hypotenuse 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, because it is the common section of two planes DCH, GCH, each perpendicular 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 is equal to the inclination of these planes, (6. def. 11,) that is, to the spherical angle ABC. Therefore the sine of the hypotenuse 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 hypotenuse, 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 hypotenuse 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 hypotenuse BC is to the radius, as the cotangent of the angle A BC 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 must pass through the pole of BD, (13. 18. 1. Theod. Sph.) And since AC is perpendicular to BD, the plane of the circle AC is perpendicular to the plane of the circle BAD, and therefore AC must also pass through the pole of BAD: wherefore, the pole of the circle BAD is in the point F, where the circles AC, DE meet. The arches FA, FD are therefore quadrants, and likewise the arches BD, BE: Therefore, in the triangle CEF, right angled at the point E, CE is the complement of the hypotenuse BC of the triangle ABC, EF is the complement of the arch ED, which is the measure of the angle ABC, and FC, the hypotenuse of the triangle CEF, triantangenter radius, in the trias the comarch A 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 EF is to the tangent of the angle ECF opposite to it, that is, in the triangle ABC, the cosine of the hypotenuse 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 hypotenuse 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 hypotenuse 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 equo, the cosine of the hypotenuse 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 hypotenuse. 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 hypotenuse BC to the cotangent of the side AB, adjacent to ABC or as) the tangent of the side AB to the tangent of the hypotenuse BC, since the tangents of two arches are reciprocally proportional to their cotangents. (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 hypotenuse 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 is to the cotangent of the hypotenuse BC, as the tangent of the side A B, adjacent to the angle ABC, is 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 hypotenuse is to the cosine of the other side. The same construction remaining; in the triangle CEF, the sine of the hypotenuse 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 hypotenuse 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 hypotenuse CF is to the radius as the side 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 BCA. Q. E. D. OF THE CIRCULAR PARTS. Fig. 15. In any right angled spherical triangle ABC, the complement of the hypotenuse, 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 hypotenuse, and proceed towards the side BA, the parts following in order will be the complement of the hypotenuse, the complement of the angle B, the side BA, the side AC, (for the right angle at A is not réckoned 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 will either be all adjacent, or one of them will be separated from each 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 parts. In the second case, the part which is separated from each of the other two is called the Middle part, and the other two are called Opposite parts. For example, if the three parts be the complement of the hypotenuse BC, the complement of the angle B, and the side BA; since these three are adjacent to each other, the complement of the angle B is the middle part, and the complement of the hypotenuse BC and the side BA are adjacent parts: But if the complement of the hypotenuse BC and the sides BA, AC be taken; since the complement of the hypotenuse is not adjacent to either of the sides, (the complements of the two angles B and C intervening between it and the sides,) the complement of the hypotenuse BC is the middle part, and the sides BA, AC are opposite parts. 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 part, then will the complement of the angle B, and the side ac, be adjacent parts. 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, BAT, AC. The same side BA being the middle part, then will the complement of the hypotenuse, and the complement of the angle C, be opposite parts; 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, then will the complement of the hypotenuse, and the side B A be adjacent parts: and by Cor. Prop. 20. Co-S, B is to Co-T, BC, as T, BA is to the radius, and therefore R x Co-S, B=Co-T, BCXT, BA. Again, Let the complement of the angle B be the middle part, then will the complement of the angle C, and the side AC be opposite parts : But by Prop. 22, Co-S, AC is to the radius, as Co-S, B is to S, C: And therefore Rx Co-S, B=Co-S, AC XS, C. Thirdly, Let the complement of the hypotenuse be the middle part, then will the complements of the angles B, C, be adjacent parts; but by Cor. 2, Prop. 19, Co-S, BC is to Co-T, C as Co-T, B to the radius: Therefore R x Co-S, BC=Co-T, Cx Co-T, B. Again, Let the complement of the hypotenuse be the middle part, then will the sides AB, AC, be opposite parts : But by Prop. 21, Co-S, AC is to the radius, as Co-S, BC to Co-S, BA; therefore R x Co-S, BC=Co-S, BAX CO-S, AC. Q. E. D. |