23. Define a quadrantal triangle, and show how its solution may be reduced to that of the right triangle. 24. Solve the quadrantal triangle whose sides are: a = 174° 12′ 49.1", b 94° 8' 20", c= 90°. 25. Solve the quadrantal triangle in which c=90°, A=110° 47' 50", B=135° 35' 34.5". 26. Given in a spherical triangle A, C, and c each equal to 90°; solve the triangle. 27. Given 460°, C=90°, and c-90°; solve the triangle. 28. Given in a right spherical triangle, A = 42° 24' 9", B=9° 4' 11"; solve the triangle. 29. In a right spherical triangle, given a = 119° 11', B=126° 54′; solve the triangle. 30. In a right spherical triangle, given c=50°, b=44°18′39′′; solve the triangle. 31. In a right spherical triangle, given A=156° 20' 30", a=65° 15' 45"; solve the triangle. 32. If the legs a and b of a right spherical triangle are equal, prove that cos a = cot A √cos c. 33. In a right spherical triangle prove that cos2 AX sin2c=sin (c-a) sin (e+a). 34. In a right spherical triangle prove that tan a cos csin b cot B. 35. In a right spherical triangle prove that sin2 A = cos2 B+ sin2 a sin2 B. 36. In a right spherical triangle prove that sin (b+c) 2 cos2 A cos b sin c. 2 37. In a right spherical triangle prove that sin (c—b) = 2 sin2 A cos b sin c. 38. If, in a right spherical triangle, p denotes the arc of the great circle passing through the vertex of the right angle and perpendicular to the hypotenuse, m and n the segments of the hypotenuse made by this arc adjacent to the legs a and b, prove that (i.) tan2 a = tan c tan m, (ii.) sin2 p=tan m tan n. § 52. SOLUTION OF THE ISOSCELES SPHERICAL TRIANGLE. If an arc of a great circle is passed through the vertex of an isosceles spherical triangle and the middle point of its base, the triangle will be divided into two symmetrical right spherical triangles. In this way the solution of an isosceles spherical triangle may be reduced to that of a right spherical triangle. In a similar manner the solution of a regular spherical polygon* may be reduced to that of a right spherical triangle. Arcs of great circles, passed through the centre of the polygon and its vertices, divide it into a series of equal isosceles triangles; and each one of these may be divided into two equal right triangles. EXERCISE XXXIII. 1. In an isosceles spherical triangle, given the base b and the side a; find A the angle at the base, B the angle at the vertex, and h the altitude. 2. In an equilateral spherical triangle, given the side a; find the angle A. 3. Given the side a of a regular spherical polygon of n sides; find the angle 4 of the polygon, the distance R from the centre of the polygon to one of its vertices, and the distance r from the centre to the middle point of one of its sides. 4. Compute the dihedral angles made by the faces of the five regular polyhedrons. 5. A spherical square is a regular spherical quadrilateral. Find the angle 4 of the square, having given the side a. * A regular spherical polygon is the polygon formed by the intersections of the spherical surface by the faces of a regular pyramid whose vertex is at the centre of the sphere. CHAPTER VIII. THE OBLIQUE SPHERICAL TRIANGLE. § 53. FUNDAMENTAL FORMULAS. LET ABC (Fig. 44) be an oblique spherical triangle, a, b, c its three sides, A, B, C the angles opposite to them, respectively. sin a sin b: sin csin A: sin B: sin C. That is, the sines of the sides of a spherical triangle are proportional to the sines of the opposite angles. In Fig. 44 the arc of the great circle CD cuts the side AB within the triangle. In case it cuts AB produced without the triangle, sin (180°— A), sin (180°—B), or sin (180° — C'), would be employed in the above proof instead of sin A, sin B, or sin C. These sines, however, are equal to sin A, sin B, and sin C, respectively, so that the Formulas [44] hold true in all cases. 2. In the right triangle BDC, by $ 49 [38], or (§ 28) Now, whence and cos a = cos p cos n = cos p cos (c—m), cos a = cos p cos e cos m + cos p sin c sin m. cos p cos m = cos b; [38] [40] Substituting these values of cos p cosm and cos p sin m in the value of cos a, we obtain cos a = cos b cos c + sin b sin c cos A and similarly, cos b = cos a cos c + sin a sin c cos B cos c = cos a cos b + sin a sin b cos C [45] [41] [43] Substituting these values of cos p sin y and cos p cos y in the value of cos A, we obtain cos A and similarly, cos B cos B cos C+ sin B sin C cos a [46] cos C cos A cos B + sin A sin B cos c Formulas [45] and [46] are also universally true; for the same equations are obtained when the arc CD cuts the side AB without the triangle. EXERCISE XXXIV. 1. What do Formulas [44] become if A=90°? if B=90°? if C90° ? if a = 90° ? if A=B=90°? if a = b = 90° ? 2. What does the first of [45] become if A=0°? if A=90°? if A180° ? 3. From Formulas [45] deduce Formulas [46], by means of the relations between polar triangles (§ 48). § 54. FORMULAS FOR THE HALF ANGLES AND SIDES. Hence, by § 30 [16] and [17], and § 31 [23], 2 sin2 A=sin(a+b−c) sin (a−b+c) cse b csc c, Now let whence, 1 (b + c − a)=s—α, (a+b−c)=s—c. |