cos b cos c + sin b sin c = cos (bc). Article 91, (d). c) cos (b 2 sin2 4= Substituting and dividing by 2, we have sin2 A (1) sin 4 (2) sin B sin (a+c-b) sin (a+b-c) But (a+c · b) = 38 -b and (a + b—c) = 18-c. (3) sin C Substituting in the value of sin2 A, and in the similar values for sin2 B and sin2 C, and extracting the square root, we have = COS a (a+c—b) sin (a + b−c). = 141. Proposition VII. The co-sine of one-half of any side of a spherical triangle is equal to the square root of the quotient obtained by dividing the co-sine of one-half the sum of the angles minus one adjacent angle into the co-sine of half the sum minus the other adjacent angle, by the product of the sines of the adjacent angles. Taking the formulas of the last article, passing to the polar triangle, making SA'B'C', substituting, reducing, and omitting the accents, we have (1) tan A (2) tan B (3) tan C 142. Proposition VIII. The tangent of one-half of any angle of a spherical triangle is equal to the square root of the quotient obtained by dividing the sine of one-half the sum of the sides minus one adjacent side into the sine of one-half the sum minus the other adjacent side, by the sine of one-half the sum of the sides into the sine of one-half the sum minus the opposite side. B) cos (S-C'). sin B sin C Dividing (1), (2), (3), article 140, respectively, by (1), (2), (3), article 138, we have cos (S cos (SA) cos (3S — C) cos (SA) cos (4S — B) 143. Proposition IX. The tangent of one-half of any side of a spherical triangle is equal to the square root of the quotient obtained by dividing minus the co-sine of one-half the sum of the angles into the co-sine of one-half the sum minus the angle opposite the side, by the co-sine of one-half the sum of the angles minus one adjacent angle into the co-sine of one-half the sum minus the other adjacent angle. Dividing (1), (2), (3), article 139, respectively, by (1), (2), (3), article 141, we have (1) tan a = tan cos S cos (S — A) cos (SB) cos (SC) Atan B cos S cos (S — B) cos (SA) cos (SC) The reciprocals of (1), (2), (3), articles 142, 143, will give formulas for co-tangents, which may be written and expressed in words. 144. Napier's Analogies. Dividing (1), article 142, by (2), we have tan A sin (sb) tan Atan B tan Atan B cos S cos (S — C) cos (SA) cos (3S — B) This, as a proportion taken by composition and division, gives Multiplying both terms of the second member ty cos A cos B, Reducing the second member by articles 89, (a), and 91, (c), tan Atan B sin (A+B) sin (38- b) sin (s-a) sin (sb) sin (s—a) B+ cos B — cos = B B tan c tan (a - b) sin (A+B) tan c ... (1) sin (A+B) : sin †(A—B) :: tan c: tan (a — b). By division and composition, we have 1 tan A tan A sin B A sin B The reciprocal of (1) X (2), article 142, gives sin s 1 sin (8 Art. 96, (11). sin s sin (8-c) sin 8 sin (sc) Reducing both members as before, we have cos (A+B) tan c cos (AB) tan (a + b) (2) cos (A+B) : cos (A—B) :: tan c: tan (a+b). Passing from (1) and (2) to the polar triangle, we have (3) sin (a+b) : sin 1(a —b) :: cot (4) cos (a+b): cos (ab) cot C: tan ≥(A—B). C: tan (A+B). 145. Proposition. In a right-angled spherical triangle, as b increases from 0° to 90°, from 90° to 180', from 180° to 270°, and from 270° to 360°, if p < 90°, h increases from p to 90°, from 90° to 180' p, decreases from 180' p to 90', and from 90° to p; if p > 90°, h decreases from p to 90°, from 90° to 180'-p, increases from 180°-p to 90°, and from 90' to p; if p= 90°, h=90° for all values of b. cos h 1. p < 90°; ... cos p is positive. cos b cos p. h If b 0, cos b 1; therefore, cos h = cos p; .. = p. As b increases from 0° to 90°, cos b is positive, and diminishes from 1 to 0; ... cos h is positive, and diminishes from cos p to 0; ... h increases from p to 90°. H p B 180° p h P As b increases from 90° to 180°, cos b is negative, and increases numerically from 0 to 1; 1; .. cos h is negative, and increases numerically from 0 to cos p; .. h increases from 90° to 180°- p, and the triangle becomes the lune HH'. н' As b increases from 180° to 270°, cos b is negative, and decreases numerically from 1 to 0; .. cos h is negative, and decreases numerically from cos p to 0; ... h decreases from 180° - p to 90°. As b increases from 270° to 360°, cos b is positive, and increases from 0 to 1; ... cos h is positive, and increases from 0 to cos p; .. h decreases from 90° to p, and the triangle becomes the hemisphere. |