188. The student will prove without difficulty: From (1), From (3), From (7), tan From (8), A+ B 2 tan 2 = sin A sin B COS cos B COS sin cos B sin tan A COS ..cos A: COS sin = = 2 2 Formulæ (20), (21), (22), (23) are known as Napier's Analogies. = = 2 a+b 2 A B sin tan B = cos c = cos a cos b. sin a sin c = 2 a+b 2 189. In the special case that one of the angles, C, is equal 90°, we derive from the foregoing formulæ the following: = b 2 A+B 2 cos A = cos a sin B; } sin b C cot = cos b sin A. tan b C cot 2' tan a tan c tan 2' tan a sin b tan b tan cos b = .. COS C = 190. DEFINITION. Two angles are said to be of the "same affection," if both are greater or both are less than 90°. They are said to be of opposite affection if one is greater and the other less than 90°. Formula (24) shows that if a and b are of the same affection, c< 90°; and if a and b are of opposite affection, c> 90°, C being a right angle. For, if a and b are both of the same affection, cos b is positive, .. cos c is positive, .. c < 90°. But if they are of opposite affection, cos a cos b is negative, .. cos c is negative; .. c> 90°. cos a • =cot Acot B. EXERCISE 1. Show that in a right-angled triangle an angle and the side opposite are of the same affection. and EXERCISE 2. Show that a side and the hypotenuse of a right-angled triangle are of the same or opposite affection, according as the included angle is less than or greater than 90°. EXERCISE 3. In any triangle, A+ B a+b 2 EXERCISE 4. If C is a right angle, show that (29) EXERCISE 5. If C is a right angle, show that 2 cos c = cos (a + b) + cos (a − b). are of the same affection. tan (c + a) tan (c — a) = tan2 § b. CHAPTER XIX SOLUTION OF SPHERICAL TRIANGLES 191. According to propositions proved in Geometry we are able to construct a spherical triangle when any three of its parts are given. Accordingly, we propose the problem, to show that if given any three parts of a spherical triangle, we are able to calculate the remaining parts. This problem is known as the solution of spherical triangles. I. Three sides. II. Two sides and included angle. III. Two sides and the angle opposite one of them. IV. Two angles and a side opposite one of them. V. One side and two angles adjacent to it. VI. Three angles. 192. The Right Triangle. Before considering the solution of the general triangle, we shall discuss the solution of the right triangle. We have the following cases: Let C be a right angle. Given: II. a, b. a, c. c, b. To calculate: c, A, B. b, A, B. a, A, B. 193. If any of the parts, when calculated, is found in terms of its sine, then the solution in general is ambiguous; since there are two angles less than 180° which have the same sine. See Case III., below. If, however, all the parts are found in terms of other ratios than their sines, or cosecants, the solution is unique. 194. DISCUSSION OF THE DIFFERENT CASES. By referring to Art. 189, the student will verify the following equations. In many instances he will find other sets of equations which will solve the problems. Given a, A; required c, b, B. sin b sin c = ; This case presents an apparent ambiguity. But by Art. 190, a and A are of the same affection. .. the solution is unique. Case III. COS C = sin a sin B = tan a tan c tan a tan A cos A cos a B, b, and c are given by their sines; hence there is, apparently at least, a series of ambiguities. (a) If sin a sin A, then a = A. = ... sin = = 1, sin b = 1, sin B = 1. Hence in this case the solution is unique. (b) There remain yet two possibilities for c; viz. : I. c and a of like affection; then b< 90°, and B < 90°. (Art. 190, Exercise 1.) cos a cos b cos c sin b sin c (Art. 194.) |