(R) V. When the two sides, containing the right-angle, are given. 1. Each angle is of the same species as its opposite side. 2. The hypothenuse is acute or obtuse, according as the sides are of the same or of different species. (S) VI. When the two angles, adjacent to the hypothenuse, are given. 1. Each side is of the same species as its opposite angle. 2. The hypothenuse is acute or obtuse, according as the angles are of the same or of different species. (T) CASE I. Given the hypothenuse and an angle, to find the side adjacent to the given angle. Given Example. In the spherical triangle ABC. A=37°25' BY NAPIER'S RULE. The complement of the angle A is the middle part, AB and the complement of AC are the extremes conjunct, or joined to the middle part. Sine Er sine ED::tang IG: tang DF. But E is radius, DI the measure of the angle A, and ED is complement, IGAC, and DFAB. Hence, rad: cos Atang AC tang AB. BY CONSTRUCTION. (Plate V. Fig. 8.) 1. With the chord of 60 degrees describe the primitive circle; and through any point a in the primitive, draw a great circle ace, making an angle BAC 37°.25′. (P. 160.) 2. Set off the hypothenuse 78°.20′ from A to m, find p the pole of the oblique circle Ace (N. 159.); join pm, and through the point c, where it cuts Ace, draw BPM; then ABC is the triangle required. To measure the required parts. 3. AB applied to a scale of chords gives the measure of AB= 75°.25'. 4. Take BC in your compasses, set one foot of the compasses on 90° on the scale of semi-tangents, and the other towards the left hand of the scale, the degrees between the points 36°.31′ will be the measure of BC. 5. Produce mp to b, and find a the pole of nPB (N. 159.); then ab applied to the scale of chords will be the measure of the angle c=81°.12. PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Given { The hypothenuse AC=61°.4′.56′′ Required the ad- Answer. AB40°.30'.20". acute. (N. 211.) 2. In the right-angled spherical triangle ABC. Given { The hypothenuse AC=113°.55′ Required the ad- Answer. AB=117°.34'. obtuse. (N. 211.) (U) CASE II. Given the hypothenuse and one angle, to find the side opposite to that given angle. Example. In the right-angled spherical The hypothenuse Ac=78°.20 The angle A =37°.25' Given { triangle ABC. Required the opposite leg BC. BY NAPIER'S RULE. A Here BC is the middle part, and the complement of a and the complement of AC are the extremes disjunct; that is, not joined to the middle part, because of the intervention of AB and the angle c. * Rad x sine BC=sine AC X sine a. or, Radius, sine of : sine AC :: sine of A : sine BC 90°. 10.00000 Here BC is acute. (N. 211) BY RULE I. وح Sine AG sine AC:: sine HG sine BC. But AG is radius, and HG is the measure of the angle A; hence, See the construction of CASE I. (T. 212.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Given {The angle A Answer. BC 50°.30'.30" acute. (N. 211.) 2. In the right-angled spherical triangle ABC. Given The hypothenuse Ac=113°.55' Required the op- Answer. BC=117°.34. obtuse. (N. 211.) posite leg BC. (W) CASE III. Given the hypothenuse and one angle, to find the other angle. Example. In the right-angled spherical triangle ABC. Given {The angle A =37. 25 f Required the angle c. BY NAPIER'S RULE. Here the three parts are connected or joined together, therefore the complement of ac is the middle part; and the complement of A and the complement of c are extremes conjunct. plement of AC, CI radius, FG the complement of the angle ▲, and DI the measure of the angle c. Hence, Cos AC rad::cot A: tang c. BY CONSTRUCTION. See the construction of CASE I. (T. 212.) PRACTICAL EXAMPLES. I. In the right-angled spherical triangle ABC. Given The angle A } Required the 61°.50.29") angle c. Answer. The angle c-47°.54.20" acute. (N. 211.) 2. In the right-angled spherical triangle ABC. The hypothenuse AC-113°.55' Required the The angle A = 31.°51' Answer. The angle c=104 obtuse. (N. 211.) Given { } angle c. (X) CASE IV. Given the hypothenuse and a leg, to find the angle adjacent to that given leg. Given Example. In the right-angled spherical triangle ABC. BY NAPIER'S RULE. The three parts are here connected; therefore the complement of a is the middle part, AB and the complement of AC are the extremes conjunct. * Rad x cos Acot AC × tang. AB. (log cot AC + log tang AB)-10=log cos A. or, Radius, sine of : cot AC ::tang AB 90° 10.00000 =78°.20' 9.31489 Here the angle A is acute. (O. 211.) BY RULE I. Tang GI tang DF:: sine EI: sine ED. But GI AC, DF AB, EI is the radius, and ED the complement of the angle A. Hence, Tang AC tang AB:: rad: cos A. 1. With the chord of 60° describe the primitive circle, and through the centre p draw CPe, and arb at right angles to it. 2. Set off the hypothenuse, by a scale of chords, from c to m, and draw the small circle mam parallel to the right circle arb. (Z. 162.) 3. Take the complement of AB (viz. 14°.35′) from the scale of semi-tangents, with which as a radius and the centre p describe a small circle parallel to the primitive (Z. 162.) intersecting the circle mam in a. 4. Through the three points cae draw a great circle, then ABC is the triangle required. To measure the required parts. 5. The angles a and c (G. 164.) are 37°.28′ and 81°.11'. 6. BC, applied to a scale of chords, will be 36°.35′. Given PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. The hypothenuse AC-61°.4.56" To find the angle The side or leg, AB-40°.30′.20" JA, adjacent to AB. Answer. The angle A=61°.50.29". acute. (0.211.) 2. In the right-angled spherical triangle ABC. Given { The hypothenuse AC=78°.20 To find the adjacent angle A. Answer. The angle a=113°.18′. obtuse. (O. 211.) (Y) CASE V. Given the hypothenuse and a side or leg, to find the angle opposite to that given leg. Example. In the right-angled spherical triangle ABC. Given { BY NAPIER'S RULE. opposite to AB. The three parts are here disjoined by the intervention of the angle A; therefore AB is the middle part, the complement of AC and the complement of c are the extremes disjunct. Rad x sine AB sine DC X sine c. or, Sine Aç : radius, sine of :: sine of AB E G I Here the angle c is acute. (O. 211.) BY RULE I. Sine AC sine CG::sine AB sine GH. and GH is the measure of the angle c. Sine AC radius:: sine AB sine c. BY CONSTRUCTION. But CG is radius, Hence, See the construction of CASE IV. (X.215.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Given The hypothenuseAc=61°.4.56" Required the angle The side or leg AB-40°.30.20" S c, opposite to AB. Answer. The angle c=47°.54.20", acute. (O. 211.) 2. In the right-angled spherical triangle ABC. Given {The AB=1170.34) G opposite to AB. The hypothenuse AC 78°.20 Required the angle Answer. The angle c=115°.9'. obtuse. (O. 211.) (Z) CASE VI. Given the hypothenuse and one side or leg, to find the other leg. Example. In the right-angled spherical triangle ABC. Given {The side |