BY NAPIER'S RULE. Here the complement of AC is the middle part, because it is not joined either to AB or BC, the angles a and c coming between them: hence, AB and BC are extremes disjunct, viz. not joined to the middle part. * Rad x cos AC COS AB X COS BC. (10+ log cos Ac)-log cos AB=log cos BC. 75°.25' radius, sine of - 90° or, Cos AB 9.40103 10.00000 = 9.30582 Here BC is acute. (O. 211.) BY RULE I. Sine BH sine CG::sine FB: sine CF. But BH is the com plement of AB, CG the complement of AC, FC the comple ment of BC, and FB radius. Hence, Cos AB COS AC::rad: cos BC. BY CONSTRUCTION. See the construction of CASE IV. (X. 215.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. The hypothenuse AC 61°.4.56" Required the Given {The side, or leg, AB-40°.30.20") other leg BC. Answer. BC 50°.30'.30". acute. (O. 211.) 2. In the right-angled spherical triangle ABC. Given Answer. BC=115°.55'. obtuse. (O. 211.) Required the side pc. (A) CASE VII. Given a side and the angle adjacent, or joining to it, to find the side opposite to the given angle. Example. In the right-angled spherical triangle ABC. The side or leg AB 75°.25' To find the side BC, The adjacent angle A=37°.25') Given { BY NAPIER'S RULE. opposite to A. Here the three parts are joined together, because the right angle B is not regarded in the rule; therefore AB is the middle part, the complement of A, and BC are the extremes conjunct. * Rad x sine AB = cot A X tang. BC. ... (10+ log sine AB)-log cot A= log tang BC. BY RULE I. Sine AH sine AB::tang GH tang BC. and GH is the measure of the angle a. But AH is radius Hence, Rad sine AB::tang/A: tang BC. BY CONSTRUCTION. (Plate V. Fig. 11.) 1. With the chord of 60 degrees describe the primitive circle, and through the centre P draw apa, and mnr at right angles to it. 2. Set one foot of your compasses on 90 degrees in the line of semi-tangents, extend the other towards the beginning of the scale, till the degrees between them be equal to the angle A 37°.25', and apply this extent from m to n. 3. Through the three points ana draw a great circle. 4. Set off AB=75°.25′ from a to B, by a scale of chords, and draw BP cutting the oblique circle in c, then ABC is the triangle required. To measure the required parts. 5. Bc measured by the scale of semi-tangents (as in Case I.) will be 36°.31'. 6. The angle c measured (G. 164.) will be 81°.12′. 7. The hypothenuse AC measured (C. 163.) will be 78°.20'. PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Given { Answer. BC 50°.30.30". acute. (P. 211.) 2. In the right-angled spherical triangle ABC. Given { The side AB 540.22' To find the leg BC, The adjacent angle A=136°.40' opposite to A. Answer. BC=142°.31'. obtuse. (P. 211.) (B) CASE VIII. Given a side and its adjacent angle, to find the angle opposite to the given side. Example. In the right-angled spherical triangle ABC. Given The side AB =75°.25' To find the angle c, opposite to AB. Its adjacent angle a=37°.25′ BY NAPIER'S RULE. Here the complement of the angle c is the middle part; and the complement of A, and AB are the extremes disjunct, being neither of them joined to the middle part. AF the complement of AB, ID the measure of the angle a; and FG the complement of GH, which measures the angle c. Hence, Radius cos AB::sine A: cos c. BY CONSTRUCTION. See the construction of CASE VII. (A. 217.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Given The side AB =40°.30'.20" Required the angle 1 Its adjacent angle A=61°.50'.29′′ Š c, opposite to AB. Answer. The angle c=47°.54′.20". acute. (P. 211.) 2. In the right-angled spherical triangle ABC. Given The side AB 1170.34 Required the angle c, Its adjacent angle a=31°.51′ ) opposite to AB. Answer. The angle c=104°.8'. obtuse. (P. 211.) (C) CASE IX. Given a side and its adjacent angle, to find the hypothenuse. Example. In the right-angled spherical triangle ABC. 75°.25' Required the hypo Its adjacent angle A=37°.25' BY NAPIER'S RULE. thenuse AC. 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 X tang. AB. ..(10+log cos A)-log tang AB-log cot AC. or, Tang AB : rad, sine of =75°.25' Here ac is acute. (P. 211.) BY RULE I. X B H Sine FH sine FG::tang BH tang co. But FH is radius, FG is the complement of the angle A, BH the complement of AB, and CG the complement of AC. Hence, Rad cos A:: cot AB cot AC. BY CONSTRUCTION. See the construction of CASE VII. (A. 217.) PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Given { = 50°.30'.30" Required the hy Its adjacent angle A= 47°.54.20" Answer. Ac=61°.4.56". acute. (P. 211.) pothenuse AC. 2. In the right-angled spherical triangle ABC. Given 1170.34" Required the hypothenuse AC. Its adjacent angle A = 31.51′ Answer. AC 113°.55'. obtuse. (P. 211.) (D) CASE X. Given a side and its opposite angle, to find the side adjacent to that angle. Example. In the right-angled spherical triangle ABC. Given {Its opposite angle a=37°.25′ BY NAPIER'S RULE. adjacent to A. The right angle в having no concern in the rule, the three parts are joined together; therefore AB is the middle part, the complement of A, and BC are the extremes conjunct. * Rad x sine AB=cot A × tang. BC. (log cota + log tang BC) -10=log sine AB. or, Radius, sine of 90° 10.00000 9.98580 A H Here AB is ambiguous. (Q. 211.) BY RULE I. Tang HG tang BC::sine AH sine AB. But HG is the measure of the angle a, and AH radius. Hence, tang A tang BC:: rad: sine AB. BY CONSTRuction. (Plate V. Fig. 10.) 1. With the chord of 60 degrees describe the primitive circle, draw Apa, and mner at right angles to it. 2. Set one foot of your compasses on 90° degrees in the line of semi-tangents, extend the other towards the beginning of the scale till the degrees between the two points be equal to the angle A 37°.25', and apply this extent from m to n. 3. Through the three points ana draw a great circle. 4. Take the complement of BC (viz. 53°.29′) from the scale of semi-tangents, with which as a radius and the centre P, describe a small circle oc parallel to the primitive (Z. 162.), meeting the great circle ana in c. 5. Through c and P draw BCP, then ABC is the triangle required. This problem is ambiguous, for ABC and aвс have the same data. To measure the required parts. 6. The angle c (G. 164.) will be 81°.12'. 7. The side AC (C. 163.) is 78°.20′. 8. The side AB applied to a scale of chords will be 75°.25′. PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. =40°.30'.20" To find the other Given {Its opposite angle A=47°.54.20" side. Answer. AB 50°.30′.30" or 129°.29.30". ambiguous. (Q. 211.) 2. In the right-angled spherical triangle ABC. Given =51°.30′ Its opposite angle A =58°. To find the other side. Answer. AB 50°.10', or 129°.50'. ambiguous. (Q. 211.) (E) CASE XI. Given a side and its opposite angle, to find the adjacent angle. Example. In the right-angled spherical The side BC 36°.31' triangle ABC. Given Its opposite angle A=37°.25 To find the angle c, BY NAPIER'S RULE. adjacent to BC. The complement of A is the middle part; and BC and the complement of c are the extremes disjunct, being neither of them joined to A, the middle part. |