PROP. XXIII. Fig. 16. In spherical triangles, whether right angled or oblique angled, the sines of the sides are proportional to the sines of the angles opposite to them. First, Let ABC be a right angled triangle, having a right angle at A; therefore, by Prop. 18, the sine of the hypotenuse BC is to the radius (or the sine of the right angle at A) as the sine of the side AC to the sine of the angle B. And in like manner, the sine of BC is to the sine of the angle A, as the sine of AB to the sine of the angle C; wherefore (11. 5.) the sine of the side AC is to the sine of the angle B, as the sine of AB to the sine of the angle C. Fig. 17, 18. Secondly, Let BCD be an oblique angled triangle, the sine of either of the sides BC will be to the sine of either of the other two CD, as the sine of the angle D opposite to BC is to the sine of the angle B opposite to the side CD. Through the point C, let there be drawn an arch of a great circle CA perpendicular to BD; and in the right angled triangle ABC, (18. of this, the sine of BC is to the radius, as the sine of AC to the sine of the angle B; and in the triangle ADC (by 18. of this) and by inversion, the radius is to the sine of DC as the sine of the angle D to the sine of AC: Therefore, ex æquo perturbato, the sine of BC is to the sine of DC, as the sine of the angle D to the sine of the angle B. In the same manner, it may be proved, that the sine of BC is to the sine of AB, as the sine of A is to the sine of C. Therefore, “ in spherical,” &c. Q. E. D. PROP. XXIV. Fig. 17, 18. In oblique angled spherical triangles, a perpendicular arch being drawn from any of the angles upon the opposite side, the cosines of the angles at the base are proportional to the sines of the segments of the vertical angle. Let BCD be a triangle, and the arch CA perpendicular to the base BD; the cosine of the angle B will be to the cosine of the angle D, as the sine of the angle BCA to the sine of the angle DCA. For by 22. the cosine of the angle B is to the sine of the angle BCA as (the cosine of the side AC is to the radius ; that is, by Prop. 22. as) the cosine of the angle D to the sine of the angle DCA; and, by permutation, the cosine of the angle B is to the cosine of the angle D, as the sine of the angle BCA to the sine of the angle DCA. Q. E. D. PROP. XXV. Fig. 17, 18. The same things remaining, the cosines of the sides BC, CD, are proportional to the cosines of BA, AD, the segments of the base. For by 21, the cosine of BC is to the cosine of BA, as (the cosine of AC to the radius ; that is, by 21, as) the cosine of CD is to the cosine of AD: Wherefore, by permutation, the cosines of the sides BC, CD are proportional to the cosines of the segments of the base BA, AD. "Q. E. D. PROP. XXVI. Fig. 17, 18. The same construction remaining, the sines of the segments of the base BA, AD, are reciprocally proportional to the tangents of B and D, the angles at the base. For by 17, the sine of BA is to the radius, as the tangent of AC to the tangent of the angle B; and by 17, and inversion, the radius is to the sine of AD, as the tangent of D to the tangent of AC: Therefore, ex æquo perturbato, the sine of BA is to the sine of AD, as the tangent of D to the tangent of B. Q. E. D. PROP. XXVII. Fig. 17, 18. The same construction remaining, the cosines of the segments of the vertical angle are reciprocally proportional to the tangents of the sides. For by Prop. 20, the cosine of the angle BCA, is to the radius, as the tangent of CA is to the tangent of BC; and by the same Prop. 20, and by inversion, the radius is to the cosine of the angle DCA, as the tangent of DC to the tangent of CA: Therefore, ex æquo perturbato, the cosine of the angle BCA is to the cosine of the angle DCA, as the tangent of DC is to the tangent of BC. Q. E. D. LEMMA I. Fig. 19, 20. In right angled plane triangles, the hypotenuse is to the radius, as the excess of the hypotenuse above either of the sides to the versed sine of the acute angle adjacent to that side, or as the sum of the hypotenuse, und either of the sides, to the versed sine of the exterior angle of the triangle. Let the triangle ABC have a right angle at B; AC will be to the radius as the excess of AC above AB to the versed sine of the angle A adjacent to AB; or as the sum of AC, AB to the versed sine of the exterior angle CAK. With any radius DE, let a circle be described, and from D the centre let DF be drawn to the circumference, making the angle EDF equal to the angle BAC, and from the point F, let FG be drawn perperidicular to DE: Let AH, AK be made equal to AC, and DL to DE; DG therefore is the cosine of the angle EDF or BAC, and GE its versed sine: And because of the equiangular triangles ACB, DFG, AC or AH is to DF or DE, as AB to DG: Therefore (19. 5.) AC is to the radius DE as BH to GE, the versed sine of the angle EDF or BAC: And since AH is to DE, as AB to DG, (12. 5.) AH or AC will be to the radius DE as KB to LG, the versed sine of the angle LDF or KAC. Q. E. D. PROP. XXVIII. Fig. 21, 22. In any spherical triangle, the rectangle contained by the sines of any two sides, is to the square of the radius, as the excess of the versed sines of the third side or base, and the arch which is the excess of the sides, is to the versed sine of the angle opposite to the base. Let ABC be a spherical triangle, the rectangle contained by the sines of AB, BC will be to the square of the radius, as the excess of the versed sines of the base AC, and of the arch, which is the excess of AB, BC to the versed sine of the angle ABC opposite to the base. |