BY RULE I. Sine CF sine CD::sine FG sine ID. But CF is the com plement of BC, CD radius, FG the complement of the angle a, and ID the measure of the angle c. Hence, Given Cos BC rad::cos A: sine c. BY CONSTRUCTION. See the construction of CASE X. (D. 220.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. 'The side BC =40°.30.20" Required the adIts opposite angle a=47°.54′.20" jacent angle c. Answer. Angle c=61°.50′.29′′ or 118°.9′.31". ambiguous. (Q. 211.) 2. In the right-angled spherical triangle ABC. The side BC =42°.12′ To find the other Given {Its opposite angle a 48°. 12'} angle c. Answer. Angle c = 64°. 35'. or 115°. 25. ambiguous. (Q. 211.) (F) CASE XII. Given a side and its opposite angle, to find the hypothenuse. Example. In the right-angled spherical triangle ABC. Given Its opposite angle A=37°.25' S nuse AC. BY NAPIER'S RULE. Here BC is the middle part; the complement of A, and the complement of AC, are the extremes disjunct. * Rad x sine BC=sine AC × sine a. ... (10+ log sine BC)-log sine A=log sine ac. or, Sine A 37°.25' rad, sine of 90° 36°.31′ ::sine BC 9.77456 G Here AC is ambiguous. (Q. 211.) BY RULE I. Sine GH sine BC::sine AG : sine AC. sure of the angle A, and AG radius. Hence, Sine A sine BC::rad: sine ac. But GH is the mea Given BY CONSTRUCTION. See the construction of CASE X. (D. 220.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. The side BC =40°.30.20" To find the hyIts opposite angle A=47°.54.20" pothenuse Ac. Answer. Ac=61°.4.56" or 118°.55.4". ambiguous. (Q. 211.) 2. In the right-angled spherical triangle ABC. Given { The side BC 11°.30' To find the hypothe nuse AC. Its opposite angle a=23°.30′ ƒ (G) CASE XIII. Given the two sides, or legs, to find an angle. Example. In the right angled spherical triangle ABC. Given The side AB=75°.25′ ̄ To find the angle a. BY NAPIER'S RULE. The three parts are here connected, because the right angle B is not regarded; therefore AB is the middle part, the complement of A, and BC are the extremes conjunct. * Rad x sine ABCot A X tang BC. •.• (10+log sine AB) — log tang AC=logcota. Tang BC =360.31′ rad, sine of - 90° 9.98578 10.11631 Sine AB sine AH::tang BC tang GH. But AH is radius, and GH is the measure of the angle a. Hence, Sine AB rad::tang BC: tang A. BY CONSTRUCTION. (Plate. V. Fig. 7.) 1. With the chord of 60 degrees describe the primitive circle, and through the centre P draw aв. 2. Set off BC 36°.31′ from a scale of chords, and AB=75°. 25′. (D. 163.) Or set one foot of the compasses on 90 degrees on the line of semi-tangents, and extend the other towards the beginning of the scale, till the distance between the points be 75°.25'; this distance applied from B to A will give the point A. 3. From c through P draw cre, 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'. Given PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. { Answer. A=47°.54.20". acute. (R. 212.) 2. In the right-angled spherical triangle ABC. Given {} The side BC=142°31'} To find the angle A. Answer. A 136°.40'. obtuse. (R. 212.) (H) CASE XIV. Given the two sides, or legs, to find the hypothenuse. Example. In the right-angled spherical triangle ABC. Given {The side BC=36°31′ BY NAPIER'S RULE. nuse AC. 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. * Rad x cos AC COS AB X COS BC. (log cos AB+log cos BC) -10=log cos AC. or, Rad, sine of 90° =75°.25' COS BC =36°.31' : COS AB : COS AC =78°.20' 10.00000 9.40103 9.90509 9.30612 E Here Ac is acute. (R. 212.) BY RULE I. H Sine FB sine FC:: sine BH sine CG. But FB is radius, cr the complement of BC, BH the complement of AB, and CG the complement of ac. Hence, Rad cos BC::COS AB Cos ac. BY CONSTRUCTION. See the construction of Case XIII. (G. 223.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Given { The side AB 50°.30'.30" To find the hypothe- Answer. AC 61°.4'.56", acute. (R. 212.) nuse AC. 2. In the right-angled spherical triangle ABC. Given { The side AB 54°.22 To find the hypothenuse Answer. AC=117°.33', obtuse. (R. 212.) AC. (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′ Given The angle c=81°.12′ BY NAPIER'S RULE. Required the Here the complement of the angle a is the middle part, and · BC and the complement of c are the extremes disjunct, being separated from the middle part by the hypothenuse ac. * Rad x cos ACOS BC X sine c. (10+log cos A)-log sine c=log cos BC. or, sine c -81°.12' rad sine of 90° ::cos LA37°.25' E 9.99486 10.00000 9.89995 Here BC is acute. (S. 212.) B ཡ BY RULE I. 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 ccos A::rad: cos BC. BY CONSTRUCTION. (Plate V. Fig. 12.) 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 semi-> tangents, extend the other towards the left hand, till the degrees contained between the points of the compasses be equal to (81°12') the 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 a 77°.25', and centre p describe B an arc; with the secant of the ▲ A, and centre p crosses it in o; with o as a centre and radius op, describe the great circle Авр. 5. Then ABC is the triangle required, right-angled at B. To measure the required parts. 6. The hypothenuse AC, measured by a scale of chords, =78°.20'. 7. The sides BC and ac (C. 163.)=36°.31', and 75°.25′. PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Required the Answer. AB50°.30.30", acute. (S. 212.) 2. In the right-angled spherical triangle ABC. Required the side BC. 117°.34', obtuse. (S. 212.) (K) CASE XVI. Given two angles, to find the hypothenuse. Given BY NAPIER'S RULE. nuse Ac. Here the complement of AC is the middle part, and the complements of a and c are the extremes conjunct. * Rad x cos AC cot A x cot c. •.' (log cot a+log cot c) — 10=log cos ac. or, Rad, sine of ::cot / C · Here AC is acute. (S. 212.) BY RULE I. Tang ID tang FG:: sine cr: sine CG. But ID is the measure of the angle C, FG the complement of the angle a, ci radius, and co the complement of Ac. Hence, tangc cotrad: cos Ac.* BY CONSTRUCTION. See the construction of CASE XV. (I. 225.) |