Sph. Trig. PROP. XXIII. TN any spherical triangle, as the rectangle contained by the fines of two fides, is to the rectangle contained by the fines of half the fum and half the difference of the bafe and the excess of the fides, fo is the fquare of the radius to the fquare of the fine of half the angle oppofite to the base. Let ABC be a spherical triangle, of which the fide AC is greater than AB; and make AD equal to AC, and AE to AB; the rectangle contained by the fines of BA, AC is to the rectangle contained by the fine of half the fum of CB, BD, and the fine of half the difference of the fame CB, BD, as the fquare of the radius to the fquare of the fine of half the angle BAC. Defcribe through C, D and B, E the great circles CFD, BGE, and bife&t the angle BAC by the great circle AGF, and join BC, BD, DC, BE; and let O be the centre of the sphere, and join OF, OG, meeting CD, BE in H, K, and join HK, and a 11. 1. draw a BL perpendicular to CD. And because in the spherical triangles CAF, DAF, the two fides CA, AF, are equal to the two DA, AF, and the angle CAF is equal to DAF; therefore the b 6. s. T. bafe CF is equal to DF », and the angle CFA to AFD; there- c 3. Def. dicular to the plane CFD: and which bifects the arch CFD, dCor.30.3. bifects CD at right angles d DUL ; e4.Def.11. therefore CD is perpendicular to the plane AGF, and DHK is therefore a right angle: For the F fame reafon, BE is perpendicular to the plane AGF; therefore f 6. 11. BK is parallel to DH: and BL, HK are parallel, because BLH, LHK are right angles; therefore BL HK is a parallelogram, 34. I and BK is equal to LH ; and because the fides CB, BD of the triangle CBD are placed in equal circles CB, ABD, the rectangle CH, HL, or CH, BK, is equal to the rectangle contained K LEM. by half the fum and half the difference of the arches CB, BD *; but in the right angled fpherical triangle AGB, the fine of AB 20. S. T. is to BK the fine of BG, as the radius to the fine of BAG! S. T. half of BAC; and in the right angled triangle ACF, the fine of of AC is to CH the fine of CF, as the radius to the fine of FAC Sph. Trig. half of BAC1; that is, as the fine of AB to BK; therefore the rectangle CH, BK is fimilar to the rectangle contained by the fines of AB, AC: Wherefore, as the rectangle contained by the fines of BA, AC is to the rectangle CH, BK contained by the fines of half the fum and half the difference of the arches CB, BD, fo is the fquare of the radius to the square of the fine of half the angle BAC ". IN any spherical triangle, the rectangle contained by the fines of two fides, is to the rectangle contained by the fine of half the fum of the three fides, and the fine of its excefs above the bafe, as the fquare of the radius to the fquare of the cofine of half the angle oppofite to the base. Let ABC be a spherical triangle; the rectangle contained by the fines of two fides AB, AC is to the rectangle contained by the fine of half the fum of the three fides, and the fine of the excefs of the faid half fum above the bafe BC, as the fquare of the radius to the fquare of the cofine of half the angle BAC. Produce BA, CA to D, E, and make AD equal to AC, and AE to AB; and defcribe the great circles CFD, BGE; and let the great circle GAF bifect the angle CAD, and meet the great circles CFD, BGE in F, G; and join EB, BC, CD, DB; and draw BL perpendicu lar to DC; and let O be the centre of the fphere, and join OF, OG, meeting CD, BE, in H, K; and join HK. It may be demonftrated, as in the preceding propofition, that C HL is equal to BK; and that the rectangle contained by the fines of BA, AC is to the rectangle BK, CH, or CH, HL, as the fquare of the radius to the fquare of the fine of half the angle BAE or CAD: but, becaufe CD is bifected m 22.6. in Sph. Trig. in H, the rectangle CH, HL is equal to the rectangle contained by the fines of half the fum, and half the difference of the arches BD, BC, and the fum of BD, BC is the fum of the three fides, and the excess of the half of BD above half of BC is the excefs of half the fum of the three fides above BC; therefore the rectangle CH, HL is equal to the rectangle contained by the fines of half the sum of the three fides, and of its excefs above BC: alfo, because the angles CAB, CAD are equal to two right angles, the half of CAB and the half of CAD are together equal to a right angle; therefore the fine of half CAD is the cofine of half CAB: But the rectangle contained by the fines of BA, AC, is to the rectangle CH, HL as the fquare of the radius to the fquare of the fine of half the angle CAD; therefore the rectangle contained by the fines of BA, AC, is to the rectangle contained by the fines of half the fum of the three fides, and of its excess above the base BC, as the fquare of the radius to the fquare of the cofine of half the angle BAC. COR. Because the rectangle contained by the fines of half the fum of the three fides, and of its excefs above BC, is to the rectangle contained by the fines of BA, AC as the fquare of the coline of half BAG to that of the radius; and that the rectangle contained by the fines of BA, AC, is to the rectangle contained by the fines of half the fum, and half the difference of the base and the excefs of the fides, as the square of the radius to the 8 22. 5. fquare of the fine of half BAC; therefore, by equality 2, the rectangle contained by the fines of half the fum of the three fides, and of its excefs above the bafe, is to the rectangle contained by the fines of half the fum, and half the difference of the base, and the excess of the fides, as the fquare of the cofine of half the angle BAC to the square of its fine; that is, as the fquare of the radius to the fquare of the tangent of half the b2. Cor. angle BAC‘. Def. 10. SOLUTION of the CASES of OBLIQUE-ANGLED GENERAL PROPOSITION. IN any spherical triangle, of the three fides and three angles, any three being given, the other three may be found. The The feveral cafes of this propofition may be refolved by the Sph. Trig. help of the three general proportions, together with the 22d, and any one of the fubfequent propofitions; as in the following Table, in which ABC is any spherical triangle; and the perpendicular AD either falls within the triangle, or meets the bafe BC produced beyond C. The cases referred to, are those of the preceding Table. Cafes Given. Songht. Solution. I AB, AC and C the angle Sin. AC: fin. AB: : fin. B: fin. C. If the fum of BA, AC [AB, AC and}BC the third/R : cos. B : : tan. AB : tan. BD B oppofite fide. (cafe 2.) and cos. AB: cos. If CD be not lefs than DB, their fum is CB; if CD be Cafes. Sph. Trig. Cafes. Given. Sought. Solution. JAB, AC and A the angle R: cos. AB:: tan. B: cot BAD, B oppofite contained 3 to AC. by the fides. (cafe 3.), and tan. AC: tan. AB:: cos. BAD: cos. DAC. If B be acute, DAC and AC are of the fame affection, otherwise they are of different affection. If DAC be not lefs than BAD, their fum is BAC: if DAC be lefs than BAD, but their fum not lefs than 180°, their difference is BAC. In other cafes BAC is ambiguous. B, C and AB, AC the fide Sin. C: fin. B:: fin. AB: fin. AC. If the fum of B and C be less than 180o, and B lefs than C, AC is acute: or if the fum of B and C be greater than 180°, and B greater than C, AC is obtufe. In other cafes, AC is ambiguous. R: cos. AB:: tan. B: cot. BAD, (cafe 3), and cos. B: : cos. C:: fin. BAD: fin. DAC, (p. 3.), which is lefs than BAD, if B, C be of diff. affection, or lefs than the fupplement of BAD, if B and C be of the fame affection: In other cafes it is ambiguous. When B and C are of the fame affection, BAC is the fum of BAD, DAC, otherwife it is their difference. Cafes. |
