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. ADAB AĎ + AB. T. : ; that is, 2 2 As the line of the sum of the two perpendiculars, is to the fine of their difference; so is the tangent of half the sum of the two bases, to the tangent of balf their difference. PROP. XXVI. The produet of the square of radius and the co-fine of the base of any Spherical triangle ABC, is equal to the product of the fines of the two sides and the co-fine of the vertical angle, together with the produkt of radius and the co-fines of the same sides. For let AD be perpendicular to BC; then, since co-f. BD = S.CBXS.CD+co-f.CBXco-f.CD rad. (by Cor. 1. to Prop. 2.) it is D evident, that cof. BD: co-f. CD :: A B S.CBXS.CD+co f.CBXco-f.CD rad. S. CB x S. CD : co-f. CD:: + co-f. CB : rad. (by Co-f. CD rad. mult. each term by S. CD T.CD But. :) 00-8.CD CO-L.CD rad. S. CB x T. CD therefore our proportion will be rad. co-f. + co-f. CB : rad. (: : co-f. BD: co-f. CD):: 00-1. AB : co-f. AC (by Corol, to Theor. 2.) whence, by multiplying means and extremes, we have co-f. . S. CB x co-f. AC X T. CD AB x radius + rad. co-f. AC X co-f. BC. But (by Theor. 1.) radius : Co-f, C X T. AC co-f. C::T. AC: T. CD – rad. co-f. C X S. AC (by Corol. 1. Prop. 1.) which last co-f. AC being substituted for its equal, we shall have, co-f. AB X rad. = S. CA X S. CB X co-f. C + rad. co-f. AC X Co-f. BC; from whence, if each term be multiplied by radius, the truth of the proposition will appear manifest. There is another way of demonstrating this proposition, from the orthographic projection of the sphere; but that is a subject which neither room, nor inclination, will permit me to treat of here. If AE be the sum, and AF the difference, of the two sides of a Spherical triangle ABC, and V be put to denote the versed sine of the vertical angle, and R the radius ; then will V R?* CO-. AF-C0-f. AB S. AC X S. BC Co-f. AB co-7. AE S. AC X S. BC It appears from the last E Prop. that co-f. AB x R is =CO-. AC X co-f. BC + S. AC X S. BC X co-f. C Rin which, for co-f. C let its equal R - V be substituted, and then we shall have co-fine AB X R= co-fine AC x co-fine BC + fine AC x sine BC B S. ACX S. BC XV : but the R sum of the two former of the three last terms is = co-f. AF XR (by Cor. 1. to Prop. 2.); therefore it will be co-s. AB X R = co-f. AF XR S. AC X S. BC X V and consequently V = R R? x co-f. AF Co-f. AB Which is the first S. AC X S. BC case. Again, because S. AC X S. BC is = 4RX Co-f. AF — Co-f. AE (by Corol. 3. to Prop. 2.) we 2R x co-f. AF-CO-f. AB shall, also, have V = co-f. AF — Co-f. AE Which is the second case. Moreover, since RX Co-f. AF-CO-. AB is - S. AB + AF X S. 2 2 AB AF (by the same) it follows that V is likewise 2R x S. AB + AF X S. AB — AF S. AC X S. BC 2. E. D. COROL Hence, because IR X V is = the square of the sine of C (by Prop. 1.) it follows that fq. S. ;C= R? X S. ŽAB + AF X S. ZAB - AF S. AC X S. BC From whence we have the following theorem, for solving the 11th case of oblique triangles, where the three sides are given, to find an angle. As the rectangle of the fines of the two fides, ina cluding the proposed angle, is to the rečtangle under tbe fines, of balf the base plus balf the difference of the fides, and half the base minus balf the difference of the sides ; fo is the square of radius, to the square of the fine of half the required angle. COROLLARY 2. Roxco-f.AF-co-f.AB Moreover, because V is = S. AC X S. BC we shall have R: S. AC X S. BC :: V: co-f. AF - CO-S. AB; which gives the following theorem, for finding a side, when the opposite angle, and the other two sides, are given. As the Square of radius, is to the rectangle of the fines of the two fides including the given angle; so is the versed fine of that angle, to the difference of the co-fines (or versed fines) of the difference of those fides, and the side required. COROL COROLLARY 3. Lastly, because V = 2R x Co-L AF CO-. AB co-f. AF - Co-f. AE we shall, by transforming the equation, and putting W for (2R - V) the versed fine of BCE (the supplement of the vertical angle) have co-f. 2R X co-f, AB W X Co-I. AF and the V 2R X co-f. ABV x co-f. AE co-fine AF = W From whence the sides themselves may be determined, when their sum, or difference, is given, with the base and vertical angle. AE= ERRATUM. |