and extract the square root of both members, we have, That is, the cosine of one-half of either angle of a spherical triangle, is equal to the square root of the sine of one-half of the sum of the three sides, into the sine of one-half this sum minus the side opposite the angle, divided by the rectangle of the sines of the adjacent sides. If we subtract Equation (1), of the preceding article, member by member, from the number 1, and recollect that, A = 180° - a', b = 180° — B', c = 180° — C', Is = 270° — †(A'+B'+C'), is is — a = 90° — 1 (B' + C'' — A'). Substituting these values in (3), Art. 81, and reducing by the aid of the formulas in Table III., Art. 63, we find, cos (A'+B'+C') cos (B'+ C' —A') sin a' = Placing sin B' sin C" ¿(A'+B'+ C') = 4S; whence, }(B'+C'-A') = {S— A'. In a similar way, we may deduce from (4), Art. 81. 83. From Equation (1), Art. 80, we have, sin A cos A+ cos B cos C = sin B sin C cos a = sin C sin b cos a sin a (1.) since, from Proportion (1), Art. 78, we have, Also, from Equation (2), Art. 80, we have, Adding (1) and (2), and dividing by sin C, we obtain, 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 ("), and, by the similar Formulas (3) and (5), of Art. 67, Equation (7) becomes, tan (AB) sin (a — b) (9.) These last two formulas give the proportions known as the first set of Napier's Analogies. cos (a+b): cos(a-b) cot: tan(A+B). (10.) sin (a+b) : sin (a−b) :: cot10: 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 triangle, as expressed in Art. 80, we obtain, cos (A+B) cos(A-B) tanje: tan (a+b). (12.) sin (A+B) sin (A-B) :: tanje: : tan (a−b). (13.) 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, as explained in Art. 30. SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 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., P. XIX., Gen. Scholium). First Case. Let A be the given 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 A b B B' from C, draw the arc CB' perpendicular to ABA'. 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: |