tion will be S. CBX T. CD rad. +co-s. CB: rad. (: : co-s. BD : co-s. CD): : co-s. AB : co-s. AC (by Cor. to Theor. 2.); whence, by multiplying means and extremes, we have co-s. S. CB x co-s. AC x T. CD AB x radius = rad. + co-s. ACX co-s. BC. But (by Theor. 1.) radius: co-s. C:: T. AC: T. Prop. 1. p. 14.), which last being substituted for its equal, we shall have, co-s. AB x rad. = S. CA × S. CB x co-s. C rad. +co-s. AC x co-s. 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. PROP. XXVII. If AE be the sum, and AF the difference, of the two sides of a spherical triangle ABC (fig. 39.), and V be put to denote the versed sine of the vertical angle, and R R2 x co-s. AF the radius; then will V = Co-s. AB S. AC X S. BC 2R × S. AB + ¦AF × S. JAB — AF S. AC × S. BC It appears from the last Prop. (fig. 40.) that co-s. ABXR is S. AC x S. BC x co-s. C R ; in co-s. AC x co-s. BC + which, for co-s. C let its equal R-V be substituted, and then we shall have co-sine AB x R = co-sine AC x coS. ACX S. BC × V sine BC+sine AC x sine BC R : but the sum of the two former of the three last terms is = co-s. AFR (by Cor. 1. to Prop. 2.); therefore it will be co-s. AB x R = co-s. AF × R. and consequently V = S. AC X S. BC × V R is the first case. Again, because S. AC x S. BC is = 1R X co-s. AF also have V = co-s. AE (by Cor. 3. to Prop. 2.), we shall 2R x co-s. AF CO-s. AB second case. Moreover, since Rx co-s. AF. Which is the - CO-s. AB Hence, because R x V is the square of the sine of C (by Prop. 1.), it follows that sq. S. C = R2 × S. 1AB + 1AF × S. ¦AB — AF 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 sines of the two sides, including the proposed angle, is to the rectangle under the sines of half the base plus half the difference of the sides, and half the base minus half the difference of the sides, so is the square of radius to the square of the sine of half the required angle. COROLLARY II. Moreover, because V = R2 x co-s. AF shall have R2: S. AC × S. BC : : V : co-s. 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 sines of the two sides including the given angle, so is the versed sine of that angle to the difference of the co-sines (or versed sines) of the difference of those sides, and the side required. CO-s. AE shall, by transforming the equation, and putting W for (2R-V) the versed sine of BCE (the supplement of the vertical angle), have co-sine AE = 2Rx co-s. AB- W x co-s. AF V and the co-s. AF = From whence the sides 2R x co-s. AB-V x co-s. AE W themselves may be determined, when their sum or difference is given, with the base and vertical angle. co-s. AB-co-s. AE; whence, by multiplying both sides ACB × S. AC × S. BC, by Prop. 1. and Cor. 3. Prop. 2.) = R2 × R × co-s. AB x S. co-s. AE R2 × S. AB+ AE 2 (by Prop. 3.) R2 x S. 2 the following analogy; As the rectangle of the sines of the two sides is to the square of the radius, so is the rectangle of the sines of half the sum of the three sides, and of the excess of that half sum above the base, to the square of the co-sine of half the vertical angle. |