Making these changes, and reducing, we have, Given the hypothenuse and one of the acute angles, to find the remaining parts. 38. The other acute angle may be found by subtracting the given one from 90° (Art. 23). The sides about the right angle may be found by Formulas (7) and (8). a B Applying logarithms to Formula (7), remembering that the logarithm of R is equal to 10, we have, Applying logarithms to Formula (8), we have, Ans. B 42° 56′ 50′′, b = 510.31, and c = = 548.255 a = 439, and B = 27° 38′ 50′′, to find 3. Given a = 125.7 yds., and B = 75° 12′, to find the other parts. Ans. C′ = 14° 48′, b = 121.53 yds., and c = 32.11 yds. 4. Given a = 325 ft., and C 27° 34', to find the other parts. Ans. B = 62° 26', c = 150.4 ft., and b = 288.1 ft. CASE II. Given one of the sides about the right angle and one of the acute angles, to find the remaining parts. 39. The other acute angle may be found by subtracting the given one from 90°. The hypothenuse may be found by Formula (7), and the unknown side about the right angle, by Formula (8). Applying logarithms to Formula (7), we have, log clog a log sin C-10; whence, Applying logarithms to Formula (8), we have, 2. Given c = 358, and B = 28° 47′, to find C, a, 3. Given 6 = 152.67 yds., and C find the other parts. Ans. B = 39° 41′ 28′′, 4. Given c = 379.628, B, α, and b. c = 183.95, and a = 239.05. c Ans. B = 50° 33′ 44′′, ・ α = 597.613, and b = 461.55. CASE III. Given the two sides about the right angle, to find the re maining parts. 40. The angle at the base may be found by Formula (12), and the solution may be completed as in Case II. EXAMPLES. 1. Given b = 26, and c = 15, to find C, B, and a. OPERATION. Applying logarithms to Formula (12), we have, As in Case II., log a log c+ (a. c.) log sin C; ·log a (a. c.) log sin C log a⚫ = • 1.176091 (15) 1.477362 .. a = 30.017. Ans. C = 29° 58′ 54′′, B = 60° 01′ 06′′, and a = 30.017. 2. Given 6 = 1052 yds., and c = 347.21 yds., to find B, C, and a. B = 71° 44′ 05′′, C′ = 18° 15′ 55′′, and a = 1108.05 yds. B = 45° 33′ 42′′, C′ = 44° 26′ 18′′, and a = 144.256. |