COROLLARY. B Hence it follows, that, in right angled spherical triangles ABC, DBC, having the same perD pendicular BC, the lines of the bases will be to each other, inversely, as the tangents of the angles at the bases : For s radius : finé AB :: tang. A : tang. BC? since radius : sine DB : : tang. D: tang. BCS we shall (by reasoning as in Cor. 1. Theor. 1.) have Şine AB : sine DB :: tang. D: tang. A. In any right-angled Spherical triangle it will be, as radius is to the co-fone of the bypothenuse, so 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: co tang. A. 2. E. D. LE M M A. As the sum of the fines of two unequat arches is to their difference, so is the tangent of half the Sum of those arches to the tangent of half their difference : and, as the sum of the co-lines is to their difference, so is the co-tangent of half the sum of the arches to the tangent of half the difference of the same arches. For, 1 L D For, let AB and R AC bethetwo pro posed arches, and Jet BG and CH be their fines, and OG and OH their co. Si fines : moreover, B В let the arch BC be equally divided in D, so that CD may be half the dif. H F GA Pe ference, and AD half the fum, of AB and AC: let the radii OD and OC be drawn, and also 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 pera pendicular to AO, and let the latter meet EC (produced) in l; lastly, draw QDK perpendicular to DD, meeting OA, OC and OI (produced) in Q, L and K. Because CD=BD, it is manifest that OD is not only perpendicular to the chord BC, but bisects it in É; whence, also, EF bisects HG; and therefore CH + BG = 2EF, and CH – BG = 2CS; also OG + OH = 20F, and OG --OH = 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 20F (OG + OH): 2HF (OG-OH):: EI: EC :: DK (the co-tang. of AD): DL (the tang. DC). 2. E. D. T HÈ 0 REM VI. In any spherical triangle ABC it will be, as the co-tangent of half the sum of the two fides is to the tangent of half their difference, so is the co-tångent of half the base to the tangent of the distance (DE) of the perpendicular from the middle of the base. DEMON *2 C E FD DEMONSTRATION. Since co-fine AC : co-fine BC :: co-fine AD : co-sine BD (by Cor. to Theor. 2.) there fore, by composition and B division, co-fine AC + co-sine BC : Co-sine AC co-fine BC :: co-fine AD + cosine BD : cofine AD - co-fine BD. But by the preceding lemma) co-fine AC + co-sine BC : co-fine AC co-fine BC :: co-tang. AC+BC AC-BC : tang and co-fine AD + co-fine BD : co-fine AD AD+BD co-fine BD:: co-tang. of AE AD - BD DE ; whence, by equality, co-tang. AC+BC AC-BC : tang :: Co-tang. AE: tang. DE. 2 2 (AD+BP) : tang (AD 2 2 COROLLARY. 2 2 Since the last proportion, by permutation, be AC+BC comes co-tang. : Co-tang. AE :: tang. AC-BC - : 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 AC + BC AC - BC tang. AE : tang. :: tang. : tang. DE; or, that the tangent of half the base, is to the tangent of half the sum of the sides, as the tangent of balf the difference of the sides, to the tangent 2 2 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 sum of the angles at tbe base, is to the tangent of half their difference, so is the tangent of half the vertical angle, to the tangent of the angle which the perpendicular CD makes with the line CF bise{ting the vertical angle. (See the preceding figure.) DEMONSTRATION. It will be (by Corol. to Theor. 3.) co-line A : cosine B :: sine ACD : fine BCD, and therefore, cofine A + co-sine B : co-line A - Co-sine B :: sine ACD + fine BCD: sine ACD - fine BCD. But B +A B-A (by the lemma) co-tang. : tang. (co-fine A + co-fine B : co-fine A - Co-fine B :: fine ACD + fine BCD : fine ACD — sinę BCD) tang. ACF : tang. DCF. 2. E. D. 2 2 B The solution of the cases of right-angled Spherical triangles. Cafe site leg Given Sought Solution. The hyp. The oppo- As radius : fine hyp. AC :: I JAC and one sine A : sine BC (by the forangle A BC mer part of Theor. 1.) The hyp. The adja- As radius : co-fine of A :: 2 JAC and one cent leg tang. AC : tang. AB (by the angle A AB latter part of Theor. 1.) The hyp. The other As radius : co-fine of AC 3 AC and one angle C :: tang. A : co-tang. C (by angle A Theor. 5.) The hyp. The other As co-fine AB : radius :: 4 AC and one leg BC co-fine AC:co-fine BC (by leg AB Theor. 2.) The hyp. | The oppo- As fine AC: radius :: fine 5 ACand one fite angle AB : fine C (by the former leg AB С part of Theor. 1.) The hyp. The adja- As tang. AC : tang. AB :: 6 AC and one cent angle radius : co-fine A (by leg AB A Theor. 1.) One leg The other As radius : sine AB :: tanAB and the leg BC gent A : tangent BC (by 7 adjacent angle A Theor. 4.) Case |