(266.) Problem. To express the surface of a spherical triangle, in terms of any two of its sides and of the included angle. If sin 1". be taken for an unit, then, the same notation being used as in the preceding article, and S, S', S", being put for the sides of the spherical triangle, T = A+ A+ A" - 180 (Art. 265.) -cot T=tan ž ( A + A'+A′′) (Introd. 13. 19.) (S'-S") tan A cos(S'+S")+cot A cos (267.) COR. Hence, two spherical triangles are equal to one another, if their vertical angles be equal, and if the tangents of the halves of the sides containing the equal angles, be reciprocally proportionals. And, by * By consulting a book of logarithms, it will be found that the logarithm of sin 1" is, also, the logarithm of П 180 means of this property, on a given sphere a spherical triangle may be described, on a given arch, that shall be equal to a given triangle, of the sphere, and shall have one of its angles equal to the angle of the given triangle. Also, an isosceles spherical triangle may be described, which shall be equal to a given triangle, and shall have its vertical angle, equal to the vertical angle of that given triangle. PROP. III. (268.) Problem. To express the surface of a spherical triangle, in terms of its sides. Let the same notation be used as in Art. 265. 266. and let P be put for the semi-sum of the three sides of the triangle. If, therefore, the square root be taken of the product of the two last equations, 2 [sin P sin (P— S) sin (P— S′) sin (P – S′′)] (Introd. 15.) sin S sin S" cot ... cot. T= S' cot S" + cos A (Art. 266.) 1 + cos S+cos S' + cos S" 2[sin P sin (P-S) sin (P-S') sin (P-S")]' (269.) COR. If the spherical triangle (T) be equilateral, (270.) SCHOLIUM. The following remarkable expression for the surface (T) of a spherical triangle, which may easily be deduced from what has been premised, was first given by Lhuillier; This form is manifestly analogous to the equation which exhibits the surface of a plane rectilineal triangle, in terms of its sides, and which might be deduced from the spherical form, by supposing the sphere's radius to be infinitely great. It serves, also, to shew, that when the sides of a spherical triangle are very small, compared with the radius of the sphere, its surface is nearly equal to that of a plane rectilineal triangle, contained by sides that are equal, in length, to the sides of the spherical triangle, each to each. For (Introd. 35.) a circular arch, when it is very small, in comparison of the radius, may be taken for its tangent, without any considerable error: and, in that case, Lhuillier's equation becomes & T= √[P. (P-S). (P-S'). (P-S")], or T = [P (P-S). (P-S"). (P-S")] ; which also (Introd. 34.) expresses the surface of a plane rectilineal triangle, of which the sides are S, S' and S". PROP. IV. (271.) Problem. Two sides of a quadrilateral spherical figure being at right angles to its base, to express in terms of those two sides, and of the base, the surface of the figure; all its sides being arches of great circles of the sphere. Let the two sides BD, CE, of the quadrilateral spherical figure BDEC, which is bounded by arches of great circles of the sphere, be at right angles to the base DE: It is required to express the surface of the figure BDEC, in terms of the arches BD, CE and DE. Produce BD and EC, until they meet in A: then (Art. 51. 36.) DA and EA are quadrants; and (Art. 54.) the side DE is the measure of the spherical angle A. But (Art. 265.) if () be taken be taken as an unit, and if Q 180. Δ ABC = LA + ▲ B+ ▲ C−180; .. A ADE - ▲ ABC=Q= 4 B+ 4 C-180; · · | Q = 1⁄2 (B + C) If, then, B be put for the base, and H, H', for the two sides that are at right angles to it, (272.) SCHOLIUM. The problem, which has been solved in Art. 271, may often be advantageously employed in determining the number of square miles contained in any large tract of the Earth's surface. For, when the latitudes, and the longitudes, of several points, in the boundary of that region, are known, the whole space may be very conveniently divided into quadrilateral spherical figures, having, each of them, two sides perpendicular to the base; and the calculation may then be carried on with great facility. Thus, let BFGC be the space to be computed; let the points B, F be in the circle of latitude ABD; and |