log c = log b log blog sin C+ (a. c.) log sin B 10; log sin C Ans. B = 45° 13′ 55′′, C = 112° 09′ 05′′, and c = 281.785. Ans. B' 134° 46′ 05′′, C = 22° 36′ 55′′, and c = 116.993. 2. Given A = 32°, a = 40, and b = 50, to find Ans. B 8° 56′ 05′′, C = 152° 11′ 42′′, c = 39.611 yds. Given two sides and their included angle, to find the remaining parts. circle meeting AB in I, and the prolongation of AB in E. Draw CI and EC, and through I draw IH parallel to EC. Because ECI is an angle inscribed in a semicircle, it is a right angle (B. III., P. XVIII., C. 2); and consequently, both CE and I are perpendicular to CI. The angle EAU being external to the triangle ABC, is equal to the sum of the opposite interior angles, that is, equal to C plus B; the angle EAC being also external to the isos- · celes triangle AIC, it is equal to twice the angle AIC : hence, twice the angle AIC is equal to C plus B, or, The angle ICB is equal to AIC diminished by the angle IBC (B. I., P. XXV., C. 6); that is, ICH = (C+B) - B = (C — B). From the two right-angled triangles ICE and ICH, have (Formula 3, Art. 37), EC IC tan † (C + B), and IH = IC tan ( C · we · B); hence, from the preceding equations, we have, after omitting the equal factor IC (B. II., P. VII.), EC: IH :: tan (C+B) : tan (C – B). The triangles ECB and IHB being similar, their homologous sides are proportional; and because EB is equal to AB+ AC, and IB to AB AC, we shall have the proportion, : ABAC. EC : IH :: AB + AC: AB Combining the preceding proportions, and substituting for AB and AC their representatives c and b, we have, In any plane triangle, the sum of the sides including either angle, is to their difference, as the tangent of half the sum of the two other angles, is to the tangent of half their difference. The half sum of the angles may be found by subtracting the given angle from 180°, and dividing the remainder by 2 the half difference may be found by means of the principle just demonstrated. Knowing the half sum and the half difference, the greater angle is found by adding the half difference to the half sum, and the less angle is found by subtracting the half difference from the half sum. solution is completed as in Case I. Then the EXAMPLES. 1. Given c = 540, b = 450, and A = 80°, to find B, C, and a. OPERATION. = c+ b = 990; c − b = 90 ; † (C+B) = †(180° — 80o) — 50°. Applying logarithms to Formula (14), we have, = log (c-b) + log tan (C + B) + (a. c.) log (c + b) log sin (CB) 10; log a = log clog sin A+ (a. c.) log sin C 2. Given B, c = 1686 yds., b = 960 yds., and A = 128° 04′, C, and α. C, 18° 21′ 21′′, C = 33° 34′ 39", a = 2400 yds. to find B, Ans. B 3. Given α = 18.739 yds., b = 7.642 yds., and C = 45° 18′ 28", to find A, B, Ans. A 112° 34' 13", B = 22° 07′ 19", c = 14.426 yds 87° 03′ 48′′, to find A, b = 289.3 yds., and c= 534.66 yds. C = α = 16.9584 ft., 5. Given b = 11.9613 ft., and C = 60° 43′ 36", to find A, B, and C. Ans. A 76° 04' 10", B = 43° 12′ 14′′, c = 15.22 ft. 6. Given a = 3754, b = 3277.628, and C 57° 53' 17", to find A, B, and C. Ans. A 68° 02′ 25′′, B = 54° 04′ 18", c = 3428.512. CASE IV. Given the three sides of a triangle, to find the remaining parts.* 46. Let ABC represent any plane triangle, of which BC is the longest side. Draw AD perpendicular to the base, dividing it into two segments BD and BD. The angles may be found by Formula 109, and 110, Mensuration. C D or (B), Lemma. Pages |