FORMULE FOR THE AREA OF A RIGHT TRIANGLE. 111. CASE I. Given the hypotenuse and an acute angle. Denoting the area by K, we have by Geometry, Whence, That is, 2K c2 sin A cos Ac2 sin 24 (Art. 74). = 4 K = c2 sin 2 A. = (36) (37) CASE II. Given an angle and its opposite side. CASE III. Given an angle and its adjacent side. CASE IV. Given the hypotenuse and another side. Since a2+b2= c2, we have CASE V. Given the two sides about the right angle. In this case, 2 K= ab. EXAMPLES. (44) 112. 1. Given c= 10.36, B= 75°; find the area. By (37), Whence, log (4K)= 2 log c + log sin 2 B. log c=1.0153; multiply by 2 = 2.0306 2 B=150°; log sin 9.6990 — 10 log (4K)= 1.7296 ... 4 K=53.65, and K= 13.41. Note. To find log sin 150°, take either log cos 60° or log sin 30°. (See page 10 of the explanation of the tables.) Find the areas of the following triangles : 2. Given A= 19° 36', a = 22.17. VIII. GENERAL PROPERTIES OF TRIANGLES. 113. In any triangle, the sides are proportional to the sines of their opposite angles. There will be two cases, according as the angles are all acute (Fig. 1), or one of them obtuse (Fig. 2). In each case let CD be drawn perpendicular to AB. The above results may be expressed more compactly as follows: a b = с = sin C sin A sin B 114. In any triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. Formula (45) of Art. 113 may be put in the form absin A: sin B. Whence, by composition and division, a+b: a − b = sin A+ sin B : sin A — sin B, 115. Since A + B = 180° — C', we have tan (A+B) = tan (90° C) cot C (Art. 14). = Thus formula (48) may be put in the form 116. In any triangle, the square of any side is equal to the sum of the squares of the other two sides, minus twice their product into the cosine of their included angle. CASE I. When the included angle A is acute. There will be two cases, according as the remaining angles are both acute (Fig. 1), or one of them obtuse (Fig. 2). In each case let CD be drawn perpendicular to AB. But, Also, Therefore, BD2 + CD2 = AD2 + CD2 + c2 − 2 c × AD. BD2 + CD2 = a2, and AD2 + CD2 = b2. |