points be 75°.25'; this distance applied from B to a will give the point A. 3. From c through P draw CPe, and through the three points CAe describe a circle, then ABC is the triangle required. To measure the required parts. 4. The angles c and A (G. 164.) will be found to be 81°.12", and 37°.25'. 5. The hypothenuse Ac (C. 163.) will be 78°.20". (H) CASE XIV. Given the two sides, or legs, to find the hypothenuse. Example. In the right-angled spherical triangle ABC. Given (The side AB=75.25). To find the hypotheThe side BC=36°.31' Inul Se AC, BY NAPIER’s RULE. The complement of the hypothenuse Ac is the middle part, being separated from the sides by the angles A and c; and AB and Bc are the extremes disjunct. * Sine FB : sine FC::sine BH : sine CG. But FB is radius, CF the complement of Bc, BH the complement of AB, and cq the complement of Ac. Hence, • Rad : cos BC::cos AB : cos AC. TBY CONSTRUCTION. See the construction of Case XIII. (G. 223.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABc. The side AB = 50°.30.30° N. To find the hypotheThe side BC=40°.30'.20" Ilulse AC, o Answer. Ac-61°.4/.56", acute. (R. 212.) 2. In the right-angled spherical triangle ABC. The side AB= 54°.22 U. To find the hypothenuse The side Bc- iso} AC. : Answer. Ac=117°.33', obtuse. (R. 212.) (I) CASE XV. Given two angles, to find a side or leg. Example. In the right-angled spherical triangle ABc. The angle A=37°.25° Required the The angle c-81°. 12' side BC. Given Given Given By NAPIER’s RULE. Here the complement of the angle A is the middle part, and BC and the complement of C are the eatremes disjunct, being separated from the middle part by the hypothenuse Ac. $ D 12 Sine ID : sine FG::sine cD : sine cF. But ID is the measure of the angle c, FG the complement of HG which is the measure of the angle A, cD radius, and CF the complement of BC. Hence, sine c : cos A: ; rad : cos BC." 1. With the chord of 60° describe the primitive circle, through the centre P draw cre, and rnm at right angles to it. 2. Set one foot of your compasses on 90° in the line of semitangents, extend the other towards the left hand, till the degrees contained between the points of the compasses be equal to (81°12') the Z c, and apply this extent from m to n. 3. Through the three points cne draw a great circle, and find its pole p. (N. 159.) 4. With the tangent of the ZA 77°.25', and centre P describe O an arc; with the secant of the ZA, and centrep crosses it in O : with o as a centre and radius op, describe the great circle ABP. 5. Then ABC is the triangle required, right-angled at B. 1. In the right-angled spherical triangle ABC. Answer. AB = 50°.30.30", acute. (S. 212.) 2. In the right-angled spherical triangle ABC. The angle c-31°.5 ..} side BC. Answer. Bc3=11.7°.34, obtuse. (S. 212.) (K) CASE XVI. Given two angles, to find the hypothenuse. Example. In the right-angled spherical triangle ABC. Given Given Given BY NAPIER’s RULE. Here the complement of Ac is the middle part, and the complements of A and c are the extremes conjunct. E P BY RULE I. Tang ID : tang FG ::sine cI : sine ca. But ID is the measure of the angle C, FG the complement of the angle A, cI radius, and cq the complement of Ac. Hence, tang Z c : cotz ::rad : cos Ac.” By construction. e PRACTICAL ExAMPLEs. 1. In the right-angled spherical triangle ABc. Given CHAP. VII. II. PRACTICAL RULES Fort solving THE DIFFERENT CASEs of RECTILATERAL, OR QUADRANTAL, SPHERICAL TRIANGLES, WITH THEIR APPLICATION BY LOGARITHMS. (L) : 1. When one side of a spherical triangle is 90°, or a quadrant, it is called a quadrantal triangle. 2. If two sides of a quadrantal triangle be each 90°, the triangle will be isosceles, each of the angles at the base will be 90°, and the base itself will be the measure of the vertical angle. If the three sides be each 90°, the three angles will be each 90°. - * 3. A quadrantal spherical triangle may be changed into a right-angled spherical triangle, and the contrary. (X. 138.) (M) RULE 1." Subtract the angle opposite to the quadrantal side from 180°, and call the remainder the hypothenuse of a new triangle. The angle opposite to the quadrantal side must C then be eonsidered as a right angle, and the two remaining angles must be represented by the sides of the quadrantal triangle which are opposite to them. The sides and angles of the right-angled spherical triangle, will represent the angles and sides of the quadrantal triangle. T. Thus in the annexed figure, where Ac is supposed to be a quadrant, the triangle ABC may be considered as a right-angled triangle, whose hypothenuse Ac- 180°– Z B, the base ABZ. C., and the perpendicular BC=the ZA; et contra. (N) RULE II. 1. In the triangle ABc, if one of the sides Ac be a quadrant, and another side BC less than a quadrant. Produce the side Bc till it becomes a quadrant, and ADB will be a right-angled triangle; wherein AD is the measure of the angle c, DB is the complement of BC, and AE is the hypothenuse. 2. In the triangle AEC, where one of the sides Ac is a quadrant, and another side ce greater than a quadrant. Subtract 90° from CE, and the remainder will be the side DE of the right-angled triangle ADE; AE is the hypothenuse, and AD is the measure of the angle c. The same reasoning will apply whether AB and AE be less or greater than quadrants, for in either case ADB and ADE will be right-angled triangles. (O) CASE I. Given a quadrantal side, its adjacent and opposite angle, to find the rest. The quadrantal side Ac- on Required the Given {; adjacent Z. c- 42°. 12' sides AB, BC Its opposite Z. B= 1 15:20 J and the A. A. By RULE II. (N. 228.) Because Ac and cd are quadrants, the angles C The angles ABC and ABD, being supplements B . to each other, may be considered as the same | 2/. angle, since they have the same sine, tangent, , !?...D &c. A "" BY RULE 1. (M. 227.) Here the supplement of the angle B64°40', must represent the hypothenuse ac, the quadrantal side Ac-the right angle b. The angles C and a must represent the sides ab and be, and the sides BC and AB must represent the angles a and c. |