sin C, we obtain, Adding (1) and (2), and dividing by sin C, taken first by composition, and then by division, gives, Dividing (4) and (5), in succession, by (3), we obtain, But, by Formulas (2) and (4), Art. 67, and Formula (E"), and, by the similar Formulas (3) and (5), of Art. 67, cos (a+b): cos(a-b) :: cot C: tan (A+B). (10.) sin(a+b) sin (a-b): cot C: cot tan†(A−B). (11.) If in these we substitute the values of a, b, C, A, and B, in terms of the corresponding parts of the polar triangie, as expressed in Art. 80, we obtain, cos (A+B) cos(A-B) :: tanic: tan (a+b). (12.) sin †(A+B) : sin †(A–B) :: tan fc: tan (a−b). (18.) the second set of Napier's Analogies. In applying logarithms to any of the preceding formulas, they must be made homogeneous, in terms of R, 88 explained in Art. 30. SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 84. In the solution of oblique-angled triangles six differ. ent cases may arise: viz., there may be given, I. Two sides and an angle opposite one of them. III. Two sides and their included angle. IV. Two angles and their included side. V. The three sides. VI. The three angles. CASE I. Given two sides and an angle opposite one of them. 85. The solution, in this case, is commenced by finding the angle opposite the second given side, for which purpose Formula (1), Art. 78, is employed. As this angle is found by means of its sine, and because the same sine corresponds to two different arcs, there would seem to be two different solutions. To ascertain when there are two solutions, when one solution, and when no solution at all, it becomes necessary to examine the relations which may exist between the given parts. Two cases may arise, viz., the given angle may be acute, or it may be obtuse. We shall consider each case separately (B. IX., P. XIX., Gen. Scholium). First Case. Let A be the given angle, and let a and be the given sides. Prolong the arcs AC and AB till they meet at A', forming the lune AA'; and A B B' from C, draw the arc CB' perpendicular to ABA'. A' From C, as a pole, and with the arc a, describe the arc of a small circle BB. between A and If this circle cuts ABA', in two points A', there will be two solutions; for if C be joined with each point of intersection by the arc of a great circle, we shall have two triangles ABC, both of which will conform to the conditions of the problem. If there is no point of intersection, or if there are points of intersection which do not lie between A and A', there will be no solution. From Formula (2), Art. 72, we have, sin CB' sin b sin A, from which the perpendicular, which will be less than 90°, will be found. Denote its value by p. By inspection of the figure, we find the following relations: 1. When a is greater than p, and at the same time less b, there will be two solutions. than both b and 180° 2. When a is greater than p, and intermediate in value between b and 180° — b; or, when a is equal to p, there will be but one solution. If a = b, and is also less than 180° b, one of the points of intersection will be at A, and there will be but one solution. or, when 8 is 3. When a is greater than P, and at the same time greater than both b and 180° - b; less than P, there will be no solution. Second Case. Adopt the same construction as before. In this case, the perpendicular will be greater than 90°, and greater also than any other arc CA, CB, CA', that can be drawn from C A B B A' to ABA'. By a course of reasoning entirely analogous to that in the preceding case, we have the following principles: 4. When a is less than P, and at the same time greater than both b and 180° - b, there will be two solutions. 5. When a is less than P, and intermediate in value between b and 180° to P, there will be but one solution. b; or, when a is equal p, and at the or, same time when a is 6. When a is less than less than both b and 180° - b; greater than p. there will be no solution. Having found the angle or angles opposite the second side, the solution may be completed by means of Napier's Analogies. We see at a glance, that a > P, since p cannot exceed A; α we see further, that b is less than both and 180°-b; hence, from the first condition there will be two solutions. Applying logarithms to Formula (1), Art. 78, we have, (a. c.) log sin a+ log sin blog sin A- 10 log sin R = (a. c.) log sin a B = 45° 21′ 01′′, and B = 134° 38′ 59′′ From the first of Napier's Analogies (10), Art. 83, we find, (a. c.) log cos į (a−b) + log cos į (a+b) + log tan } (A + B) −10 =log cot C. |