PRACTICAL EXAMPLES. 1. In the right-angled spherical triangle ABC. Given { The angle c=61°.50′.29" To find the hypo Answer. Ac=61°.4.56", acute. (G. 172.) thenuse AC, 2. In the right-angled spherical triangle ABC. The angle A=31°.51'} Given { The angle c=104°.8′ S Required the hypothenuse AC. Answer. Ac=113°.55', obtuse. (G. 172.) CHAP. IV. GENERAL RULES FOR SOLVING THE DIFFERENT CASES OF RECTILATERAL, OR QUADRANTAL, SPHERICAL TRIANGLES. (Z) 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. (Z. 135.) (A) RULE I. C 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 then be considered 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. Thus in the annexed figure, quadrant, the triangle ABC may triangle, whose hypothenuse AC and the perpendicular BC the A where AC is supposed to be a be considered as a right-angled 180°-4B, the base ABC; A; et contra. Eb 2 (B) RULE (B) 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. (C) CASE I. Given a quadrantal side, its adjacent and opposite angle, to find the rest. The quadrantal side Ac=90°.00′) Given Its adjacent 4c= Its opposite B Required the 42°.12' 115°.20' sides AB, BC and the 44. BY RULE II. (B. 188.) Because AC and CD are quadrants, the angles CAD and CDA are each of them a right angle (Z. 187.), and AD is the measure of the angle c. The angles ABC and ABD, being supplements to each other, may be considered as the same angle, since they have the same sine, tangent, &c. 1. To find AB, by Case 12, right-angled spherics. sine ABD sine AD sine c:: rad: sine AB. Here AB is acute=48°. B ....D A 2. To find BD, and hence BC, by Case 10, right-angled spherics. rad: cot ABD::tang AD=tang 4C: sine BD. Here BD is acute 25°.25', and its complement BC=64°.35′′. BAD, by Case 11, right-angled spherics. c: rad :: cOS ABDcos 64°.40': sine BAD. 35°.17', and its complement=BAC 54°.43′. BY RULE I. (A. 187.) Here the supplement of the angle B= 64°.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 bc, and the sides BC and AB must represent the angles a and c. DA B 1. To find the Lc, in the triangle abc, by Case 5, right angled spherics. sine ac sine suppt. B: rad: : sin ab=sin c; sin ▲c=sin AB. Here the 4c or AB is acute=48°. 2. To find the La, in the triangle abc, by Case 4, rightangled spherics. rad: cot ac=cot suppt. B:: tang ab=tang c: cos a=cOS BC. Here cosine of La=64°.35' acute. L 3. To find bc, in the triangle abc, by Case 6, right-angled spherics. cos ab=cos <c: rad :: cos ac cos suppt. B: cos bc=cos A. Here be is acute=54°.43′. BY CONSTRUCTION. (Plate V. Fig. 13.) 1. With the chord of 60 degrees describe the primitive circle; and through the centre P draw aPD, ad cre at right angles to it. 2. Set one foot of your compasses on 90 degrees on the line of semi-tangents, extend the other towards the beginning of the scale, till the degrees between them be equal to the angle c= 42°.12', and apply this extent from D to A. 3. Through the three points cae draw a great circle, then AC will represent the quadrantal side. 4. With the tangent of the complement of the angle B, and centre c, describe an arc; with the secant of the complement of the same angle, and centre A, cross it in o: with o as a centre, and radius oa, describe the great circle Bab. Then ABC is the triangle required. To measure the required parts. The sides AB and BC (C. 158.) will be 48° and 64°.35', NOTE. The right-angled spherical triangle abc (Plate V. fig. 14.) into which the quadrantal triangle ABC is transformed, may be constructed and measured exactly in the same manner as Case IV. of right-angled spherical triangles was constructed and measured. See Plate V. fig. 9. (D) CASE II. Given a quadrantal side, and the other two sides, to find the rest. The quadrantal side Ac= 90° * Given The side AB The side BC =115°. 9' =113°.18' Required the angles A, B, and c. *This example was formed from Example 2, Case VI. of right-angled spherics. The B was made equal to the supplement of the hypothenuse, the two sides AB and BC were made equal to the angles of the right-angled triangle; and hence the legs of the right-angled triangle become angles in the quadrantal triangle. In a similar manner the other examples were made. BY BY RULE II. (B. 188.) Make e/CD=AC, then CAD, CDA, and ADB are each of them a right angle, and AD measures the angle c (Z. 187, and M. 130.); hence, if we take 90° from BC we get 23°.18' BD; and ADB is a right-angled triangle. B=78°.20′. 1. To find the LB, by Case 4, right-angled spherics. 117°.34'. 3. To find the DAB, by Case 5, right-angled spherics. BY RULE I. (A. 187.) Here the quadrantal side Ac must represent the right angle b; AB the LC; and BC the La; ac the supplement of the 4B, ab the 2c, and bc the LA. C DAB+ <CAD= D A 1. To find the hypoth. ac, by Case 16, right-angled spherics. Rad:cota=cot BC:: cot 4 ccot AB: cos ac sup. B=78°.20′. Here ac or suppt. 4B is acute (G. 172.); therefore 4B is obtuse =101°.40'. 2. To find ab, by Case 15, right-angled spherics. Sine Asin BC:rad :: cos c= cos AB:cosab= =cos 4c=62°.26'. Here abc is obtuse (G. 172.), and therefore=1170.34'. 3. To find bc, by Case 15, right-angled spherics. Sine C sin AB: rad :: cos acos BC: cos bc=cos ▲ A=64°.5′. Here bc A is obtuse (G. 172.), and therefore=115°.55'. If the two preceding cases be thoroughly understood, there can be no difficulty in solving those which follow; for which reason, they are given as practical exercises. (E) CASE III. Given a quadrantal side, and its two adjacent angles, to find the rest. The quadrantal side AC=90° Required AB, Given The angle The angle c=42°.12′) angle B. Answer. AB=48°, BC=64°.35', and the angle B=115°.20′. (F) CASE (F) CASE IV. Given a quadrantal side, one of the other sides, and the angle comprehended between them, to find the rest. The quadrantal side Ac= 90° Given The side The angle CAB Required EC, AB 115° 9' and the angles Answer, BC=113°.18', the angle B=101°.4', c=117°.34'. (G) CASE V. Given a quadrantal side, its adjacent angle, side opposite to that angle, to find the rest. and The quadrantal side AC=90° Given Its adjacent angle c=42°.12 Required BC, A and B. And the opposite sideAB=48°.00′ Answer. BC=64.35', the angle A-54°.43′ and B=115°.20'. But the required parts are ambiguous, and therefore are either acute or obtuse. (H) CASE VI. Given the quadrantal side, one of the other sides, and an angle opposite to the quadrantal side, to find the rest. The quadrantal side AC= 90° Given The side AB The angle B =115°. 9' Required BC, and the angles A and c. Answer. BC=113°.18', the angle A=115°.55', and c= 117°.34′. CHAP. V. INVESTIGATION OF GENERAL RULES FOR SOLVING THE DIFFERENT CASES OF OBLIQUE SPHERICAL TRIANGLES, BY DRAWING A PERPENDICULAR FROM THE VERTICAL ANGLE UPON THE BASE, PROPOSITION XXV. Shewing the manner of applying BARON NAPIER's rules to oblique spherical triangles, from which several useful corol laries are deduced. (I) When the three given parts do not follow each other in a regular order, viz. when an unknown part intervenes, a perpendicular should always be drawn from the end of a given side, and opposite to an adjacent given angle. But when the three given quantities follow each other without the intervention of an unknown quantity, the perpendicular should be drawn in such a manner, as to fall not only from the end of a given side and opposite to an adjacent given angle, but likewise |