and writing AG GF GD GF AE = GD. for the proportion AG : GD :: AE EH AD, derived from similar triangles, and GF = GD. HD we get tan c cot a; A third demonstration is effected by exhibiting the particular results of Props. 17, 18, 19, when the angle A is a right angle, which will give six of the rules; and then by combining these results the remaining four are immediately obtained. Since A 90°, sin A = 1, cos A = 0, radius being unity. Therefore, first (Prop. 17), sin b: sin a: sin B: 1; whence sin b = sin a sin B (1); similarly, sin c = sin a sin C (2). Secondly, Since by Prop. 18, cos a = cos b cos c + sin b sin c cos A; we have cos a = cos b cos c (3). Thirdly, By Prop. 19, cos A + cos B cos C = sin B sin C cos a, therefore, cos B cos C = sin B sin C cos a, cos α = cot B cot C (4), = also, cos B+cos A cos C sin A sin C cos b, similarly, cos C = sin B cos c (6). Fourthly, Having b as the middle part only once, we must have it again. therefore, cos B = = sin c cos a sin a cos c = tan c cot a (9); similarly, cos C = tan b cot a (10). These rules are so important, that we do not deem an apology necessary for having given three different forms of the demonstration of them. It is worthy of remark that the radius is merely a multiplier of the sine of the middle part, which can immediately be substituted from the knowledge of the fact, that it requires a rectangle to be equal to a rectangle; which, on the hypothesis that cosines and tangents are lines, the product of two cosines or two tangents amounts to. In practice, however, the radius is usually considered to be unity. † PROP. XXI. If, from an angle of a spherical triangle there be drawn a perpendicular to the opposite side or base, the rectangle contained by the tangents of half the sum, and of half the difference of the segments of the base, is equal to the rectangle contained by the tangents of half the sum, and of half the difference of the two sides of the triangle. Let ABC be a spherical triangle, and let the arc CD be drawn from the angle C, at right angles to the base AB; tan (BD+AD) tan (BD-AD)=tan (a+b) tan † (a—b). By Napier's rules, R cos a=cos CD cos BD, R cos b-cos CD cos AD, therefore, cos b: cos a: cos AD: cos BD, and cos b-cos a: cos b+cos a: cos AD-cos BD: cos AD+ cos BD, therefore, tan (a+b) tan (a−—b): R2:: tan (AD+BD) tan (BD-AD): R2 (Pl. Tr. Art. 25). consequently, tan (BD+AD) tan (BD-AD)=tan (a+b) tan (a-b). Q. E. D. COR. Since when the perpendicular CD falls within the triangle, BD+AD=AB the base; and when it falls without, BD-AD= AB; therefore, in the first case, tanc tan (BD-AD)=tan (a+b) tan † (a−b); SECTION II. PROBLEM I. Solution of right-angled spherical triangles. This problem has sixteen cases, the solutions of which, when the radius is unity, are contained in the following table, where ABC is any spherical triangle, right angled at A. They are all derived from Napier's rules. A TABLE for determining the affections of the sides and angles found by the preceding rules. AC and B of the same affection, If BC 90°, AB and B of the same affection, otherwise different, If BC 90°, C and B of the same affection, otherwise different, AB and C are of the same affection, (15 Cor.) 2 (15) 3 (14) .4 (15 Cor.) (14) If AC and C are of the same affection, BC_90° otherwise When BC 90°, AB and AC of the same; otherwise of different affection, AC and B of the same affection, When BC 90°, AC and C of the same; otherwise of different affection, BC 90°, when AB and AC are of the same affection, (15 Cor. 1) 13 B and AC of the same affection, (14) 14 (14) 14 AB and C of the same affection, (14) 15 AC and B of the same affection, (14) 15 The cases marked ambiguous are those in which the thing sought has two values, and may either be equal to a certain angle, or to the supplement of that angle. Of these there are three, in all of which the things given are a side, and the angle opposite to it; and, accordingly, it is easy to show that two right-angled |