. A treatise on plane and spherical trigonometry, and its applications to astronomy and geodesy, with numerous examples . = colog sin a = log sin A = log sin B = .\ B = 64°264, B = 115° 33 56. *(« + &) = 45° 0 0. l(a-6) = - - 2° 15 0. i(A + B) = 60° 28 2. i(A-B)=- - 3° 58 2. J(A + B) = 86° 158. i(A-B) = - - 29° 31 58. Since both values of B are such that A — B, A — B, anda — b, are all negative, there are two solutions, by the aboveRule. (1) When B = 64° 26 4. log sin £ (a - 6) = -colog sin \ (a+b) = cot i(A - B) = log tan-
. A treatise on plane and spherical trigonometry, and its applications to astronomy and geodesy, with numerous examples . = colog sin a = log sin A = log sin B = .\ B = 64°264, B = 115° 33 56. *(« + &) = 45° 0 0. l(a-6) = - - 2° 15 0. i(A + B) = 60° 28 2. i(A-B)=- - 3° 58 2. J(A + B) = 86° 158. i(A-B) = - - 29° 31 58. Since both values of B are such that A — B, A — B, anda — b, are all negative, there are two solutions, by the aboveRule. (1) When B = 64° 26 4. log sin £ (a - 6) = -colog sin \ (a+b) = cot i(A - B) = log tan-= C = 38° 40 48. C = 77° 21 36. log sin i(A + B) =$(A-B)= log tan |(a - b) = log tan-= ° 2 c 2: :26°1740. c =53° 35 20. CASE III. 311 (2) WhenB = 115°3356. log sin £(«-&) = sin | (a+6) = cot J(A-B) = log tan--= 5 2 C 5°3550i. •. c=n°ir40£. logsin|(A + B) =|(A-B) = logtan|(a- 6) = log tan-= 6 2 .-. - = 4°3247i. 2 4 .-. c=9° 534|.. ^4ns. B = 64° 26 4, C =77° 2136, c =53° 3520;B= 115° 3556, C = H°1140J-, c= 9° 534£. Otherwise thus: Let fall theperpendicular CD; denote ADby m, the angle ACD by , andCD by p. Then we have cos A = tan m cot b; .% tan m = cos A tan b (1) cos b = cot A cot = cos b tan A Again, cos a = cos (c — m) cosp; cos 6 = cos m cos p. .-. cos (c — m) = cos a cos m -r- & . Also, cos (C — cj>) = cot a tanp; cos <£ = cot b tanp. •% cos (C — ) = cot a tan 6 cos <£ . sin b (2) (3) Lastly, sinB: sin A sma (4) (5) The required parts are given by (1), (2), (3), (4), (5). Ex. 2. Given a = 73°4938, 6=120°5335, A = 88°5242:find B, C, a Ans. B = 116° 4448, C = 116° 4448, c = 120° 5535. 312 SPHERICAL TRIGONOMETRY. 212. Case IV. — Given two angles, A, B, and the side aopposite one of them; to find b, c, C. This case reduces, by aid of the polar triangle, to thepre
Size: 2031px × 1231px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1900, booksubjecttrigono, bookyear1902
