36. Problem. To two legs are known. The legs given. solve a right triangle, when the [B. p. 40.] Solution. Given (fig. 4.) the legs a and b, to solve the tri angle. First. The angles are obtained from (4) log. tang. A log. cotan. B = log. a+ (ar. co.) log. b. Secondly. To find the hypothenuse, we have by (28) h = √(a2 + b2). (31) Thirdly. An easier way of finding the hypothenuse is to make use of (23) or (25) ha cosec. A a sec. B ; (32) or, by logarithms, log. h = log. a log. cosec. A = log. a+ log. sec. B. 37. EXAMPLES. 1. Given the hypothenuse of a right triangle equal to 49.58, and one of the acute angles equal to 54° 44′; to solve the triangle. Solution. The other angle = 90° 54° 44' 35° 16'. Then making h=49.53, and A = 54° 44′; we have, by (21) and (22), 2. Given the hypothenuse of a right triangle equal to 54.571, and one of the legs equal to 23.479; to solve the triangle. Solution. Making h = 54.571, a = 23.479; we have, by (27), 3. Given the two legs of a right triangle equal to 44.375, and 22.165; to solve the triangle. * To avoid negative characteristics, the logarithms are retained as in the tables, according to the usual practice with the logarithms of decimals, as in B., p. 29. Examples of right triangles. Solution. Making a = 44.375, b=22.165; we have, 4. Given the hypothenuse of a right triangle equal to 37.364, and one of the acute angles equal to 12° 30'; to solve the triangle. Ans. The other angle 77° 30′ 8.087 5. Given one of the legs of a right triangle equal to 14.548, and the opposite angle equal to 54° 24'; to solve the triangle. 6. Given one of the legs of a right triangle equal to 11.111, and the adjacent angle equal to 11° 11'; to solve the Examples of right triangles. 7. Given the hypothenuse of a right triangle equal to 100, and one of the legs equal to 1; to solve the triangle. Ans. The other leg = 99.995 8. Given the two legs of a right triangle equal to 8.148, and 10.864; to solve the triangle. Sine of the sum of two angles. CHAPTER IV. GENERAL FORMULAS. 38. The solution of oblique triangles requires the introduction of several trigonometrical formulas, which it is convenient to bring together and investigate all at once. 39. Problem. To find the sine of the sum of two angles. Solution. Let the two angles be BAC and B'AC (fig. 9), represented by the letters M and N. At any point C, in the line AC, erect the perpendicular BB'. From B let fall on AB' the perpendicular BP. Then represent the |