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 perpendi cular 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, and 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 perpendicular 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. Let D be the centre of the sphere, and let AD, BD, CD, be joined, and let the sines AE, CF, CG of the arches AB, BC, AC be drawn; let the side BC be greater than BA, and let BH be made equal to BC; AH will therefore be the excess of the sides BC, BA; let HK be drawn perpendicular to AD, and since AG is the versed sine of the base AC, and AK the versed sine of the arch AH, KG is the excess of the versed sines of the base AC, and of the arch AH, which is the excess of the sides BC, BA: Let GL likewise be drawn parallel to KH, and let it meet FH in L, let CL, DH be joined, and let AD, FH meet each other in M. Since therefore in the triangles CDF, HDF, DC, DH are equal, DF is common, and the angle FDC equal to the angle FDH, because of the equal arches BC, BH, the base HF will be equal to the base FC, and the angle HFD equal to the right angle CFD: The straight line DF therefore (4. 11.) is at right angles to the plane CFH: Wherefore the plane CFH is at right angles to the plane BDH, which passes through DF (18. 11.) In like manner, since DG is at right angles to both GC and GL, DG will be perpendicular to the plane CGL; therefore the plane CGL is at right angles to the plane BDH, which passes through DG: And it was shown, that the plane CFH or CFL was perpendicular to the same plane BDH; therefore the common section of the planes CFL, CGL, viz. the straight line CL, is perpendicular to the plane BDA (19. 11.) and therefore CLF is a right angle: In the triangle CFL, having the right angle CLF, by Lemma I. CF is to the radius as LH, the excess, viz. of CF or FH above FL, is to the versed sine of the angle CFL; but the angle CFL is the inclination of the planes BCD, BAD, since FC, FL are drawn in them at right angles to the common section BF: The spherical angle ABC is therefore the same with the angle CFL; and therefore CF is to the radius as LH to the versed sine of the spherical angle ABC; and since the triangle AED is equiangular (to the triangle MFD, and therefore) to the triangle MGL, AE will be to the radius of the sphere AD (as MG to ML; that is, because of the parellels as) GK to LH: the ratio therefore which is compounded of the ratios of AE to the radius, and of CF to the same radius; that is, (23. 6.) the ratio of the rectangle contained by AE, CF to the square of the radius, is the same with the ratio compounded of the ratio of GK to LH, and the ratio of LH to the versed sine of the angle ABC; that is, the same with the ratio of GK to the versed sine of the angle ABC; therefore the rectangle contained by AE, CF, the sines of the sides AB, BC, is to the square of the radius as GK, the excess of the versed sines AG, AK, of the base AC, and the arch AH, which is the excess of the sides to the versed sine of the angle ABC, opposite to the base AC. Q. E. D. *COR. Hence, 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 sum of the sides is to the coversed sine of the angle opposite to the base. LEMMA II. FIG. 23. The rectangle contained by half the radius, and the excess of the versed sines of two arches, is equal to the rectangle contained by the sines of half the sum, and half the difference of the same arches. Let AB, AC be any two arches, and let AD be made equal to AC the less: the arch DB therefore is the sum, and the arch CB the difference of AC, AB: Through E the centre of the circle, let there be drawn a diameter DEF, and AE joined, and CD likewise perpendicular to it in G; and let BH be perpendicular to AE, and AH will be the versed sine of the arch AB, and AG the versed sine of AC, and HG the excess of these versed sines: Let BD, BC, BF be joined, and FC also meeting BH in K. Since therefore BH, CG are parallel, the alternate angles BKC, KCG will be equal; but KCG is in a semicircle, and therefore a right angle; therefore BKC is a right angle; and in the triangles DFB, CBK, the angles FDB, BCK, in the same segment are equal, and FBD, BKC are right angles; the triangles DFB, CBK are therefore equiangular; wherefore DF is to DB, as BC to CK, or HG; and therefore the rectangle contained by the diameter DF and HG is equal to that contained by DB, BC; wherefore the rectangle contained by a fourth part of the diameter, and HG is equal to that contained by the halves of DB, BC: But half the chord DB is the sine of half the arch DAB, that is, half the sum of the arches AB, AC; and half the chord of BC is the sine of half the arch BC, which is the difference of AB, AC. Whence the proposition is manifest. |