each other, inversely, as the tangents of the angles at the bases: For radius: finé AB:: tang. A: tang. BCZ fince radius: fine DB:: tang. D: tang. BC we fhall (by reasoning as in Cor. 1. Theor. 1.) have Sine AB fine DB:: tang. D: tang. A. THEOREM V. In any right-angled fpherical triangle it will be, as radius is to the co-fine of the hypothenufe, fo is the tangent of either angle to the co-tangent of the other angle. For (CEF being as in the last) it will be, as radius fine CE :: tang. C: tang. EF (by Theorem 4.) that is, radius: co-fine AC: tang. C: cotang. A. 2, E. D. LEMMA. As the fum of the fines of two unequal arches is to their difference, fo is the tangent of half the Sum of thofe arches to the tangent of half their difference: and, as the fum of the co-fines is to their difference, fo is the co-tangent of half the fum of the arches to the tangent of half the difference of the fame arches. For, half the fum, of AB and AC: let the radii OD and OC be drawn, and alfo the chord CB, meeting OE in E and OA (produced) in P; draw ES parallel to AO, meeting CH in S, and EF and OK perpendicular to AO, and let the latter meet EC (produced) in I; laftly, draw QDK perpendicular to OD, meeting OA, OC and OI (produced) in Q, L and K. Because CD = BD, it is manifeft that OD is not only perpendicular to the chord BC, but bifects it in E; whence, alfo, EF bifects HG; and therefore CH+ BG2EF, and CH — BG — 2CS; also OG + OH = 2OF, and OGOH = 2HF: but 2EF (CH+ BG): 2CS (CH-BG) :: EF: CS:: EP: EC (by 14. 4.):: DQ (the tangent of AD): DL (the tangent of DC, by 20. 4.) And 2OF (OG + OH): 2HF (OGOH):: EI: EC :: DK (the co-tang. of AD): DL (the tang. DC). 2. E. D. In any fpherical triangle ABC it will be, as the co-tangent of half the fum of the two fides is to the tangent of half their difference, fo is the co-tangent of half the bafe to the tangent of the distance (DE) of the perpendicular from the middle of the bafe. *2 DEMON DEMONSTRATION. C EFD Since co-fine AC: co-fine BC :: co-fine AD co-fine BD (by Cor. to Theor. 2.) there fore, by compofition and B divifion, co-fine AC + co-fine BC co-fine AC co-fine BC co-fine AD + cofine BD: cofine AD- co-fine BD. But (by the preceding lemma) co-fine AC + co-fine BC: co-fine AC co-fine BC:: co-tang. AC+BC 2 .: tang. AC-BC 2 and co-fine AD + co-fine BD : co-fine AD co-fine BD:: co-tang. of AE DE - BD AC+BC 2 : tang, (AD+BD) 2); whence, by equality, co-tang. AC-BC DE. : tang. :: co-tang. AE: tang. 2 2 : tang. DE, and it is proved, in p. 13. that the tangents of any two arches are, inversely, as their co-tangents; it follows, therefore, that tang. DE; or, that the tangent of half the bafe, is to the tangent of half the fum of the fides, as the tangent of half the difference of the fides, to the tangent tangent of the distance of the perpendicular from the middle of the base. THEOREM VII. In any spherical triangle ABC, it will be, as the co-tangent of half the fum of the angles at the bafe, is to the tangent of half their difference, fo is the tangent of half the vertical angle, to the tangent of the angle which the perpendicular CD makes with the line CF bifecting the vertical angle. (See the preceding figure.) DEMONSTRATION. It will be (by Corol. to Theor. 3.) co-fine A: cofine B fine ACD: fine BCD; and therefore, cofine A+ co-fine B: co-fine A-co-fine B:: fine ACD+ fine BCD: fine ACD-fine BCD. (by the lemma) co-tang. B+A BA : tang. 2 But (co-fine A+ co-fine B: co-fine A-co-fine B:: fine ACD+fine BCD: fine ACD — finę BCD) tang. ACF: tang. DCF. 2. E. D. |