PROP. XVIII. FIG. 13. IN right angled spherical triangles, the sine of the hypothenuse is to the radius, as the sine of either side is to the sine of the angle opposite to that side. Let the triangle ABC be right angled at A, and let AC be either of the sides; the sine of the hypothenuse BC will be to the radius as the sine of the arch AC is to the sine of the angle ABC. Let D be the centre of the sphere, and let CG be drawn perpendicular to DB, which will therefore be the sine of the hypothenuse BC; and from the point G let there be drawn in the plane ABD the straight line GH perpendicular to DB, and let CH be joined; CH will be at right angles to the plane ABD, as was shown in the preceding proposition of the straight line FA: Wherefore CHD, CHG, are right angles, and CH is the sine of the arch AC; and in the triangle CHG, having the right angle CHG, CG is to the radius as CH to the sine of the angle CGH (1. Pl. Tr.): But since CG, HG are at right angles to DGB, which is the common section of the planes CBD, ABD, the angle CGH will be equal to the inclination of these planes (6. def. 11). that is, to the spherical angle ABC. The sine therefore of the hypothenuse CB, is to the radius as the sine of the side AC is to the sine of the opposite angle ABC. Q. E. D. COR. Of these three, viz. the hypothenuse, a side, and the angle opposite to that side, any two being given, the third is also given by Prop. 2. PROP. XIX. FIG. 14. IN right angled spherical triangles, the cosine of the hypothenuse is to the radius as the cotangent of either of the angles is to the tangent of the remaining angle. Let ABC be a spherical triangle, having a right angle at A, the cosine of the hypothenuse BC will be to the radius as the cotangent of the angle ABC to the tangent of the angle ACB. Describe the circle DE, of which B is the pole, and let it meet AC in F, and the circle BC in E; and since the circle BD passes through the pole B of the circle DF, DF will also pass through the pole of BD. (13. 15. 1. Thead. Sph.) And since AC is perpendicular to BD, AC will also pass through the pole of BD; wherefore the pole of the circle BD will be found in the point where the circles AC, DE meet, that is, in the point F: The arches FA, FD, are therefore quadrants, and likewise the arches BD, BE: In the triangle CEF, right angled at the point E, CE is the complement of the hypothenuse BC of the triangle ABC, EF is the complement of the arch ED, which is the measure of the angle ABC, and FC the hypothenuse of the triangle CEF, is the complement of AC, and the arch AD, which is the measure of the angle CFE, is the complement of AB. But (17. of this) in the triangle CEF, the sine of the side CE is to the radius, as the tangent of the other side is to the tangent of the angle ECF opposite to it, that is, in the triangle ABC, the cosine of the hypothenuse BC is to the radius, as the cotangent of the angle ABC is to the tangent of the angle ACB. Q. E. D. COR. 1. Of these three, viz. the hypothenuse and the two angles, any two being given, the third will also be given. COR. 2. And since by this proposition the cosine of the hypothenuse BC is to the radius, as the cotangent of the angle ABC to the tangent of the angle ACB. But as the radius is to the cotangent of the angle ACB, so is the tangent of the same to the radius (Cor. 2. def. Pl. Tr.); and, ex æquo, the cosine of the hypothenuse BC is to the cotangent of the angle ACB, as the cotangent of the angle ABC to the radius. PROP. XX. FIG. 14. IN right angled spherical triangles, the cosine of an angle is to the radius, as the tangent of the side adjacent to that angle is to the tangent of the hypothenuse. The same construction remaining; in'the triangle CEF, (17. of this) the sine of the side EF is to the radius, as the tangent of the other side CE is to the tangent of the angle CFE opposite to it; that is, in the triangle ABC, the cosine of the angle ABC is to the radius as (the cotangent of the hypothenuse BC to the cotangent of the side AB, adjacent to ABC or as) the tangent of the side AB to the tangent of the hypothenuse, since the tangents of two arches are reciprocally proportional to their cotangent. (Cor. 1. def. Pl. Tr.). COR. And since by this proposition the cosine of the angle ABC is to the radius, as the tangent of the side AB is to the tangent of the hypothenuse BC; and as the radius is to the cotangent of BC, so is the tangent of BC to the radius; by equality, the cosine of the angle ABC will be to the cotangent of the hypothenuse BC, as the tangent of the side AB, adjacent to the angle ABC, to the radius. PROP. XXI. FIG. 14. IN right angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the hypothenuse is to the cosine of the other side. The same construction remaining; in the triangle CEF, the sine of the hypothenuse CF is to the radius, as the sine of the side CE to the sine of the opposite angle CFE (18. of this): that is, in the triangle ABC, the cosine of the side CA is to the radius as the cosine of the hypothenuse BC to the cosine of the other side BA. Q. E. D. PROP. XXII. FIG. 14. IN right angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the angle opposite to that side is to the sine of the other angle. The same construction remaining; in the triangle CEF, the sine of the hypothenuse CF is to the radius, as the sine of the side EF is to the sine of the angle ECF opposite to it; that is, in the triangle ABC, the cosine of the side CA is to the radius, as the cosine of the angle ABC opposite to it, is to the sine of the other angle. Q. E. D. OF THE CIRCULAR PARTS. Fig. 15. In any right angled spherical triangle ABC, the complement of the hypothenuse, the complements of the angles, and the two sides, are called The circular parts of the triangle, as if it were following each other in a circular order, from whatever part we begin: Thus, if we begin at the complement of the hypothenuse, and proceed towards the side BA, the parts following in order will be the complement of the hypothenuse, the complement of the angle B, the side BA, the side AC (for the right angle at A is not reckoned among the parts), and lastly, the complement of the angle C. And thus at whatever part we begin, if any three of these five be taken, they either will be all contiguous or adjacent, or one of them will not be contiguous to either of the other two: In the first case, the part which is between the other two is called the Middle part, and the other two are called Adjacent extremes. In the second case, the part which is not contiguous to either of the other two is called the Middle part, and the other two Opposite extremes. For example, if the three parts be the complement of the hypothenuse BC, the complement of the angle B, and the side BA; since these three are contiguous to each other, the complement of the angle B will be the middle part, and the complement of the hypothenuse BC and the side BA will be adjacent extremes: But if the complement of the hypothenuse BC and the sides BA, AC be taken: since the complement of the hypothese is not adjacent to either of the sides, viz. on account of the complements of the two angles B and C intervening between it and the sides, the complement of the hypothenuse BC will be the middle part, and the sides BA, AC opposite extremes. The most acute and ingenious Baron Napier, the inventor of Logarithms, contrived the two following rules concerning these parts, by means of which all the cases of right angled spherical triangles are resolved with. the greatest ease. RULE I. The rectangle contained by the radius and the sine of the middle part, is equal to the rectangle contained by the tangents of the adjacent parts. RULE II. The rectangle contained by the radius and the sine of the middle part, is equal to the rectangle contained by the cosines of the opposite parts. These rules are demonstrated in the following manner : First, Let either of the sides, as BA, be the middle Fig. 16. part, and therefore the complement of the angle B, and the side AC, will be adjacent extremes. And by cor. 2. prop. 17. of this, S, BA, is to the Co-T, B, as T, AC is to the radius, and therefore RxS, BA=Co-T, BxT, AC. The same side BA being the middle part, the complement of the hypothenuse, and the complement of the angle C, are opposite extremes; and by Prop. 18. S, BC is to the radius, as S, BA to S, C; therefore RxS, BA=S, BC XS, C. Secondly, Let the complement of one of the angles, as B, be the middle part, and the complement of the hypothenuse, and the side BA will be adjacent extremes: And by Cor. Prop. 20. Co-S, B is to Co T, BC, as T, BA is to the radius, and therefore Rx Co-S, B=Co-T, BC×T, BA. Again, Let the complement of the angle B be the middle part, and the complement of the angle C, and the side AC will be opposite extremes: And by Prop. 22. Co-S, AC is to the radius, as Co-S, B is to S, C: And therefore RxCo-S, B-Co-S, AC xS, C. Thirdly, Let the complement of the hypothenuse be the middle part, and the complements of the angles B, C, will be adjacent extremes: But by Cor. 2. Prop. 19. Co-S, BC is to Co-T, B as Co-T, B to the radius: Therefore R x Co-S, BC Co-T, Cx Co-T, C. Again, Let the complement of the hypothenuse be the middle part, and the sides AB, AC will be opposite extremes: But by Prop. 21. Co-S, AC is to the radius, as Co-S, BC to Co-S, BA; therefore Rx Co-S, BC=Co-S, BAX Co-S, AC. Q. E. D. |