We take the smaller value of b, for the reason that A, being greater than B, requires that a should be greater than b. Applying logarithms to Proportion (12), Art. 83, we have, log tan c log cos (A + B) + log tan (a + b) we have, sin ɑ : sin c :: sin A: sin C, log sin c + log sin A+ (a. c.) log sin a 10; The smaller value of C is taken, for the same reason as before. 2. Given A = 50° 12', B = 58° 08′, and a b, c, and C. a = 62°42′, CASE III. Given two sides and their included angle. 87. The remaining angles are found by means of Napier's Analogies, and the remaining side, as in the preceding cases. log tan (A-B) = log sin (a - b) + log cot C - 10; we have, ž (a — b) = 26° 12′ 20′′, and, lọg cos (a — b) log cot C (a. c.) log cos (a + b) The greater angle is equal to the half sum plus the half difference, and the less is equal to the half sum minus the half difference. Hence, we have, A 27° 31' 44", and B = 5° 17' 58". Applying logarithms to the Proportion (13), Art. 83, we have, log tan c = log sin (A + B) + log tan (a - b) + (a. c.) log sin (A — B) — 10; Ans. A = 120° 59′ 47′′, B = 33° 45′ 03′′, c = 43° 37′ 38′′. = 84° 14' 29", b = 44° 13′ 45′′, and to find A and B. Ans. A = 130° 05′ 22′′, B = 32° 26' 06". CASE IV. Given two angles and their included side. 88. The solution of this case is entirely analogous to that of Case III. Applying logarithms to Proportions (12) and (13), Art. 83, and to Proportion (11), Art. 83, we have, log tan (a+b) log tan (a - b) log cos (A – B) + log tan c + (a. c.) log cos †(A + B) — 10; log sin (AB) + log tan c +(a. c.) log sin (A + B) — 10; log cot C = log sin (a + b) + log tan (A – B) + (a. c.) log sin †(a — b) — 10; The application of these formulas are sufficient for the solution of all cases. Ans. C 121° 36′ 12′′, a = 40° 0′ 10′′, b = 50° 10′ 30′′. and b. CASE V. Given the three sides, to find the remaining parts. 89. The angles may be found by means of Formula (3), Art. 81; or, one angle being fonnd by that formula, the other two may be found by means of Napier's Analogies. EXAMPLES. 1. Given a = 74° 23', b = 35° 46′ 14′′, and c = 100° 39', to find A, B, and C. Applying logarithms to Formula (3), Art. 81, we have, log cos 4 = 10 + [log sin s + log sin (†s — a) A 10+ is + (a. c.) log sin b + (a. c.) log sin c - 20]; [log sin s + log sin (†s — a) + (a. c.) log sin b + (a. c.) log sin c], for Using the same formula as before, and substituting B A, b for A2 and a for b. b, and recollecting that 1 s − b = 69° 37′ 53′′, we have, Using the same formula, substituting C for and α for C, recollecting that sc have, A, с for A, 4° 45′ 07′′, we |
