ log cos J (5 -(7) = 9.90074 log sec* (^+(7) = 0.35807 log tan J a =9.97159 log tan J (5 + c) =0.23040 i(6 + c) = 120° 28^ 6/>'. logsin J(5-C) = 9.78226 log CSC ^5 + C) = 0.04636 log tan J a =9.97159 log tan 1(6 -c) =9.80021 lih-c) =32° 15M5/^ log sin J (5 +(7) = 9.95364 log sec J (6 -c) =0.07283 log cos ^ a =9.86322 log cos J ^ =9.88969 J ^ = 39° 7^ 54^^. I = 152° 43^ 51^^. c = 88° 12^ 21^^ A = 78° 15^ 48^^ 5. Given find J. = 125°4F44^^, a = 128°31M6^^, (7= 82° 47^ 35^^, c = 107°33^20^^ 5= 52° 37^ 57^^; 5= 55°47^40^^' H^ + C) = 104° 14^39.5^^. i(A-C)= 21° 27^ 4.5^^. J 6 =26° 18^ 58.5^^ log cos J (^-(7) = 9.96883 log sec i{A + C) = 0.60896 log tan J 6 = 9.69424 log tan J (a + c) =0.27203 i{a + c) =118° 7^33^^ log sin J (J. + C) = 9.98644 log sec f (a -c) =0.00743 log cos J 5 =9.95248 log cos J 5 =9.94635 i ^ = 27° 53^ 50^^, ^ 142 TRIGONOMETRY. logsin J(A-C) = 9.563i3 log CSC -I (^+ (7) = 0.01356 log tan 16 = 9.69:124 log tan i (a -c) =9.27093 Ha - c) = 10° -ZV IS'\ a = 128° 31M6^^. c = 107° 33^ 20^^. B = 55° 47^ 40^^ 1. Given find a= 73°49^38^^ 5=116°42^30^^ 6 = 120° 53^ 35^^, c = 120°57^27^^ A= 88° 52^2^^; C=116°47^ ¥\ log sin A = 9.99992 log sin b = 9.93355 log CSC a = 0.01753 log sin B = 9.95100 .5= [180°- (63° 17^ 30^0] = 116° 42^ 30^^ ^ {B+A) = 102° 47^ 36^^. ^{B-A)= 13° 54' 54^^. i(5 + a)= 97°2F36.5'^ l{b-a)= 23° 3V 58.5^'. log sin J {B+A) = 9.98908 log CSC J (5-^) = 0.61892 log tan Hb-a) = 9.63898 log tan J c =10.24698 ^c= 60°28'43.5'^ c=120°57'27'^ log sin ^{b + a) = 9.99641 logcscH&-a) =0.39873 log tan ^ {B-A) = 9.39401 log cot J C Exercise XXVII. Page 128. find ^1 = 120° 47M5'^ B.^ = 59° 12' 15'^ Ci= 55° 42' 8''. 9.78915 58° 23^ C=116°47' 4''. Given a = 150° 57' 5", 6 = 134° 15' 54", A = 144° 22' 42" ; Co = 23° 57' 17.4", C, C^= 29° 97° 42' 55", 8' 39". A > 90°, (a + &) > 180°, a > 6 ; hence two solutions. I. sin -Sj = sin .A sin 6 CSC a. log sin A log sin b colog sin a log sin 5i = 9.76524 = 9.85498 = 0.31377 = 9.93399 ^1=120° 47' 45". B^= 59° 12' 15". ^ia-b) = 8° 20' 35.5". \{a + b) = 142° 36' 28.5". i(^-^i)= 11° 47' 28.5". 1{A-B^)= 42° 35' 13.5". log sin i (a + 6) = 9.78338 log esc l{a-b) = 0.83833 log tan J (.4-5i)= 9.31963 log cot J (7i = 9.94134 }Ci = 48°51'27.7". Ci = 97° 42' 55.4". TEACHERS EDITION. 143 log sin i{a + b) = 9.78338 log sin A - 9.99592 log CSC Ha - ^) = 0.83833 log sin h = 9.99605 log tan ^ {A-B^)^ 9.96338 colog sin a = 0.00803 log cot ^ (72 = 10.58509 log sin B = 0.00000 ^0^ = 14° 34^ 19.6^^ B = 90°. C\ = 29° 8^ 39^^. tan c = cos A tan h. i{A + 5i) = 132° 35^ 13.5^^. cot C= tan J. cos h. i {A + B^) = 101° 47^ 28.5^^. log cos J. = 9.13499 log tan h = 10.86812 log sin H^+^i)= 9-86703 log tan c = 10.00311 colog sin ^{A-B^)= 0.68963 c = 45° 12^ 19^^. log tan Ha- ^) =9.16629 log tan Hi =9.72295 Hi = 27° 5F 4^^. Cj = 55° 42^ 8^^. log tan A log cos h log cot C = 0.86092 = 9.12793 = 9.98885 C=45°44^ log sin I {A+B.,)= 9.99074: colog sin H^-^2)=C>.16960 log tan Ha -6) =9.16629 log tan H2 = 9-32663 H2 = 11° 58^ 38.7^^. C2 = 23° 57^ 17.4^/. 3. Given find a = 79° 0^54.5^^ J5 = 90°, & = 82°17^ 4^^ c = 45°12a9^^ ^ = 82° 9^25.8^^; C=45°44^ 4. Given a = 30° 52^ 36.6^^, b = 31° 9' W, A = 87° 34^ 12^' ; show that the triangle is impossible. From [43] sin ^ = sin ^ sin b esc a. log sin A log sin b log CSC a log sin B = 9.99961 = 9.71378 = 0.28972 = 0.00311 sin 5 =1.009. impossible, since sin 5> 1. Exercise XXVIII. Page 129. 1. Given ^ = 110°10^ 5=133°18^ a = 147° 5^32^^; (7= 70° 20^ 40^^. sin 6 = sin a sin B esc A. find J = 155° 5^8^^ c= 33° 1^36^^ log sin a log sin B colog sin A log sin b = 9.73503 = 9.86200 = 0.02748 = 9.62451 h = 155° 5^ 18^^ 144 TRIGONOMETRY. ^ {B + A) = 121° 4:4:\ ^{B-A)= 11° 34^ 1(5- a) = 3°59^53^^ ^{b + a) = 151° 5^ 25^^. logsmH^+^) =9.92968 colog sin H^-^) = 0.69787 log tan I (6 -a) =8.84443 log tan J c = 9.47198 ^ c = 16° 30^ 48^^ c = 33° 1^36^^. colog sin ^{b — a) = 0.15663 log sin J (5 + «) = 9.68433 log tan J (5-^) =9.31104 log cot J (7 = 9.15200 J C = 35° 10^ 20^^. C=70° 20M0^^ 2. Given find ^ = 113°39^2K^ 5 = 124° 7^20^^ .S = 123°40nS^^ c = 159°53^ 2^^ a= 65° 39^ 46^^; (7= 159° 43^ 35^^. log sin a = 9.95959 log sin B = 9.92024 colog sin J. =0.03812 log sin b = 9.91795 b = 124° 7^ 20^^ i (5 + ^) = 118° 39^ 49.5^^ iiB-A)= 5° 0^ 28.5^^. i (5 - a) = 29° 13^ 52^^. 1 (5 + a) = 94° 53^ 33^^. log sin 1 (5+^) = 9.94422 colog sin H-S-^) = 1.05901 log tan H6 - a) = 9.7478 9 log tan J c =10.75112 Jc= 79°56^51^^ c = 159° 53^ 2^^. log sin ^- (5 + a) = 9.99842 colog sin i{b-a) = 0.31128 log tan i(^-^) =8.94264 log cot I C = 9.25234 1 (7= 79°5F47.7^^ (7 = 159° 43^ 35^^ 3. Given find J. = 100° 2^1.3^^ 5= 90°, B= 98°30^28^^ c=147°41M3^^ a= 95°20^38.7^^;C=148° 5^33^^ log sin a log sin 5 log CSC A log sin b = 9.99811 = 9.99519 = 0.00670 = 0.00000 b = 90°. 1{A + B)^ 99° 16^ 19.7^^. i{A-B)= 0° 45^51.7^^ Ha - &) = 2° 40^ 19.4^^. log sin UA+B) = 9.99428 colog sin U^-B) = 1.87484 log tan ^{a-b) = 8.66904 log tan ^c =10.53816 Jc= 73° 50^ 51. 7^^ c=147°4F43^^ log sin ^ (a + 5) = 9.99953 colog sin i{a-b) = 1.33144 log tan J (^-5) =8.12520 log cot J C =9.45617 1(7= 74°2^46.3^^ C=148°5^33^^ 4. Given A = 24° 33^ 9^^ B = 38° 0^ 12^^ a= 65° 20^ 13^^; show that the triangle is impossible. TEACHERS EDITION, 145 cot X = cos a tan B. log cos a log tan B log cot X = 9.62042 = 9.89286 = 9.51328 X = 71° 56^ 30^^ sin {c — x) = cos A sec B sin x. log cos A log sec ^ log sin X log sin (c — x) sin (c — x) = 9.95884 = 0.10349 -- 9.97806 = 0.04039 = 1.0974. Since sin {c — x)';> 1, the angle C is impossible. .-. the triangle is impossible. Exercise XXIX. Page 131. 1. Given find ^ = 116°44'49'^ a = 120° 55^ 35^^ A = 116°44M9^^ B = 63° 15^ 14^^. 6= 59° 4^25^^ B= 63° 15^4 '^ C= 91° 7' 2V^, c = 106° 1.0'' 22^^; C= 91° 7^21^^ a = 120° 55^ 35^^ b= 59° 4^25'^ c = 106° 10^ 22^^ 2 s = 286° 10^ 22^^ s = 143° 5M1^^. 2. Given find a= 50° 12^ 4^^ A= 59° 4^28'^ 6 = 116°44^48^^ B= 94°23a2^^ c = 129° IF 42^^; (7=120° 4^52^^ s-a^ 22° 9^ 36^^ s-b= 84° 0M6/^. a= 50° 12^ 4^' s-c = 36° 54^ 49^^ 6 = 116° 44^ 48^^ log sin (s - a) = 9.57657 log sin (s - 5) = 9.99763 log sin (s-c) = 9.77859 c = 129° 11^ 42^^ 2s = 296° 8^34^^ s = 148° 4^ 17^^ log CSC s = 0.22141 s_a= 97°52M3^^ log tanV = 19.57420 s - 5 = 31° 19^ 29^^ log tan r = 9.78710. s - c = 18° 52^ 35^^. tan J -4 = tan r esc (s — a). log tan J ^ =10.21053 log tan J 5 = 9.78948 log sin {s-a) = 9.99589 log sin (s - 6) = 9.71591 log tan J (7 = 10.00851 log sin (s-c) = 9.50992 ^A= 58° 22^ 24.8^^ log CSC s = 0.27666 ^B= 31° 37.2^. logtan^r =9.49838 J (7= 45°33M0.8^^ log tan r = 9.74919. 146 TRIGONOMETEY. log tan J ^ =9.75330 log tan ^ 5 = 0.03328 log tan i (7 = 0.23927 1 A = 29° 32^ Wr ^B= 47° IK 36^^ J C = 60° 2' 26^^ A = 59° 4^ 28^^ B = 94° 23^ 12'r C= 120° 4^ 52^^, 3, Given find a = 131° 35^ 4^^ 2l = 132°14^2F^ 6 = 108°30a4^^ ^=110°10^40^^ c= 84°46^34^^; C= 99°42^24^^ a = 131° 35^ 4^^ 5 = 108° 30^ 14^^ c = 84° 46^ 34^' 2s = 324°51^52^^ s = 162° 25' 56'^ s-a=^ 30° 50' 52''. s-b= 53° 55' 42". s-c = 77° 39' 22". log sin (s — a) log sin (s — b) log sin (s — c) log CSC s log tan^ r log tan r 9.70991 9.90756 9.98984 0.52023 10.12754 10.06377. log tan J A = 0.35386 log tan J 5 = 0.15621 log tan ^ (7 = 0.07393 iA= 66° 7^ 10.6". I 5 = 55° 5' 20". I C = 49° 51/ 12". A = 132° 14' 21". ^=110° 10' 40". C= 99° 42' 24". 4. Given find a = 20°16'38", A= 20° 9' 54", 6 = 56° 19' 40", B = 55° 52' 31", c = 66° 20' 44" ; C = 114° 20' 17". a = 20° 16' 38" b = 56° 19' 40" c = 66° 20' 44" 2s = 142° 57' 2" s = 71° 28' 31". s-a= 51° 11' 53". s-b= 15° 8' 51". s-c= 5° 7' 47". log sin (s — a) log sin (s — b) log sin (s — c) log CSC s log tan^ r log tan r 9.89172 9.41715 8.95139 0.02311 8.28337 9.14168. log tan M = 9.24996 log tan J 5 = 9.72453 log tan i (7 = 10.19029 J J. = 10° 4' 56.8". J 5 = 27° 56' 15.5". f (7= 57° 10' 8.6". A = 20° 9' 54". .5 = 55° 52' 31". a= 114° 20' 17". TEACHERS EDITION. 147 • Exercise XXX. Page 132. 1. Given find ^ = 130°, a = 139°2F22^^ £=110°. 6 = 126° 57^ 52^^ C= 80°; c= 56°51M8^/. A = 130° B = 110° c = 80° 2^ = 320° s = 160°. S- A = 30°. S- B = 50°. S- -C = 80°. log cos S = 9.97299 log sec (^S^- -A) = 0.06247 log sec {8- -B) = 0.19193 log sec {S- -c) = 0.76033 log tan^ E = 10.98772 log tan B = 10.49386. log tan ^ a = 10.43139 log tan ^ b = 10.30193 log tan ^ c la = 9.73353 = 69°40M1^^ JS = 63° 28-' 56'^. i^ = 28°25^54'^ a = 139° 2V 22''. h = 126° 57^ 52'^. c - 56°5F48''. 2. Given find A = 59°5Wl(y^, «i=128°42^29'^ i? = 85«36^50^^ h= 64° 2^47^^ c=^d°mn(y^; c= 128° 42' 29'^. A = 59° 55^ 10^^ B = 85° 36^ 50^^ C = 59° 55^ 10^^ 2^=205° 27^ 10^^ S == 102° 43' 35'\ iS-A= 42° 48^ 25^^. S-B= 17° 6M5^^. S-C^ 42°48^25^^ log cos S = 9.34301 log sec {S-A) =0.13451 log sec (S-B) =0.01967 log sec {S-C) =0.13451 log tan^i? = 9.63170 log tan E = 9.81585. log tan I a log tan I b log tan I c = 9.68134 = 9.79618 = 9.68134 ^ a = 25° 38^ 45.5^^. ^6=32° y23.6^^. ^ c =^ 25° 38^ 45.5^^ a=-51°17^3F^. J = 64° 2M7^^. c = 51°17^3F^. 3. Given find A = 102° W 12'', a = 104° 25^ 9", B= 54° 32' 24:", b= 53° 49^ 25^^, C= 89° 5^46^^; c= 91° 44' 24". A = 102° 14^ 12^' B = 54° 32^ 24^^ C= 89° 5M6^^ 2 ^=245° 52^ 22^/ ^S'=122° 50^ 11^^, 148 TRIGONOMETRY. 8-A- = 20° 41^ 59^^. A = 4° 23^ 35^^ S-B- = 68° 23^ 47^^ B = 8° 28^ 20^^ S-C = = 33° 50^ 25^^ C = 172° 1*7^ 56^^ log cos S log sec {S log sec {S log sec {S log tan^iJ = 9.73536 ~A)= 0.02898 -B) = 0.43394 • - (7) = 0.08061 = 0.27889 2/S' = S= S-A = S-B = S-C==- 185° 9^51^^ 92° 34^ 55.5^^. 88° 11^ 20.5^^ 84° 6^ 35.5^^. -(79° 43^ 0.5^0- log tan a = 0.13945. log cos S = 8.65368 log tan ^ ( log tan ^ < z =0.11047 b = 9.70551 log sec {S — log sec {S — log sec (-S* — A) = 1.50029 B) = 0.98876 C) = 0.74833 log tan ^ c = 0.05885 log tan^i^ = 11.89106 la= 52° 12^ 34.6^^. log tan E = 10.94553. ih= 26° 54^ 42.5^^. ic= 48° 52^ 12^^ a = 104° 25^ 9^^ 6 = 53° 49^ 25^^. c= 97°44^24^^ log tan J a log tan J b log tan ^ c = 9.44524 = 9.95677 = 10.19720 la= 15° 34^ 35.5^^ J6= 42° 9^ 11.5^^ 4. Given find Jc= 57°34^58^^ A= 4° 23^ 35^^ a= 31° 9^1^^ a= 31° 9MF^. ^= 8° 28^ 20^^, b= 84°18^23^^ b = 84° 18^ 23^^ C = 172°17^ 56^^; c = 115° 9^56^^ c = 115° 9^56^^ Exercise XX> [I. Page 135. 1. Given find log 26159 = 4.41762 A = 84° 20^ 19^^ ^=26159^^ colog 648000 = 4.18842-10 B = 27° 22^ C = 75° 33^; 40^^ i^= 0.12685 i22. = ^+5+C-180°. log 3.14159 logi^ = 0.49715 = 9.10319-10 A = = 84° 20^ 19^^ i^= 0.12685 i^^ B = = 27° 22^ 40^^ G = - 75° 33^ 187° 15^ 59'^ 2. Given find 180° a= 69° 15^ 6 6 = 120° 42^ 47 '', ^=216°40a8^^. E = - 7° 15^ 59^^ = 26159^^ c= 159° 18^ 33 TEACHERS EDITION. 149 a = 69° 15^ 6 = 120° 42^ c = 159° 18^ 6'^ 47// 2 s = 349° 16^ s = 174° 38^ s-a = 105° 23^ s-b= 53° 55^ s-c= 15° 19^ ^s = 87° 19^ -^ (s - a) = 52° 41^ Hs-6)= 26° 57^ 1 (s - c) = 7° 39^ 26^/ 13^^. 26^^ 40^^. 6.5^^ 33.5^^ 43^^ 50^^. log tan ^s =11.32942 log tan |(s- a) =10.11804 log tan J(s-6)= 9.70645 logtan ^(s-c)= 9.12893 logtan^J.E' =10.28284 log tan i~E = 10.14142. IE= 54° 10^ 4.6^^ -£;=216° 40^ 18^^. 3. Given find a= 33° 1^45^^ ^=133° 48^ 53^^. 6 = 155° 5a8^^ (7 = 110° 10^; tan m = tan a cos C. (^ 54) cose = cos a seem cos (6— m). (^54) log tan a = 9.81300 log cos c = 9.53751 log tan m = 9.35051 m= 167° 22^ 6-m = -(12°16^42^0- log cos a = 9.92345 log sec m = 0.01064 log cos (6 - m) = 9.98995 log cos c = 9.92404 c = 147° 5^ 30^^ a= 33° 1M5^^ h = 155° 5^ 18^^ c = 147° 5^ 30^^ 2s = 335° 12^ 33^/ s = 167° 36^ 16.5^^. s — a = 134° 34^ 31.5^^. s-6 = 12° 30^ 58.5^^. s — c = 20° 30^ 46.5^^. ^s = 83° 48^ 8.25^^ Hs - «) = 67° 17^ 15.75^^ Hs-6) = 6° 15^ 29.25^^ Hs-c) = 10° 15^ 23.25^^ log tan J s = 0.96419 log tan J (s- a) =0.37824 log tan ^(s- 6) =9.04005 log tan i{s — c) = 9.25755 log tan^ IE 9.64003 log tan iE 9.82002. i E = 33° 27^ 13^^/ E = 133° 48^ 53^^. 4. Find the spherical excess of a triangle on the earth's surface (re- garded as spherical), if each side of the triangle is equal to J.°. Given a, 6, and c each = 1°; then 2s = 3°. ^s = 45^ 1° 30^ ^(^s-a) = 15^ Hs-6) = Hs-c) = = 8.11696 s s-a= 30^ s-h= 30^ 8-c= 30^. log tan J s log tan J (s- a) = 7.63982 logtan J(s-6)= 7.63982 log tan J (s-c) = 7.63982 logtanH^ =11.03642 log tan i ^ = 5.51821. i.£'=6.814^^ j5'=27.25^^ 15^. 15^. 150 TRIGONOMETRY. Exercise XXXII. Page 148. 1. Find the dihedi-al angle made by the lateral faces of a regular ten- sided pyramid ; given the angle A = 18°, made at the vertex by two adjacent lateral edges. About the vertex of the pyra- mid, describe a sphere. It will intersect the lateral surface, forming a regular spherical decagon, of which the sides = 18°, being measured by the plane angles at the centre. Pass a plane through and A and C, forming an isosceles spheri- cal triangle ABC. Then (by Prob. 3, Ex. XXIII.) given side a = 18°, n = 10, to find angle A of the polygon. ■ 1 A I A 180° •sm ^ A = sec ^ A cos -• log cos 18^ colog cos 9° = 9.97821 = 0.00538 log sin M = 9.98359 74° 21^ A = 148° 42^ iA 2. Through the foot of a rod which makes the angle A with a plane, a straight line is drawn in the plane. This line makes the angle B with the projection of the rod upon the plane. What angle does this line make with the rod ? Let CO be a straight line, making the angle A with the plane OH; IF Sb straight line passing through the foot of CO, making the angle B with the projection DO of CO upon the plane GH. It is required to find the angle COI== X. AVith a radius equal to unity, from as a centre, construct the spherical triangle DCI. Then i = A, c=B, d=COI=x CEK=-Tt. angle. Since OE is the projection of CO on the plane OH; CE drawn from C to E, is perpendicular to OE. .'. CDI =rt. triangle. By Formula [37], cos d = cos i cos c. .•. cos X = cos A cos B. 3, Find the volume V of an ob- lique parallelopipedon ; given the three unequal edges a, b, c, and the TEACHERS EDITION, 151 three angles I, m, n, which the edges make with one another. /\ r \ ">-<^ \¥^, \ a N^ __, -^■-"^A C B Let ^5 be a parallelopipedon, and I, m, and n, the angles which the unequal edges a, b, and c make. We are to find V, the volume. Let w = the inclination of the edge to the plane of a and b. V= area base X altitude. Area base =^ bz, when z = a sin I. Altitude = X, when a? = c sin w. .'. V= abc sin I sin w. Suppose a sphere to be constructed having for its centre the vertex of the trihedral angle whose edges are a, b, and c. The spherical triangle whose vertices are the points where a, b, and c meet the surface has for its sides I, m, n ; and vj = perpen- dicular arc from side I to the oppo- site vertex. Let L, M, iV denote the angles of the triangle. Then by [38] and [47], sin w = sin m sin N = 2 sin m sin J iVcos | N, Or if s = l{l-\-m^-n), sin w = - — Vsin s sin(s— Z) sin(s— m) sin(s— n) sin? .-. V= 2a6c Vsinssin(s— Z) sin(s— m) sin(s— ri) 4. The continent of Asia has nearly the shape of an equilateral triangle, the vertices being the East Cape, Cape Romania, and the Promontory of Baba. Assuming each side of this triangle to be 4c5UO geographical miles, and the earth's radius to be 3440 geographical miles, find the area of the triangle : (i.) regarded as a plane triangle ; (ii.) regarded as a spherical triangle. Area = J (base x altitude). Altitude = V48002 - 24002. = V 17280000. log \/l7280000= 3.61877 log 2400 =3.38021 log area =6.99898 Area = 9976500. (ii.) 180° a, b, and c 4800° 60 2 s = 240° Hs- -a)= 20° i(s- -6)= 20° Hs- -c)= 20° log tan J s = 10.23856 log tan 1 (s - a) = 9.56107 log tan l{s-b) = 9.56107 log tan l{s-c) = 9.56107 logtan^i^ = 8.92177 1^=16° 7^ ^.V\ J5'=64° 28^32.5^^ = 232112.5^^ 152 TRIGONOMETRY. logi: log- '6-18000 logi?2 logi^ = 5.36570 = 4.68557 = 7.07312 = 7.12439 i^= 13316560. 5. A thip sails from a harbor in latitude I, and keeps on the arc of a great circle. Her course (or angle between the direction in which she sails and the meridian) at starting is a. Find where she will cross the equator, her course at the equator, and the distance she has sailed. Let NESW be the earth, WCE the equator, iVand Sthe nortli and south poles. Let A be the point from which the ship starts, AFD the parallel of latitude the ship starts from, and AB the great circle of its course. Then BAE=a = course of ship. log sin \ A \A = 10.00000 - = 90°. -10 AE=l = latitude of its starting- A = 180°. place. Arc a = 180°. BE = TO = place of crossing the equator. B = course at equator. AB = c? = distance sailed. By Napier's rule, sin I — tan m cot a ; whence tan m = sin I tan a. cos B = cos I sin a. cot d = cot I cos a. 6. Two places have the same latitude I, and their distance apart, measured on an arc of a great cir- cle, is d. How much greater is the arc of the parallel of latitude be- tween the places than the arc of the great circle ? Compute the results for I = 45°, d = 90°. In isosceles spherical triangle ABO sin J J. = sin * dcsc{dO°-l) = sin J c?sec I. Let ?=45°, d=90°. log sin J d = 9.84949 log sec I = 0.15051 TEACHERS EDITION. 153 Arc K = a X cos Z = iaV2 = 90°\/2. 90° V2 - 90° = 90°( V2 - 1). 7. The shortest distance d be- tween two places and their latitudes I and V are known. Find the dif- ference between their longitudes. Let C represent the north pole, A the position of one city, B the position of the other. Then AB = d, and if m represent the longitude of one city, mV that of the other, I the latitude of one city, and V that of the other, angle Cwill be equal to {m-m^), and 5(7 will equal 90°- Z, and ^Cwill be equal to 90° -^^ Therefore we have an oblique spherical triangle with three sides given to find the angle C. Now from Formula [44], cos c = cos a cos h + sin a sin h cos C; then by substituting, cos d = cos (90°- I) cos (90° - V) + sin (90° - I) X sin (90°— V) cos (m — m^) ; or cos d = sin I sin V + cos I cos V cos (m — m^). .'. cos (w — m^) = (cos (i — sin Z sin V) X sec I sec V. 8. Given the latitudes and longi- tudes of three places on the earth's surface, and also the radius of the earth ; show how to find the area of the spherical triangle formed by arcs of great circles passing through the places. Let A, 5, and G represent the positions of three places on the earth's surface. I 62 shows how to find the dis- tance between two places when the latitudes and difference in longitude are given. In this case we have the latitudes given, and also the longitudes ; so that we can find the difference in longitude. Let GH = equator ; then, from | 54, in triangle ABC, tan m = cot a cos 6, and from § 62, cos BC= sin a sec m sin (5+m), and the same with the distance be- tween the other places. Therefore, we have given the three sides of a spherical triangle, to find the area. 154 TRIGONOMETRY. By [51], tan^ \ E= tan I s tan i (s — a) X tan i{s — h) tan J (s — c). Then, since we have the radius of the sphere given (= R) and the spherical excess = E, from Formula F=^TrR\ 180= 9. The distance between Paris and Berlin (that is, the arc of a great circle between these places) is eq'ial to 472 geographical miles. The latitude of Pans is 48° 50^ 13'^; that of Berlin, 52° 30^ 16^^. When it is noon at Paris what time is it at Berlin ? Let AO represent the latitude of Paris, and BK the latitude of Berlin. Then C represents the dif- ference in longitude. 6M = 6 = 41° 9M7^^ CB = a=2>T 29M4^^ AB==c= 7° 52^ (472 H- 60) 2s= 86° 31^31'/ s = 43° 15^ 45.5^^ s-a= 5° 46^ 1.5^^ s-b= 2° 5^58.5>'>'. s - c = 35° 23^ 45.5^^ tan'^ J C= cscs sin (s — a) sin (s — b) esc (s — c). log CSC s = 0.16409 log sin (s- a) = 9.00210 log sin (s- 5) = 8.56391 log CSC (s-c) = 0.23716 log tannic =17.96726 tan J (7 = 8.98363 1(7=5° 30^ 2^^ (7-11° 0^4^^ - 660. Difference of time = Y^5(660) minutes = 41 min. Time at Berlin, 12 h. 44 mm. 10. The altitude of the pole be- ing 45°, I see a star on the horizon and observe its azimuth to be 45° ; find its polar distance. iVP=45°. FM^p. .:PZ = 4:5° = 1. PZM^a. We have given two parts of the triangle, a = 45°. (90° - = 45°. cosp = cos a cos I. cos a = \/|. cos I = V^. .'. cosp = J. p = 60°. TEACHEES EDITION. 155 11. Given the latitude I of the observer, and the declination d of the sun ; find the local time (appar- ent solar time) of sunrise and sunset, and also the azimuth of the sun at these times (refraction being neg- lected). When and where does the sun rise on the longest day of the year (at which time d= -\- 23° 27^) in Boston {I = 42° 2V), and what is the length of the day from sunrise to sunset? Also, find when and where the sun rises in Boston on the shortest day of the year (when (^=- 23° 270, and the length of this day. To find the hour angle t when the sun is on the horizon. PM:= 90° - d. ^Q = 90°. .-. PQ = 90° - Z -h 90° = 180° - 1. Then in triangle PMQ, by [39], cos QPM== tan PQ cot Pif, or cos t = tan(180°-r)cot(90°-tf), cos t = ~ tan I tan d. Also to find azimuth a. In triangle PMQ, MQ is meas- ured by angle QZM= 180° - a. Then, by [37], cos PM= cos PQ cos MQ. cos (90° - d) = cos (180° - Z) cos (180° - a). sind = — cos I (— cos a), or cos a = sin c? sec I, Now cos t = — tan d tan I. Time of sunrise = 12 o'clk a.m. 15 Time of sunset 15 o'clk P.M. log tan d = 9.63726 log tan I = 9.95977 log cos t = 9.59703 t = 66° 42^ 26^^ 12 - — = 7 h. 33 min. 10 sec. 15 15 = 4 h. 26 min. 50 sec. .*. shortest day = 2 X 4 h. 26 min. 50 sec. = 8 h. 53 min. 40 sec. Again, cos a = sin c? sec I. log sin d = 9.59983 log sec I = 0.13133 log cos a = 9.73116 a = 57° 25^ 15^^ .-. a^=122° 34^45^^ Longest day = 12 hrs. -h[(7h. 33min. lOsec. -4h. 26 min. 50 sec.)] = 15 h. 6 min. 20 sec. 156 TRIGONOMETKY. 12. AVlien is the solution of the problem in Example 11 impossible, and for what places is the solution impossible ? d has for its maximum value 23° 27^. Suppose I = 66° Sy. Then tan (180° - Z) = - tan ^ = — cot d. Formula cos t = — tan I tan d becomes cos t^ — cot d tan d = -1. .-. i=180°; that is, the sun just appears in the south on shortest day. For places within the Arctic cir- cle, I > 66° 33^, and —tan I numeri- cally greater than — cot d. Hence — tan I tan c? > — 1 (numerically), or cos t = ± 1+, which is not possible. Hence the sun may fail to rise during 24 hours. 13. Given the latitude of a place and the sun's declination ; find his altitude and azimuth at 6 o'clock A.M. (neglecting refraction). Com- pute the results for the longest day of the year at Munich {I = 48° 9^. FZM= a. FZ=dO°-l PJf=90° -d=p. ZPM= t. ZM= 90° - h. I = 48° 9^ Sun's declination on longest day, 23° 27^. By Napier's Rules, sin h =&hxl sin d. log sin I = 9.87209 log sin d = 9.59983 log sin h = 9.47192 Altitude = A = 17° 14^ 35^^ By Napier's Rules, cot a = cos 7 tan d. log cos I = 9.82424 log tan d = 9.63726 log cot a = 9.46150 Azimuth = a = 73° 5V W\ 14. How does the altitude of the sun at 6 a.m. on a given day change as we go from the equator to the pole ? At what time of the year is it a maximum at a given 'place ? (Given sin A = sin Z sin d.) The farther the place from the equator, the greater the sun's alti- tude at 6 A.M. in summer. At the equator it is 0°. At the north pole it is equal to the sun's declination. At a given place, the sun's altitude TEACHERS EDITION". 157 at 6 A.M. is a maximum on the longest day of the year, and then sin h = sinl sin e (where e = 23° 27''). 15. Given the latitude of a place north of the equator, and the dec- lination of the sun ; find the time of day when the sun bears due east and due west. Compute the results for the longest day at St. Peters- burg {I = 59° 560- Let NESWhQ the horizon, ^ the zenith, NZS the meridian, WZE the prime vertical, WAE the equi- noctial, P the elevated pole, M the position of the sun when due east, MB its declination, and ZPM its hour angle. Then we have the right spherical triangle PZM, with PM and PZ known, to find ZPM. Given I and c?, to find U PM= 90° - d. PZ=dO°-l From ^ 48, Case II., cos 5= tana cote. Substitute in this equation, cos t = tan PZ cot PM, or cos t = tan (90°- 1) cot (90°- d). .'. cos t = cot I tan d. And from ^ 65, the times of bear- ing due east and west are 12 A.M. and — p.m., 15 15 respectively. Since the day given is the longest day of the year, the declination of the sun = 23° 27^ .*. we have given d= 23° 27'' and I = 59° 56^ to find t. Now cos t = cot I tan d. log cot I = 9.76261 log tan cZ= 9.63726 log cos t = 9.39987 t = 75° 27^ 24^^ .*. 12 = 6 hrs. 58 min. A.M., 15 and — = 5 hrs. 2 min. p.m. 15 16. Apply the general result in Example 15 (cos t = cot I tan d) to the case when the days and nights are equal in length (that is, when d = 0°). Why can the sun in sum- mer never be due east before 6 a.m., or due west after 6 p.m. ? How does the time of bearing due east and due west change with the decli- nation of the sun ? Apply the gen- eral result to the cases where I 1 ; therefore this case is impossible. If 1 = d, then cost=l, and t = 0°; that is, the times both coincide with noon. The explanation of this result is, that the sun at noon is in the zenith ; hence, on the prime vertical, at the pole, Z = 90°, cosi= 0°, i = 90°; therefore the sun in sum- mer always bears due east at 6 a.m. and due west at 6 p.m. 17. Given the sun's declination and his altitude when he bears due east; find the latitude of the ob- server. N In the figure let Z be the zenith of observer, F the elevated pole, M the position of the sun. Then ZM= 90° - h. FM= 90° - d. FZ=dO°-l. Since the sun M bears due east, MZF is a right angle. /. by Napier's Kules, cos FM= cos FZ cos MZ. .'. sin d^sinl sin h. sin I = sin dcsch. 18. At a point in a horizontal plane MI^ a staff OA is fixed, so that its angle of inclination AOB with the plane is equal to the lati- tude of the place, 51° 30^ N., and the direction OB is due north. What angle will OB make with the shadow of OA on the plane, at 1 P.M. Given direction of OB due north, AOB = 51° 30^= I, and plane MN horizontal ; to find BOC. Produce OA ; it will pass through the pole. The sun being on the equinoctial, FOS = 90°, and the shadow 00 will lie in the plane of this angle. Draw OZ JL to plane MJSF; it will lie in the plane of OB and OA. SFZ = hour angle of sun at 1 p.m. = 15°. SFZ = CAB, being vertical angles. .■.CAB=\b°. ABO= 90°, since OB is the pro- jection of OA on plane MN. Arc AB = 51° 30^, being the meas- ure of plane angle AOB. Then in right spherical triangle ABC, by [41], tan BC^ tan BAGs'm AB. TEACHERS EDITION. 159 log tan 15° =9.42805 log sin 51° 30^ = 9.89354 log tan BQ = 9.32159 Arc BO = 11° 50^ 35^^ Arc -SC measures plane angle BOC. .: BOC= 11° 50^ 35^^ 19. What is the direction of a wall in latitude 52° 30^ N. which casts no shadow at 6 a.m. on the longest day of the year. The wall must lie in the line pass- ing through the sun, in order that it may cast no shadow. In the figure, PZ= 90° -I PM= 90° - d. ■ ZPD = 6 X 15°= 90°. To find irZP=a;. By [41], sin (90°- V) = tan {90° -d) cot MZP. cos I = cot d cot X. Or, cot re = cos Z tan e. log cos I = 9.78445 log tan e = 9.63726 log cot a; =9.42171 X = 75° 12^ 38^^. 20. At a certain place the sun is observed to rise exactly in the north- east point on the longest day of the year ; find the latitude of the place. When the sun rises in the north- east on the longest day of the year, a = 45°, c?=23° 27^ To find I. In the formula cos a = sin d sec I. log sec I = log cos a + log esc d. log cos 45° = 9.84949 log CSC 23° 27^ = 0.40017 log sec I = 0.24966 I = 55° 45^ 6^^. 21. Find the latitude of the place at which the sun sets at 10 o'clock on the longest day. Since the sun sets at 10 o'clock, the hour angle of the sun is equal to 15° X 10 = 150°. The declination of the sun is equal to 23° 27^. .'. we have in the triangle two parts given ; viz., the angle ZPM, and ifP= 90° -d In the formula, cot I = cos t cot d. t = 150°. d= 23° 27^ log cos t = 9.93753 log cot £? = 0.36274 log cot I = 0.30027 I = 63° 23^ 41^^ 22. To what does the general formula for the hour angle, in ^ 67, become when (i.) h = 0°, (ii.) I = 0° and c? = 0°, (iii.) Zor c? = 90°. 160 TEIGONOMETRY. In the general formula, § 67, let /i = 0. Then sin I i = ± [cos J(Z ^- p) X sin J (Z +p) sec I cscp]^. (11.) sin i t ^ h — cos t cos j(;+rt = .^I±£2^(U£). sin I ii+p)^^^j '-^°f^pl sec Z cscp = cos I sinp Substitute these values in the first equation, 1 — cos ^ ^ / l +cos(? + p) 2 ~^ 2 ^ Jl_-cos_(Z_+^) X cos I sinp 1 — cos i = Vl — cos^ {I +p) COS I smp 1 — cos i = sin (I + p) COS Isinp = (sin I cosp + COS I sinp) x^L_ COS I smp = tan I coip + 1. COS t = — tan Z cot j3. But _p-90°-d .*. cot j3 = tan d, and cos i = — tan I tan d sin 2- -4 = ■v/sin(s— J)sin(s— c)csc h csc c. Z = 0. d=0. FZ =90° -I =90° = e. Pif - dO°-d = 90° = 6. ^ir= 90° -?i = a. A = t. s-b = ^ (90° - A). s-c = ^ (90° - ^). csc b = csc 90° = 1. csc c = csc 90° = 1. Substitute sin ^ i = Vsin J (90°- h) sin J (90°-;^) = sin J (90° -/i). sin ^ = sin (90° - h). t=90°-h = z. (iii.) Zor(?=90°. PZ=90°-2 = 0° = a. .•. no triangle will be made. .•. answer indeterminate. Pir= 90° -cZ=0° = 6. /. no triangle formed. .*. result indeterminate. 23. What does the general for- mula for the azimuth of a celestial body, in § 68, become when t = 90° = 6 hours ? When t = 90°, m = 0, and we have a right spherical triangle with the two legs given to- find the angle opposite one of the legs. TEACHERS EDITION. 161 Then by substituting the values of the given parts in the formula, tan B = tan h esc a, we have tan a = tan (90°-c?) esc (90°-Z) or tana =cot(Zsec^. tan a cot a sec i and cot a = tan d cos Z. 24. Show that the formulas of § 69, if t = 90°, lead to the equation sin Z = sin A esc cZ ; and that if cZ = 0°, they lead to the equation cos I = sin h sec t. I. If< = 90°. From g 69, sin ^ = cos n cos MQ. sin d= cosm cos MQ. Divide (1) by (2), II. If (2=0°. From § 69, cos I = COS m sin h esc c?. (3) cos t (1) (2) sin ^ cos n sin cZ cosm but now n = ZP=90°- and m = 0°. sin h • 7 .'. = sm Z. sin d .'. sin Z = sin A esc cZ. tancZ = tanm Multiply (3) by (4), cos Z cos t cos m sin h (4) tanm cos d cos Z cos ^ = sin m sin cos d _^ But if d = 0°, PM= 90°, and m = 0. .-, cos Z cos ^ = sin h. .-. cosZ = sin h sec <. 25. Given latitude of place 52° 30^ 16^^ declination of star 38°, its hour angle 28° 17^ 15^^ find its altitude. Given P^=90°-Z. PM= 90° - d. ZPM= t ; required ZM= 90° - h. Let PQ = m. By Napier's Rules, tan m = cot d cos cZ. 162 TRIGONOMETRY. log cot fZ = 0.10719 log cos t = 9.94477 log tan m = 10.05196 m = 48° 25^ 10^^ sin h = sin (Z + m) sin d sec m, log sin (Z+m)= 9.99206 log sin d = 9.78934 log seem =0.17804 log sin h = 9.95944 - 10 h = 65° 37^ 20^. 26. Given latitude of place 51° 19'' 20'''', polar distance of star 67° 59^ 5^^ its hour angle 15° 8^ 12''''; find its altitude and its azi- muth. Given PM= polar distance of star. ZPM= hour angle of star. PZ = co-latitude of observer. Find PZM= azimuth of star, and DM its altitude. Let d=90°- PM. Let PQ = m. tan m = cot d cos t. sin h = sin {I + m) sin d sec m. tan a = sec {I + m) tan t sin m. I = 51° 19^ 20^^. c? = 90° - (67° 59^ 5^0- = 22° 0^ 55^^. t = 15° 8^ 12^^. log cot d = 10.39326 log cos t = 9.98466 ^ log tan m =10.37792 m = 67° 16^ 22^^ log sin (l+m) = 9.94351 log sin d = 9.57387 log sec m = 0.41302 log sin h = 9.93040 h = 58° 25^ 15^^. log sec {l+m) = 0,32001 log tan t = 9.43218 log sin m = 9.96490 log tana =9.71709 a = 152° 28^. 27. Given the declination of a star 7° 54^ its altitude 22° 45^ 12^^ its azimuth 129° 45^ 37^^ ; find its hour angle and the latitude of the observer. sin t = 8ma cos h sec d. = 9.88577 = 9.96482 = 0.00414 log sin a log cos h colog cos d log sin t = 9.85473 t = 45° 42^ tan m = cot d cos t. cos n = cos m sin h esc d. I =90°-(m±7i). log cot d = 10.85773 log cos t = 9.84411 log tan m = 10.70184 m = 78° 45^ 45^^. TEACHERS EDITION. 163 log cos TO log sin h log CSC d log cos n = 9.28976 = 9.58745 = 0.86187 = 9.73908 n = 56° W 39^^. m-n = 12° V Q'\ 90° _ (m - w) = 67° 58^ 54^^ .-. Z = 67° 58^ 54^^. 28. Given the longitude u of the sun, and the obliquity of the eclip- tic e = 23° 27^ ; find the declination d, and the right ascension r. In the figure let P represent the pole of the equinoctial A VB, S the position of the sun, and Q the pole of the ecliptic EVF. Then VS = u. VB = r. 8R = d. RVS=e. Then in the right triangle B VS, by [38], sin SB = sin VS X sin B VS, sin d or sm u sm e. Also by [39], cos BVS= tan B Fcot VS, or cos e = tan r cot u. tan r = tan u cos e. 29. Given the obliquity of the ecliptic e = 23° 27^, the latitude of a star 51°, its longitude 315°; find its declination and its right ascen- sion. In Fig. 47, given VT= 315° or - 45°, TM= 51°, i2Fr= 23° 27^ , to find VB^r and BM= d. In right triangle VTM, cos VM= cos FT'cos TM, and tan ili'FT^ tan MT esc FT. log cos 315° = 9.84949 log cos 51° = 9.79887 log cos VM = 9.64836 VM= 63° 34^ 36^^. log tan 51° log CSC 315° log tan MVT MVT- = 10.09163 = 0.15051 (n) = 10.24214 in) - (60° 12^ 14.5^0- In right triangle B VM, BVM=BVT+ TVM = 23° 27^-(60° 12^ 14.5^0 = - (36° 45^ 14.5^0- By [38], sin BM= sin FJf sin B VM. log sin VM = 9.95208 log sin i? FIT =9.77698 log sin BM = 9.72906 BM=d=2>2°2¥l2^'. 164 TRIGONOMETRY. Also, by [41], sin VB = tan BM cot E VM. . log tan i^lf =9.80257 logcoti^FJlf = 0.12677 (n) log sm VB = 9.92934 {n) Fi^= -(58° 11^ 43^0- .-. Fi2 = 360° -58° IIMS^^ = 301°48M7^^ 30. Given the latitude of a place 44° 50^ 14^^ the azimuth of a star 138° 58^ 43^^ and its hour angle 20°; find its declination. Given c = 90° - 44° 50' 14^' = 45° 9' 46^'. A = 138° 58' 43''. B = 20°. 1{A-B)== 59° 29' 22". 'l{A-\-B) = 79° 29' 22". J c =22° 34' 53". log cosJ(^-5) = 9.70560 colog cos jU+^) = 0.73893 log tan J c =9.61897 log tan J (a + 5) = 0.06350 Ha + &) = 49° 10' 26". log sin J {A-B) = 9.93528 colog %m^{A+B) = 0.00735 log tan J c =9.61897 log tan J(a- 6) = 9.56160 ^(a-6) = 20° 1'21.5". .-. a = 69° 11' 48". 90° - 69° 11' 48"= 20° 48' 12". 31. Given latitude of place 51° 31' 48", altitude of sun west of the meridian 35° 14' 27", its declina- tion +21° 27' ; find the local ap- parent time. Byi67, P^=90°-?, ■ Pir= 90° -(Z = p, ZM= 90° - A ; required t = ZBM. p = 68° 33'. J(Z + A+^) = 77°39'37.5". J(Z-/i+p) = 42° 25' 10.5". log log colog colog log 15 cosJ(^+P + ^)= 9.32982 sin J(Z+_p-A)= 9.82901 cosZ = 0.20614 sinp = 0.03117 2 )19.39614 sin I « = 9.69807 J ^ = 29° 55' 55.5". t = 59° 51' 51". = 3 h. 59 min. 27f sec. p.m. 32. Given latitude of place ?, the polar distance p of a star, and its altitude h : find its azimuth a. TEACHERS EDITION. 165 I Altitude = ZM= 90°- h. Co- latitude = FZ = 90°- 1. Polar distance = FM = 90° -d=p. Azimuth = FZM oi a. cos ^A= Vsin s sin (s— a) esc b esc c. Let Then sin s A^ FZM or a, a=p, b = 90°- h, c = 90°- I = sm[90°-} (l+h-p)] = cos J (h + l—p). sin (s-a) = sin [90°- H^+^+P)] = cos J(/i + ^+p). CSC b = CSC (90°— A) = sec h. esc c = esc (90°— l) = sec I. .'. cos I a = Vcos|(_p+A+^)cos|(/i+^-p)sec^secA SUEYETI.N-G. Exercise I. Page 143. 1. Required the area of a triangular field whose sides are respectively 13, 14, and 15 chains. Area = Vs{s — a) (s — 6) (s — c). s = i(13 + 14 + 15) = 21, s- 5 = 21 -14 = 7, ■ s-a = 21 -13 = 8, s-c = 21 -15 = 6. Area = V21 x 8 x 7 X 6 = VS^ x 7^ X 2* = 3 X 7 X 2^ = 84 sq. ch. = 8.4 A. = 8 A. 64 p. 2. Required the area of a triangular field whose sides are respectively 20, 30, and 40 chains. Area = V45 x 25 x 15 x 5 = V3=* X 5^ = 3 x 5 V3 x 5 = 75 VTS = 290.4737+. 290.4737 sq. ch. = 29.04737 a. = 29 a. 7.579 p. = 29 a. 7f p., nearly. 3. Required the area of a triangular field whose base is 12.60 chains, and altitude 6.40 chains. Area = J base X altitude. Area = J X 12.6 X 6.4 = 40.32 sq. ch. = 4.032 A. = 4 A. b^^j p. 4. Required the area of a triangular field which has two sides 4.50 and 3.70 chains, respectively, and the included angle 60°. Area = ^bc sin A. Area = ^ x 4.5 x 3.7 X 0.866 = 7.20945 sq. ch. = 0.7209 a. = 115/oP., nearly. 5. Required the area of a field in the form of a trapezium, one of whose diagonals is 9 chains, and the two perpendiculars upon this diag- onal from the opposite vertices 4.50 and 3.25 chains. Area = J X 9 (4.5 + 3.25) = 34.875 sq. ch. = 3.4875 a. . = 3 A. 78 p. 168 SURVEYING. 6. Required the area of the field ABCDEF (¥ig. 19), if ^^=9.25 chains, FF^=6A0 chains, BE= 13.75 chains, DD^= 7 chains, DB= 10 chains, CC^= 4 chains, and AA^=4:.15 chains. 2area^i^.£' = 6.4x9.25 = 59.2 2 area BDFA = 13.75 (4.75 + 7) = 161.5625 2zxQ2iBDC =10x4 =40 2 area ABC DBF = 260.7625 area ABCDFF = 130.38125 130.38125 sq. ch. = 13.038125 a. = 13 a. 6j\ p. 7. Required the area of the field ABCDEF (Fig. 20), if AF^--^ 4 chains, FF'= 6 chains, EE'= 6.50 chains, AE^= 9 chains, AD = 14 chains, AQ' = 10 chains, AB^= 6.50 chains, BB^= 7 chains, CC^=Q.7o chains. 2 area Tli^i^'' =4x6 =24 2areai^''^^^i^=5(6 + 6.5) = 62.5 2&YesLEE'B =6.5x5 = 32.5 2 area ABB^ = 6.5 x 7 = 45.5 2 area BCC'B^ = 3.5 (7 + 6.75) = 48.125 2areaaDC^ =6.75x4 = 27 2 area ABCDEF = 239.625 area ABCDEF = 119.8125 119.8125 sq. ch. = 11.98125 a. = 11 a. 157 p. 8. Required the area of the field AGBCD (Fig. 15), if the diagonal AC= 5, BB^ (the perpendicular from B to AC) = 1, DD^ (the perpen- dicular from D to AC)^ 1.60, EE'= 0.25, FF'= 0.25, GG'= 0.60, HM' = 0.52, KK'=OM, AE^ = 0.2, E^F^^O.bO, F^G^=0A5, G^H^=0A5, S^K'^'-Om, and ^^^ = 0.40. 2sirea,ADCB =5(1+1.6) =13. 2 area AEE' = 0.25 x 0.2 = 0.05 2 3.vesiEE'F'F =0.5(0.25 + 0.25)= 0.25 2&reiiFF'G'G =0.45(0.25 + 0.6)= 0.3825 2 area (?G^i7^^ = 0.45(0.6 +0.52) = 0.504 2 area ^.ff^ir'ir= 0.6 (0.52 + 0.54)= 0.636 2 9^vesi,KK'B =0.4x0.54 = 0.216 2 area ADCBKHGFE = 15.0385 Sivesi ADCBKHGFE = 7.51925. TEACHERS EDITION. 169 9. Required the area of the field AOBCD (Fig. 16), iiAD = Z,AC = 5, AB=Q, angle i)J.a= 45°, angle ^4C=30°, ^^^=0.75, AF^=2.2b, AE= 2.53, AG'= 3.15, EE'= 0.60, FF'=^ 0.40, and 0Q'= 0.75. 2 area ^1)05 = 3 X 5 X 0.7071 + 5 x 6 x 0.5 2q.xq2.EGB =0.75x3.47 = 25.6065 = 2.6025 2s,Yea,AI)CBGE =28.2090 2 area AEFH= 0.75 x 0.6 + 1.5(0.6 + 0.4) + 0.4 x 0.28 = 2.062 2 area ADCBGHFE area ADCBGHFE = 26.147 = 13.0735. 10. Determine the area of the field ABCD from two interior stations F and P^ if PP'= 1.50 chains, angle P^PB^ 3° 35^ P''P^ = 113° 45^, P^PD = 165° 40^ P'PC = 303° 15^ angle PP'G = 89° 35^ PP^P = 185° 30^ PP^A = 309° 15^ • PP^i) = 349° 45^ Area = A PAD + A PCD + A PPC+ A PAB. ZPP'D= 10° 15^ ZPDP'= 4° 5^ ZPP^P = 174° 30^ Z PPM = 50° 45^ ZP^P^ = 15° 30^ ZPBF= 1°55^ PD = PP' sin PP^P> sin PDF' log PP^ = 0.17609 log sin PP^D = 9.25028 colog sin PDP^ = 1.14748 log PD = 0.57385 ZPP'C- ZPCP^-- PP^ sin PPM 89° 35^, 33° 40^. PA = sin PAF^ log PP^ = 0.17609 log sin PPM =9.88896 colog sin PAP^= 0.57310 log PA = 0.63815 PC = PP'smPP'C sin PCP^ log PP^ = 0.17609 log sin PP^C = 9.99999 colog sin PCP^ = 0.25621 log PC = 0.43229 PB PP' sm PP'B sm PBP^ log PP^ = 0.17609 log sin PP^B = 8.98157 colog sin PBP^ = 1.47566 log PB = 0.63332 Z ^Pi) = 51° 55^ ZZ>PC= 137° 35^ ZPPC=60°20^ Z^PP = 110°10^ 170 SURVEYING. 2 area FAD = PDxPA sin APD. log PD = 0.57385 log PA = 0.63815 log sin APD = 9.89604 log 2 area = 1.10804 2 area P^i) = 12.825. 2 area P^P = PAxPB sin ^PP. log P^ = 0.63815 log PB = 0.63332 log sin APB = 9.97252 log 2 area = 1.24399 2areaP^P =17.538. 2 area PCZ) = PD x P(7sin PPC. log Pi) = 0.57385 log PC = 0.43229 log sin DPC = 9.82899 log 2 area = 0.83513 2 Sires. PCD =6.8412. 2 area PBC= PCx PB sin PPC. log PC = 0.43229 log PB = 0.63332 log sin PBC = 9.93898 log 2 area = 1.00459 2areaPP(7 =10.106. 2 A P^P> = 12.825 2 A PCD = 6.841 2 A PBC = 10.106 2APAB =17.538 2ABCD =47.310 ABCD = 23.655 sq. ch. 23.655 sq. ch. = 2.3655 a. = 2 A. 58^- p., nearly. 11. Determine the area of the field ABCD from two exterior stations P and P\ if PP^= 1.50 chains. angle P'PB = 41° 10^ angle PP'D = 66° 45^ P'PA = 55° 45^ PP'C = 95° 40^ P^PC = 77° 20^ PP^B = 132° 15^ P^PD = 104° 45^ PP^A = 103° 0^ Area = (A P^CP + A P'CD) - (A P^^P + A P'AD). ZP''PP = 41° 10^ ZPBP'= 6°35^ Z P^P^ = 55° 45^ ZP^PP = 104° 45^ Z PDP' = ZPAP' = P'B = PP' sin P^PP sin PBP^ log PP^ = 0.17609 log sin P^PP= 9.81839 colog sin PBP^ = 0.94063 log P'B 0.93511 8° 30^ 21° 15^ P'D = ZP^PC= 77° 20^ ZPCP'= 7° 0'. PP^ sin P^PD sin PPP^ log PP^ = 0.17609 log sin P^PD = 9.98545 colog sin PPP^ = 0.83030 log P'D = 0.99184 TEACHERS EDITION. 171 P'G FF' sin P'PQ sm PCP^ log PP' = 0.17609 log sin P^PC= 9.98930 colog sin PCP^ = 0.91411 log P^C = 1.07950 p^^ _ PP' sin P'PA sm P^P-' log PP^ = 0.17609 log P'PA = 9.91729 colog P^P'' = 0.44077 log P'A = 0.53415 ZPP^C=36°35^ ZCP''i) = 28° 55^ Z^P''P = 29°15^ Z ^P^i) = 36° 15^ 2 area P^CP = P^Cx P^B sin PP'^O. log P^C = 1.07950 log P'B = 0.93511 log sin PP^ (7 =9.77524 log 2 area 2 area P^CP 1.78985 61.639. 2 area P'^CP = P^Cx P^Psin CP^P. log P'C = 1.07950 log P'B = 0.99184 log sin CP^n = 9.68443 log 2 area = 1.75577 2 area P^CP = 56.986. 2 area PMP =P^P x P'A sin ^P^P. log P^P =0.93511 log PM = 0.53415 log sin ^P^P= 9.68897 i log 2 area 2 area P^AB 2 A P^CP 2 A P^CP = 1.15823 = 14.396. = 61.639 = 56.986 118.625 34.248 2 ABCD ABCD = 84.377 = 42.1885 2 area P^AD ^P^A x P^ D sin AP^D. log PM = 0.53415 log P^D = 0.99184 log sin ^P^P= 9.77181 log 2 area = 1.29780 2 area P^AD = 19.852. 2 A P'AB 2 A P'' J.P = 14.396 = 19.852 34.248 42.1885sq.ch. = 4.21885 A. = 4 A. 35 p., nearly. 172 SURVEYING. Exercise II. Page 152. 1. iV^. S. U. w. M.D. li.A. S.A. 1 2 3 4 5 S. 750 E. S. 150 E. S. 75^ W. N. 450 E. N. 450 W. 6.00 4.00 6.93 5.00 5.19.^ 3.54 3.67 1.55 3.86 1.80 -1-^9- 5.79 -5-80- 1.04 3.54 6.70 -6-69- 3.67 5.79 6.83 0.13 3.67 5.79 12.62 6.96 3.80 3.67 13.4520 13.4689 8.9745 48.7132 12.5280 21.647 sq. ch. = 2.1647 A. = 2 A. 26 p., nearly. 26.9209 70.2157 26.9209 43.2948 21.6474 2. m S. E. w. M.D. N.A. S.A. 1 N. 450 E. 10.00 7.07 . . . 7.07 . . . 7.07 7.07 49.9849 2 S. 750 E. 11.55 . . . 2.99 11.16 . . . 18.23 25.30 75.6470 3 S. 15° W. 18.21 . . . 17.59 . . . 4.71 13.52 31.75 558.4825 4 N.450 W. 19.11 13.51 13.52 13.52 182.6552 ..... 232.6401 634.1295 232.6401 200.74 Bq.ch.=S !0.074 A. = 20 A. 12 P., nearly 401.4894 200.7447 TEACHERS EDITION. irs 3. N. S. E. W. M.D. C5 N.A. S.A. 1 ^ N.150E. 3.00 2.90 ... 0.78 0.78 0.78 2.2620 2 N. 750 E. 6.00 1.55 5.79 -5-80- 6.57 7.35 11.3925 3 s. 150 W. 6.00 5.80 . . . 1.55 5.02 11.59 67.2220 4 N. 750 W. 5.20 1.35 5.02 5.02 6.7770 20.4315 67.2220 20.4315 23.395 sq. ch. = 5 J.3395 A L. - 2 A. 54 P., nearly. 46.7905 23.3953 4. N. S. E. W. M.D. N.A. S.A. 1 2 3 4 5 6 N. 890 45' E. S. 70 00' W. S. 28° 00' E. S. 0045'E. N. 84° 45' W. N. 2030'W. 4.94 2.30 1.52 2.57 5.11 5.79 0.00 -e-02- 0.45 -e-47- 5.76 -5-78- 2.29 -2-28- 1.34 2.58 -2-57- 4.93 -4-94- • . . 0.71 0.02 -0-03- 0.29 -0-28- 5.10 -5-09- 0.27 -0-25- 4.93 4.64 5.35 5.37 0.27 4.93 9.57 9.99 10.72 5.64 0.27 2.5380 1.5552 21.9153 13.3866 27.6576 2-94.^ A. — 2 A. I.'il p.. nparlv. 4.0932 62.9595 4.0932 58.8663 29.4332 174 SURVEYING. 5. iV. S. E. W. N 51045 W. 2.39 1.48 . . . . . . 1.88 S. 85° w. 6.47 . . . 0.56 6.45 S. 550 10' w. 1.62 0.93 1.33 1.49 . . . 9.66 1.48 0.01 N. 8. E. w. M.D. N.A. S.A. 2 3 4 5 6 1 S. W. N. 30 45' E. S. 66045'E. N". 15° E. S. 82° 45' E. S. 2oi5'E. 6.39 1.70 4.98 6.03 9.68 6.36 -6-38- 4.80 -4-81- 0.03 -0-ei- 0.67 0.77 -e-T6- 9.69 0.43 -e-42- 1.56 1.29 5.98 0.39 -9-38- 9.65 -9-66- 9.65 9.22 7.63 6.37 0.39 9.65 18.87 16.88 14.03 6.76 0.39 120.0132 67.3440 0.2895 11.3096 5.2052 3.7791 8 S.qQ A — 8 A ^1-p TiParW 187.3572 20.5834 20.5834 •»'j« 166.7738 83.3869 TEACHERS EDITION. 175 6. iV. S. E. W. S. 81O20' W. N. 76° 30' W. 4.28 2.67 0.62 0.65 . . . 4.23 2.60 0.65 0.62 6.83 0.03 N. S. JE. W. S. 7° E. S. 27° E. S. 103 30'E. N. 76^45' W. 1.79 1.94 5.35 1.70 0.39 1.78 1.73 5.26 0.22 0.88 0.98 1.65 8.77 0.39 2.08 1.65 . . . 8.38 0.43 1 2 3 4 N'. S. E. w. M.D. N.A. S.A. S. W. N. 5° E. S. 87° 30' E. S. E. 8.68 5.54 8.65 0.03 0.24 8.38 0.79 -e-76- 5.55 -6-53- 0.46 -0-43- 6.80 -6-83- 6.80 6.01 0.46 6.80 12.81 6.47 0.46 110.8065 0.2040 1.5528 3.8548 5.2597 A. = 5 A. 42 P.. nearlv. 110.8065 5.6116 5.6116 105.1949 52.597 176 SUEVEYING. 7. 2f. S. E. W. M.n. mA. S.A. 3 4 1 2 S. 5O00'E. N. 88^30' E. N. 6° 15' W. S. 81O50' W. 5.86 4.12 6.31 4.06 0.12 -0-11- 6.28 -6-27- 5.83 6-84- 0.57 -e-58- 0.53 -6-51- 4.14 -4-12- 0.67 -9.69- 4.00 -4-02- 0.53 4.67 4.00 0.53 5.20 8.67 4.00 0.6240 54.4476 3.0899 2.2800 55.0716 5.3699 5.3699 5 U85 a.= 2a. 78 P., ne< irly. 49.7017 24.8508 8. ■ iV^. S. E. r. M.D. N.A. S.A. 3 4 1 2 S. 3° 00' E. E. N. 5030'W. S. 82° 30' W. 5.33 6.72 6.08 6.51 0.03 -e-ee- 6.0s -6-65- 5.29 -5-32- 0.82 -6-85- 0.28 6.73 -6-72- 0.57 -e-58- 6.44 -6-45- 0.28 7.01 6.44 0.28 7.29 13.45 6.44 0.2187 81.7760 1.4812 5.2808 81.9947 6.7620 6.7620 J.761 A. = 3 A. 12 2 P., m iarly. 75.2327 37.6163 TEACHEES EDITION. 177 9. 1 2 3 4 5 1 iV. S. E. W. M.n. c5 N.A. S.A. N.20°00'E. N. 73° 00' E. S. 450 15'E. S. 38° 30' W. "Wanting. 4.62i 4.16i 6.18i 8.00 4.35 1.22 5.04 4.35 6.26 1.58 3.98 4.39 4.98 4.97 1.58 5.56 9.95 4.97 1.58 7.14 15.51 14.92 4.97 6.8730 8.7108 25.0488 67.4685 93.3992 1 ■ ■ 40.6326 160.8677 40.6326 ( 5.012 A. = 6 A. 2 ] P., neai ly. 120.2351 60.1175 10. N. S. E. W. M.D. c5 N.A. S.A. 6 7 8 9 1 2 3 4 5 N. 32° 00' E. s. 75050'E. S. 140 45' "W. S. 79"15'E. s. 3000'E. S. 860 45'"W. S. 37°00' W. N. 810 00'W. N. 610 00'W. 8.68 6.38 0.98 4.52 4.23 4.78 2.00 7.45 2.17 7.3 -7-3 1.1^ -1-1- 1.0 -1-0^ 3 3- 1 ?- i 3- 1.58 -1-56- 0.95 0.86 -6-84- 4.23 -4-^- 0.29 -0-27- 1.60 4.61 -4-69- 6.20 -6-19- 4.44 0.22 0.25 4.77 1.20 7.35 -7-36- 1.90 4.61 10.81 10.56 15.00 15.22 10.45 9.25 1.90 4.61 15.42 21.37 25.56 30.22 25.67 19.70 11.15 1.90 33.7913 12.7110 1.9760 24.3636 20.3015 21.9816 127.8306 7.4443 31.5200 48.4783 233.4416 48.4783 { ).248 A. = 9 A. 40 P., nea rly. 184.9633 92.48 178 SUHVEYING. Exercise HI. Page 163. 1. N. S. E. w. M. D. N.A. S.A. BC S. 60° E. 4.000 4.000 4.000 3.464 3.464 .... 6.928 2.000 3.464 • . > CD S. 30° E. 6.928 6.000 3.464 . . . 6.928 10.392 .... 62.352 DA N. 60O W. 8.000 4.000 .... 6.928 6.928 27.112, .... 27.712 69.280 27.712 20.784 sq. ch.= 2.0 784 A. = 2 A. 12| p., nearly. 41.568 20.784 2. N. S. E. W. M.D. N. A. S.A. AB BC N. N. 80°20'E. 79.86 121.13 79.860 20.338 119.410 119.410 2428.56 .... . . . 119.410 . . . CD S. 40° 00' E. 90.00 . . . 68.943 57.851 177.261 296.671 .... 20453.39 DM s. 55052'w. 100-65 . . . 59.350 .... 81.289 95.972 273.233 2696.33 16216.38 EA N. 730 41' W. 109.00 28i)95 .... 95.972 95.972 .... 5124.89 36669.77 5124.89 15772.44 jp. = 9^ A. 92 P., nearly. 31544.88 15772.44 teachers' edition. 179 Exercise IV. Page 161. 1. From the square ABCB, containing 6 a. 1 r. 24 p., part off 3 A. by a line EF parallel to AB. 6 A. 1 R. 24 E. = 64 sq. ch. ; Voi = 8 ch. = AB. 3 a. =30 sq. ch. AB 8 2. From the rectangle ABCD, containing 8 a. 1 n. 24 P., part off 2 A. 1 E. 32 p. by a line ^i^ parallel to AI)=7 ch. Then, from the remain- der of the rectangle part off 2 a. 3 r. 25 p. by a line GE parallel to IJB. 8 A. 1 R. 24 p. = 84 sq. ch. = ABCB. 2 A. 1 R. 32 p. = 24.5 sq. ch. = AEFD. 2 A. 3 R. 25 p. = 29.0625 sq. ch. = EBHQ. AE = ^^^='-^^3.5ch. AD 7 A^CD^84^^2ch. AD 7 EB =AB-AE =12- 3.5 = 8.5 ch. ^ EBHG ^2jm25 ^ 3 ^^ ^h.. nearly. EB 8.5 3. Part off 6 A. 3 e. 12 p. from a rectangle ABCD, containing 15 A. by a line EF parallel to AB ; AD being 10 ch. 6 A. 3 R. 12 p. = 68.25 sq. ch. = ABFE. 15 A. = 150 sq. ch. = ABCD. AB=^^^ = '-^ = 15ch. AD 10 ^^ = ^^^=68^ = 4.55ch. AB 15 4. From a square ABCD, whose side is 9 ch., part off a triangle which shall contain 2 a. 1 r. 36 p., by a line BE drawn from B to the side AD. 2 A. 1 R. 36 p. = 24.75 sq. ch. AB 9 180 SURVEYING. 5. From ABCD, representing a rectangle, whose length is 12.65 ch., and breadth 7.58 ch., part off a trapezoid which shall contain 7 A. 3 e. 24 p., by a line BE drawn from B to the side DO. 7 A. 3 E. 24 p. = 79 sq. ch. ABCD = 12.65 X 7.58 = 95.887 sq. ch. A BCE= 95.887 - 79 = 16.887 sq. ch. ^55. 2BCE 2x16.887 . .r- . , CE = = = 4.456 ch., nearly. BC 7.58 ^ 6. In the triangle ABC, AB = 12 ch., AC= 10 ch., and 5C= 8 ch. ; part off 1 A. 2 E. 16 p., by the line DjE" parallel to AB. 1 A. 2 E. 16 p. == 16 sq. ch. CAB = Vl5x 3x5x7 = 39.6863 sq. ch. CDE = CAB - ABED = 39.6863 - 16 = 23.6863 sq. ch. CAB : CDE :: CA^ : CD' : : CB^ : CE\ 39.6863 : 23.6863 : : 10^ : CD\ .'. CD = 7.725 ch. : : 82; CE\ .-. C^-6.18 ch. AD = CA- CD =10- 7.725 = 2.275 ch. BE = CB - CE = 8 - 6.18 = 1.82 ch. 7. In the triangle ABC, AB=2Q ch., AC = 20 ch., and BC= 16 ch. ; part off 6 A. 1 E. 24 p., by the line DE parallel to AB. 6 A. 1 E. 24 p. = 64 sq. ch. CAB = V31x5x 11x15 = 159.9218 sq. ch. CDE^ CAB - ABED = 159.9218 - 64 = 95.9218 sq. ch. CAB : CDE : : CA" : CD" : : (7^' : CE 2 159.9218 ; ; 95.9218 : : 202; cd\ 162 : ce\ . CD = 15.49 ch. . CE = 12.39 ch. AD= CA- CD = 20- 15.49 = 4.51 ch., nearly. BE = CB - CE = IQ - 12.39 = 3.61 ch., nearly. 8. It is required to divide the triangular field ABC among three per- sons whose claims are as the numbers 2, 3, and 5, so that they may all have the use of a watering-place at C; AB = 10 ch., AC= 6.85 ch., and CB = 6.10 ch. teachers' edition. 181 Since the tr' angles have the same altitude, they are to each other as their bases. Hence it is only necessary to divide the base 10 into the three parts, 2 ch., 3 ch., 5 ch. 9. Divide the five-sided field ABCHE among three persons, X, Y, and Z, in proportion to their claims, X paying ^500, Y paying! 750, and Z paying % 1000, so that each may have the use of an interior pond, at P, the quality of the land being equal throughout. Given AB = 8.64 ch., ^(7=8.27 ch., CZr= 8.06 ch., HU = 6.S2 ch., and ^^ = 9.90 ch. The perpendicular FD upon ^^-5.60 ch., FB^ upon ^C=6.08 ch., FB^^ upon CH^ 4.80 ch., FB''^ upon HE^bA^ ch., and FB^^^^ upon EA = 5.40 ch. Assume FH as the divisional fence between X's and Z's shares ; it is required to determine the position of the fences FM and PiV between X's and Y's shares and Y's and Z's shares, respectively. If Pbe joined to the vertices, the field is divided into triangles, whose bases are the sides, and the altitudes the given perpendiculars upon the sides from P. AFB = 8.64 X 2.80 = 24.1920 sq. ch. BFC = 8.27 X 3.04 = 25.1408 CFH = 8.06 X 2.40 = 19.3440 JTFE = 6.82 X 2.72 = 18.5504 EFA = 9.90 X 2.70 = 26.7300 ABCHE =113.9572 The whole area 113.9572 sq. ch. must be divided as the numbers 500, 750, 1000, or as 2, 3, 4. 2 + 3 + 4 = 9. 9 : 113.9572 : : 2 : 25.3238 sq. ch. = X's share. : : 3 : 37.9857 sq. ch. = Y's share. : : 4 : 50.6476 sq. ch. = Z's share. FH'is assumed as the line between X's and Z's shares. Since the tri- angle FHE is less than X's share by 25.3238 - 18.5504 = 6.7734 sq. ch., this difference must be taken from the triangle FEA. The area of FEM is then 6.7734 sq. ch., and the altitude Pi)^^/^ = 5.40. ,. ^J/=2P^^=2xa773_4^ 2.5087 ch. FB''^^ 5.40 PMA = FEA - FEM= 26.7300 - 6.7734 = 19.9566. sq. ch. Since Y's share is greater than FMA (19.9566) and less than FMA + FAB (44.1486), the point iV^is on AB. 182 SURVEYING. Y's share diminished by PMA equals PAN; that is, FAN= 37.9857 - 19.9566 = 18.0291 sq. ch. ^^^2P4i^^2xiM291 = 6.439 ch. FD 5.60 10. Divide the triangular field ABO, whose sides AB, AC, and BC are 15, 12, and 10 ch., respectively, into three equal parts, by fences EG and Di^ parallel to BC. ABC = \/18.5 X 3.5 X 6.5 X 8.5 == 5U81169 sq. ch. ADF = ^ of 59.81169 = 19.9372 sq. ch. AEG = I of 59.81169 = 39.8744 sq. ch. ABC : AEG :: AB'' : AE'' : : AC'' : Zg'. 59.81169 : 39.8744 : : 15^ : AE^. :. AE= 12.247 ch. : : 122; A^. .',AG= 9.798 ch. ABC : ADF: : Tb" : AD" : : ZC' : AF\ 59.81169 : 19.9372 : : 15^ : A&. .-. AD = 8.659 ch. : : 122 . Jp\ ,-,AF = 6.928 ch. 11. Divide the triangular field ABC, whose sides AB, BC, and AC are 22, 17, and 15 ch., respectively, among three persons, A, B, and C, by fences parallel to the base AB,. so that A may have 3 a., B 4 a., and C the remainder. CAB = V27 X 5 X 10 X 12 = 127.2792 sq. ch. CDG = CAB - ABGD = 127.2792 - 30 = 97.2792 sq. ch. CEF = CAB - ABFE = 127.2792 - 70 = 57.2792 sq. ch. CAB : CDG : : M^ : CG^ : : OT : C&. 127.2792 : 97.2792 : : 17^ : C^. :. CG = 14.862 ch. ^2 : : 152 . cj)\ ... CD = 13.113 ch. CAB:CEF::CB':CF^ : : GZ' : GF. 127.2792 : 57.2792 : : I72 : CF^. .-. CF= 11.404 ch. : : 152 : Oil .■.CE= 10.062 ch. ^- TEACHERS EDITION. 183 Exercise V. 1. Find the difference of level of two places from the following field notes : back-sights, 5.2, 6.8, and 4.0 ; fore-sights, 8.1, 9.5, and 7.9. 8.1 + 9.5 + 7.9 = 25.5 5.2 4-6.8+4 =16 9.5 2. Write the proper numbers in the third and fifth columns of the following table of field notes, and make a profile of the section. Station. +S. II. I. -S. H.S. Remarks. B 1 2 3 t.p. 4 5 6 6.944 20. 19.5 21.3 23.0 22.3 21.431 20.4 21.8 24.1 Bench on post 22 feet north of 0. 26.944 7.4 5.6 3.9 4.6 5.513 4.9 3.5 1.2 3.855 25.286 Surface. 3. Stake of the following notes stands at the lowest point of a pond to be drained into a creek ; stake 10 stands at the edge of the bank, and 10.25 at the bottom of the creek. Make a profile, draw the grade line through and 10.25, and fill out the columns H.O. and C, the former to show the height of grade line above the datum, and the latter, the depths of cut at the several stakes necessary to construct the drain. 184 SURVEYING. Station. B 1 2 3 4 5 6 7 8 9 10 10.25 + S. 6.000 572 H.I. -S. 10.2 5.3 4.6 4.0 6.8 7.090 3.9 2.0 4.9 4.3 4.5 11.8 H.S. 25 H.G. 20,8 20.4 20.0 19.6 19.2 18.8 18.4 18.0 17.6 17.2 16.8 16.7 0.0 5.3 6.4 7.4 5.0 5.1 6.2 8.5 6.0 7.0 7.2 0.0 Remarks. Bench on rock 30 feet west of stake 1. 10 10.25 Date Due ^ DATE DUE . 1 201-5503 PRINTED IN U.S.A. BOSTON COLLEGE 3 9031 01550242