Introducing R, applying logarithms, and reducing, (4) becomes log cos A[log 1p+log (3p—a)+a.c. log b+a.c. log c]. In like manner introduce R and apply logarithms to (5) and (6). By subtracting both members of (1) from 1 and reducing we find ( 1 p − b ) ( p −c) (≥ p − a) (≥ p—c) (9) (6)=(12) tan Cp-a) (p—b). pp-b) Pp-c) 100. Problem. To find the area of a triangle when two sides and their included angle are given. Let k denote the area of the triangle ABC, of which the two sides. b and c and their included angle A are given. (1) 2 kbp. b B a Introducing R, and applying logarithms, we have log (2 k) = log blog clog sin A-10. 101. Examples. C 1. Two sides of a triangle are 345.6 and 485, respectively, and their included angle is 38° 45' 40'; what is the area? Ans. 52468. 2. Two sides of a triangle are 784.25 and 1095.8, respectively, and their included angle is 85° 40′ 20′′; what is the area. Ans. 428470. 102. Problem. To find the area of a triangle when the three sides are given. By the last problem we find (1) kbe sin A, (2) sin A2 sin A cos A. Article 95, (5). (3) sin A= (p-b) p-c). Article 98, (7). bc (4) (5) sin A cos App―a). Article 98, (4). bc 2 √ ≥ p (≥ p − a) (≥ p—b) (≥p—c} bc (6) k = √1p(pa) (≥ p—b) (≥ p-c). 103. Examples. 1. The sides of a triangle are 40, 45, 55, required the area. Ans. 887.412. 2. The sides of a triangle are 467, 845, 756, required the area. Ans. 175508. 104. Problem. Given the perimeter and angles of a triangle, required the sides. b sin B (1) (2) a sin A C sin C a sin A (3) Adding and reducing by Articles 96, (5) and 95, (5), we have Let p (5) a+b+c, and reduce by 96, (7), we have p a 2 cos B cos C sin A psin A cos B cos C Introducing R and applying logarithms, we have log a log p+ log sin A+ = a. c. log cos B+ a. c. log cos C— 10. Similar formulas can be found for b and c. But, after a is found, b and c can be more readily found by article 69. 105. Examples. 150, A = 70°, B = 60°, C=50°, re 2. Given p31234.36, A = 35° 45', B = 45° 28', C98° 47', required a, b, c. 3. Given p Ans. a = 7985, b 9742.5, c 375, A55° 46′ 18′′, B 82° 49′ 08′′, C= 41° 24′ 34′′, required a, b, c. Ans. a 125, b = 150, c = 100. 106. Problem. Given the three sides of a triangle, to find the radius of .. (5) BOC+AOC+AOB= 1 (a + b + c) r = 1 pr. But (6) ABC= √ ≥ p (≥ p − a) (p—b) (p—c). S. N. 9. .. (7) pr = VP(pa) (p—b) (p-c). 1. The three sides of a triangle are 20, 30, 40, respectively, required the radius of the inscribed circle. Ans. 6.455. 2. The three sides of a triangle are 100, 150, 200, respectively, required the radius of the inscribed circle. Ans. 32.275. 108. Problem. Given the three sides of a triangle to find the radius of the circumscribed circle. Let O be the center of the circle, and R the radius. Let OD be perpendicular to b, then The angle 0 the angle B, since each is measured .'. (3) R = 4 k 41 p (pa) (p-b) (p-c) Prove that the formula will be the same if the center is without the triangle. |