of the other side AB is to the cosine of the hypothenuse AC. In the comp. triangle abC, R: s. Ca :: s. a: s. Cb (48), that is, R: cos BC :: cos. AB: cos. AC. PROP. IX. 57. In a right-angled spherical triangle ABC radius is to the cosine of either of the sides BC, as the sine of the angle BCA adjacent to that side is to the cosine of the other angle A. In the comp. triangle abC, R: s. Ca :: s. aCb: s. ba (48), that is, R: cos. BC :: s. BCA : cos. A. Note. By means of the preceding propositions all the cases of right-angled spherical triangles may be resolved. PROP. X. 58. In all spherical triangles, whether right or obliqueangled, the sines of the sides are proportional to the sines of the angles opposite to them. Fig. 4 and 6. Let ABC be either a right or an oblique-angled triangle; the sine of any side BC: sine of any other side AC :: sine of the angle A opposite to BC: sine of the angle B opposite to AC. First, let ABC (fig. 4) be a right-angled triangle, having a right angle at B; then and s. AC: R s. AC: R or s. B :: s. BC : s. A (48), :: s. AB: s. C; therefore s. AB: s. C:: s. BC: s. A (11. 5). Secondly, let ABC (fig. 6) be an oblique-angled triangle. Through the angular point C draw an arc of a great circle CD perp. to the op. side AB. In the right-angled triangle BCD, s. BC: R:: s. CD: s. B(48), and in the right-angled triangle ADC, s. AC: R:: s. CD: s. A; therefore s. BC: s. AC:: s. A: s. B (23. 5). In the same manner it may be proved that s. BC: s. AB :: s. A: s. C. If three of these quantities be given to find the fourth, the case will be doubtful, because the quantity is expressed by a sine, unless there be some other circumstances which may remove the ambiguity. (112 Pl. Tr.) 59. Cor. Hence s. AB+s. BC: s. ABs. BC :: s. A+s. C: s. As. C (E. 5). Consequently tan. (AB+BC): tan. (AB BC) :: tan. (A + C): tan. † (AC) (125. Pl. Tr.) 60. Remark 1. In deducing formula for the solution of oblique-angled triangles, by drawing a perp. from the vertical angle upon the opposite side, if we suppose the perp. to fall within the triangle, and the sides and angles to be less than 90°, the formula so derived will answer in all other cases. But we must attend to the algebraic signs of the quantities (110, 111 Pl. Tr.); for if the signs of the terms be once true, and then be made to vary according to the variation of the signs of the factors which compose them, they will continue true. The reader must give attention to this circumstance in some of the following demonstrations. 61. Remark 2. If the sides of a spherical triangle be conceived to decrease indefinitely, the sines or the tangents of the sides will approach in length to the sides as their limit; and the spherical triangle will approach to a plane triangle as its limit. Therefore, in a rectilinear triangle, AC : BC :: s. B : s. A, as was proved in Plane Trig. In like manner, in all cases of spherical triangles, when the sines or the tangents of the sides enter into any analogy, the sides themselves may be substituted for their sines or tangents, and the conclusions will be true in all cases of rectilinear triangles. PROP. XI. 62. In any spherical triangle ABC, if a perpendicular arc CD be drawn from any angle C upon the opposite side, or base, AB, the cosines of the segments of the base AD, DB will be proportional to the cosines of the adjacent sides AC, BC. In the right-angled triangle ADC, R: cos. DC :: cos. AD: cos. AC (56), and in the right-angled triangle BDC, R: cos. DC :: cos. BD: cos. BC; therefore cos. AD: cos. AC :: cos. BD: cos. BC, or cos. AD: cos. BD :: cos. AC : cọs. BC. PROP. XII. 63. The same construction remaining, the cosines of the angles A and B at the base will be proportional to the sines of the corresponding segments ACD, BCD of the vertical angle C. In the triangle ACD, R: cos. CD:: s. ACD: cos. A (57), and in the triangle BCD, R: cos. CD :: s. BCD : cos. B; therefore s. ACD: cos. A:: s. BCD: cos. B, or cos. A cos. B:: s. ACD: s. BCD. PROP. XIII. 64. The same construction remaining, the sines of the segments of the base AD, BD will be reciprocally proportional to the tangents of the adjacent angles A and B at the base. In the triangle ADC, R: s. AD :: tan. A : tan. DC (49), and in the triangle BDC, R: s. BD :: tan. B: tan. DC; therefore s. AD: s. BD:: tan. B: tan. A. PROP. XIV. 65. The same construction remaining, the cosines of the segments ACD, BCD of the vertical angle C will be reciprocally proportional to the tangents of their adjacent sides AC, BC. In the triangle ADC, R: cos. ACD :: tan. AC: tan. CD (53), and in the triangle BDC, R: cos. BCD :: tan. BC : tan. CD; therefore cos. ACD: cos. BCD :: tan. BC: tan. AC. PROP. XV. 66. In any spherical triangle ABC, if a perpendicular arc CD be drawn from any angle C upon the opposite side, or base, AB, the sine of the sum of the sides containing the angle C is to the sine of their difference, as the cotangent of half the sum of the segments of the vertical angle is to the tangent of half their difference. Because tan. BC: tan. AC :: cos. ACD: cos. BCD (65), tan. BC+tan. AC: tan. BC tan. AC :: cos. BCD + cos. ACD: cos. BCD ~ cos. ACD (E. 5). But tan. BC + tan. AC: tan. BC tan. AC :: s. (BC + AC): s. (BC AC) (126 Pl. Tr.), and cos. BCD + cos. ACD : cos. BCD cos. ACD:: cot. (BCD + ACD): tan. 1⁄2 (BCD ACD) (125 Pl. Tr.). Hence s. (BC+ AC) : s. (BC AC) :: cot. (BCD + ACD) or cot. C: tan. 1⁄2 (BCD ~ACD) (11.5). 2 s. 1⁄2 (BC+AC) × cos. 1⁄2 (BC+AC). 67. Cor. Because s. (BC+AC) (115 Pl. Tr.), and R = s. (BC ~ AC), s. (BC+AC) x cos. (BC+ AC): s. (BC AC) × cos. 1⁄2 (BC ~ AC) :: cot. 1⁄2 C : tan. 1⁄2 (BCD ~ ACD). = 68. Remark. If two sides and the included angle of any spherical triangle be given, we can find, by this proposition, the term BCD ACD, and consequently the angles at the vertex, ACD, BCD. Therefore when two sides and the included angle of a spherical triangle are given, the triangle may be resolved into two right-angled triangles, in each of which the hypothenuse and an angle are known; consequently the other angle and sides may be found by prop. IV and VI. Hence the side and angles of the proposed triangle will be determined. PROP. XVI. 69. If a perpendicular arc CD be drawn from an angle C of a spherical triangle ABC, to the opposite side, or base, AB, the tangent of half the sum of the segments of the base will be to the tangent of half the sum of the other two sides, as the tangent of half their difference is to the tangent of half the difference of the segments of the base; that is, tan. (BD + AD) : tan. (BC + AC) :: tan. (BC~ AC): tan. (BD AD). M ~ Because cos. BC: cos. AC :: cos. BD: cos. AD (62), cos. BC+cos. AC: cos. BC cos. AC :: cos. BD+cos. AD: cos. BD cos. AD (E. 5). But cos. BC+cos. AC: cos. BC cos. AC:: cot. (BC+AC): tan. † (BC ~ AC) (125 Pl. Tr.); therefore cot. (BC+AC): tan. (BCAC):: cot. (BD+ AD): tan. (BD ☛ AD) (11. 5), or cot. 1⁄2 (BC+AC) : cot. (BD + AD) :: tan. 1⁄2 (BC ~ AC) : tan. § (BD~ AD). But cot. (BC+AC) : cot. { (BD+AD) :: tan. 1⁄2 (BD+AD) (BC + AC) (42 Pl. Tr.). Hence tan. (BD + AD) (BC+AC) :: tan. ¦ (BC ~ AC) : tan. 1⁄2 (BD - AD). tan. = S 70. Cor. 1. When the perp. CD falls within the triangle, then BD + AD AB the base; and when it falls without, then BD-ADAB the base. Therefore, in the first case, the proportion becomes tan. AB: tan. (BC + AC) :: tan. (BC-AC): tan. (BD AD); and, in the second case, it becomes, by inversion and alternation, tan. AB: tan. (BC+AC) :: tan. (BC-AC): tan. (BD + AD). 71. Cor. 2. From the demon. cot. (BC+ AC): tan. (BC AC) :: cot. (BD+AD): tan. (BD AD); that is, in any spherical triangle, the cot. of half the sum of two sides: tan. of half their difference :: cot. of half the sum of the segments of the base made by a perp. from the opposite angle: tan. of half their difference. 72. Remark. If the three sides of a spherical triangle be given, we can find, by the analogy in the proposition, or in cor. 2, the fourth term AD BD or AD + BD, and consequently the segments of the base AD, BD. Therefore when the three sides of a spherical triangle are given, the triangle may be resolved into two right-angled triangles, in each of which two sides are known; therefore the angles may be found by prop. VII, and consequently the angles of the proposed triangle will be determined. These two theorems are ascribed to Napier, and are so well adapted to calculation by logarithms, that they are considered two of the most useful propositions in spherical trigonometry. PROP. XVII. 73. In any spherical triangle, if a perpendicular arc be drawn from an angle to the opposite side, or base, |