but one, &c. we fhall have, EF: 2 AG (AC-BC, Jee Prop. 13.): co-tang. EFI: tang. Q; and as co & F: radius: co-tang. AG: AE: 2AG (AÇBC): 2AE (ABQ, E. D. :: PROP. XX. The hypothenufe AC, and the fum, or difference, of the legs AB, BC, of a right-angled spherical triangle ABC, being given, to determine the triangle. Let AE be the fum, and AF the difference of the two legs. Because, radius co-f. AB :: co-f. BC: co-f AC (by Theor. 2.) B E therefore, co-f. AB x co-f. BC rad. x co› f. AC; but the former of thefe is rad. x co-1. AE+co-f. AF (by Corol. 3. to Prop. 2.); therefore 2 x co-f. AC co-f. AE + co-f. AF. Whence it appears, that, if from twice the co-fine of the hypothenufe, the co-fine of the given fum, or difference, of the legs, be fubtracted, the remainder will be the co-fine of an arch, which added to the faid fum, or difference, gives the double of the greater leg required. COROLLARY. Hence, if the two legs be fuppofed equal to each other (or the given difference o), then will the co-fine of the double of each, be equal to twice the co-fine of the hypothenufe minus the radius. PROP, XXI. One leg BC and the fun, or difference, of the bypothenufe and the other leg AB being given, to determine the bypothenufe (fee the last figure.) Since rad.: co-fine BC:: co-fine AB: co-fine AC (by Theor. 2.), it will be (by comp. and div.) radius +co-fine BC: rad.-co-f. BC:: co-f. AB + co-f AC: co-f. AB-co-f. AC. But the radius may be confidered as the fine of an arch of 90°, or the co-fine of o: and, therefore, fince (by the lemma in p. 30.) co-fine o + co-fine BC: co-f. o-co-f, BC +0 BC-o 2 BC::co-tang : tang. 2 ; and, co fine AB+co-f. AC: co-f. AB-co-f. AC:: co BC BC equality, that 2 2 co-tang. -tang. : : co-tang : tang. 2 AC-AB tang. of balf the given leg, is to its tangent; fo is the co-tang. of half the fum of the hypothenuje and the other leg, to the tangent of half their difference. PROP. XXII. The angle at the base and the fum, or difference, of the hypothenuse and bafe, of a right-angled fphe rical triangle being given, to determine the triangle, First, 1 B First, it will be, rad. : co-f. A: T. AC: T. AB (by Theor. 1.) and therefore rad. + co-f. A : rad. co-f. A: T. AC + T. AB: T. ACT. AB: whence, by arguing as in the laft Prop. it will appear, that, co-tang. A tang. A:: rad. + co-f. A: rad. co-f. A (:: T. AC + T. AB: T. AC-T. AB)::S. AC+ AB: S. AC-AB (by Prop. 4.). Hence it appears, that, As the co-tangent of half the given angle, is to its tangent; fo is the fine of the fum of the hypothenuse and adjacent leg, to the fine of their difference. 'PROP. XXIII. The hypothenufe AC and the fum, or difference, of the two adjacent angles being given, to find the angles. Let EC be perpendicular to BC; and then it will be, rad. co-f. AC :: T. A: E B AC::S. A+ ACE: S. A-ACE; whereof the two laft terms, by fubftituting 90°- ACB for ACE, will become S. 90°+ A-ACB (co-f. ACB -A) and S.A+ACB-90° refpectively. Whence it appears, that, As the co-tangent of half the bypothenufe, is to its tangent; fo is the co-fine of the difference of the angles at the hypothenufe, to the fine of the excess of their fum above a right-angle. COROL COROLLARY. Hence, if the angles be supposed equal, then it will be, as radius: tang. AC: tang. AC: fin. 2A-90°. PROP. XXIV. In two right-angled Spherical triangles ABC, ADE, baring one angle A common, let there be given the two perpendiculars BC, DE and the fum, or difference, of the hypothenufes AC, AE, to determine the triangles. gent of balf the fum of the two perpendiculars, is to the tangent of half their difference; fe is the tangent of half the fum of the two hypothenuses, to the tangent of half their difference. PROP. XXV. In two right-angled spherical triangles ABC, ADE, having the fame angle A at the base, let there be given the two perpendiculars BC, DE and the fum or difference of the bases AB, AD, to determine the bafes (Jee the preceding figure.) Since Since T. DE T. BC: S. AD: S. AB (by Theor. 4. and equality); therefore is T. DE + T. BC: T. DE-T. BC:: S. AD + S. AB: S. AD -S. AB; whence, (by Prop. 4. and the lemma in p. 30.) it will be, S. DE+ BC: S. DE—BC :: T. AD + AB AD-AB 2 : T. 2 ; that is, As the fine of the fum of the two perpendiculars, is to the fine of their difference; fo is the tangent of half the fum of the two bafes, to the tangent of half their difference. PROP. XXVI; The product of the fquare of radius and the co-fine of the bafe of any spherical triangle ABC, is equal to the product of the fines of the two fides and the co fine of the vertical angle, together with the product of radius and the co-fines of the fame fides. For let AD be perpendi cular to BC; then, fince co-f BD = S.CBxS.CD+co-f.CB x co-f.CD |