But, by formulas (2) and (4), Art. 67, and formula (E"), Art. 66, equation (6) becomes and, by the similar formulas (3) and (5), of Art. 67, equation (7) becomes These last two formulas give the proportions known as the first set of Napier's Analogies; viz., cos (a+b) : cos (a-b) :: cot C : tan (A+B). (10.) sin(a+b) sin (a-b) : :: cot C: 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 supplemental polar triangle, as expressed in Art. 80, we obtain cos (A+B) : cos (A-B) :: tan c: tan (a+b), (12.) In applying logarithms to any of the preceding formulas, they must be made homogeneous in terms of R, as explained in Art. 30. In all the formulas, the letters may be interchanged at pleasure, provided that, when one large letter is substituted for another, the like substitution is made in the corresponding small letters, and the reverse: for example, C may be substituted for A, provided that at the same time c is substituted for a, &c. NOTE. It may be noted that, in formulas (10) and (12), whenever the sign of the first term of the proportion is minus, the sign of the last term must, also, be minus, i. e., whenever (a+b) is greater than 90°, (A+B) must, also, be greater than 90°, and the reverse; and similarly, whenever (a + b) is less than 90°, (A + B) must, also, be less than 90°, and the reverse. SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRI ANGLES. 84. In the solution of oblique-angled triangles six different 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., Gen. S. 1). 1st Case: A < 90°. Let A be the given acute angle, and let a and b be the given sides. Prolong the arcs AC and AB till they meet at A', forming the lune AA'; and from C, draw the arc CB" perpendicular to ABA'. From C, as a pole, and with the A b B B arc a, describe the arc of a small circle BB'. If this circle cuts ABA', in two points between A and 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 and AB'C, both of which will conform to the conditions of the problem. If only one point of intersection lies between A and A', or if the small circle is tangent to ABA', there will be but one solution. 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, This per from which the perpendicular may be found. pendicular will be less than 90°, since it can not exceed the measure of the angle A (B. IX., Gen. S. 2, 1°); 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 than both b and 180° - b. there will be two solutions. 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. 3. When a is greater than p, and at the same time greater than both b and 180° — b; or, when a is less than p, there will be no solution. 101 2d Case: A > 90°. Adopt the same construction as before. the perpendicular will be greater than 90°, because it can not be less than the measure of the angle A (B. IX., Gen. S. 2, 2°): it will, also, be greater than any other arc CA, CB, CA', that can be drawn from C to ABA. course of reasoning By a en A In this case, B B B C tirely analogous to that in the preceding case, we have the following principles : same time 4. When a is less than p, and at the greater than both b and 180° - b, there will be two solu tions. 5. When a is less than p, and intermediate in value between b and 180° will be but one solution. b; or, when a is equal to p. there 6. When a is less than p, and at the same time less than both b and 180°-b; or, when a is 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. Examples. b = 82° 58′ 17′′, and A = 1. Given a = 43° 27′ 36′′, 29° 32′ 29′′, to find B, C, and c. We see that a > p, since p can not exceed A (B. IX., Gen. S. 2, 1°); we see, further, that a is less than both |