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ever treated very loosely by authors on trigonometry: some of them speaking of sides as the supplements of angles, and scarcely any of them remarking which of the several triangles formed by the intersection of the arcs DE, EF, DF, is the one in question. Besides the triangle DEF, three others may be formed by the intersection of the semi- d. circles, and if the whole circles be considered, there will be seven other triangles 23) formed. But the proposition only obtains 2^ E with regard to the central triangle (of \ each hemisphere), which is distinguished D from the three others in this, that the two of angles A and F are situated on the same side of ac, the two B and E on the same side of Ac, and the two G and D on the same side of AB.
In every spherical triangle the following proportion obtains, viz. as four right angles (or 360°) to the surface of a
hemisphere ; or, as two right angles (or 180°) to a great circle of the sphere ; so is the excess of the three angles of the triangle above two right angles, to the area of the triangle.
Let Abcbe the spherical triangle. Complete one of its sides as BC into the circle boe F, which may be supposed to bound the upper hemisphere. Prolong also, at both ends, the two sides AB, Ac, until they form semicircles estimated from each angle, that is, until BAE = ABD = cAF= Acd=180°. Then will chr-180°–BFE ; and consequently the triangle AEF, on the anterior hemisphere, will be equal to the triangle bed on the opposite hemisphere. Putting m, m', to represent the surface of these triangles, p for that of the triangle BAF, q for that of cAE, and a for that of the proposed triangle ABc. Then a and m together (or their equal a and m togerther) make up the surface of a spheric lune comprehended between the two semicircles AcD, ABD, inclined in the angle A : a and p together make up the lune included between the semicircles car, cBF, making the angle c : a and q together make up the spheric lune included between the semicircles BCE, BAE, making the angle B. And the surface of each of these lunes, is to that of the hemisphere, as the angle made by the comprehending semicircles, to two
Cor. 1. Hence the excess of the three angles of any spherical triangle above two right angles, termed technically the spherical ercess, furnishes a correct measure of the surface of that triangle. Cor. 2. If or = 3.141593, and d the diameter of the - o sphere, then is ord’. A + e .*= the area of the spherical triangle.
Cor. 3. Since the length of the radius, in any circle, is equal to the length of 57.2957795 degrees, measured on the circumference of that circle ; if the spherical ercess be mulplied by 57.2957795, the product will express the surface of the triangle in square degreest.
Cor. 4. When a = 0, then A + b + c = 180°: and when a=#s, then A+B+c=540°. Consequently the sum of the three angles of a spherical triangle is always between 2 and 6 right angles ; which is another confirmation of th: 3.
Cor. 5. When two of the angles of a spherical triangle are right angles, the surface of the triangle varies with its third angle. And when a spherical triangle has three right angles, its surface is one-eighth of the surface of the sphere.
*This determination of the area of a spherical triangle is due to Albert Girard (who died about 1633). But the demonstration now commonly given of the rule was first published by Dr. Wallis. It was considered as a mere speculative truth, until General Roy, in 1787, employ. edit very judiciously in the great Trigonometrical Survey, to correct the errors of spherical angles. See Phil. Trans. vol. 80, and the next chapter of this volume. t Excess in degrees, i. e. excess * = area "+57 2957795. Excess' - area.” 23:00 57-2957795 area in sqm. feet x 3600 T 57.2957795 - (365.1546)” The log. of the factor to these square feet is 9.326737, which is the log. employed in art. 5, Schol. Prob. 8, ch. v. following. Vol. II. 6
Remark. Some of the uses of the spherical excess, in the more extensive geodesic operations, will be shown in the fol. lowing chapter. The mode of finding it, and thence the area when the three angles of a spherical triangle are given, is obvious enough; but it is often requisite to ascertain it by means of other data, as, when two sides and the included angle are given, or when all the three sides are given. In the former case, let a and b be the two sides, c the included angle, and cot on . cot *b -F cos c —in a L.
E the spherical excess: then is cot #E =
When the three sides a, b, c, are given, the spherical excess may be found by the following very elegant theorem, discovered by Simon Lhuillier :
tan le=v(anot; to.tant**.tant**.tan =##5. The investigation of these theorems would occupy more space than can be allotted to them in the present volume.
In every spherical polygon, or surface included by any number of intersecting great circles, the subjoined proportion obtains, viz. as four right angles, or 360°, to the surface of a hemisphere; or, as two right angles, or 180°, to a great circle of the sphere; so is the excess of the sum of the angles above the product of 180° and two less than the number of angles of the spherical polygon, to its area.
For, if the polygon be supposed to be divided into as many triangles as it has sides, by great circles drawn from all the angles through any point within it, forming at that point the vertical angles of all the triangles. Then, by th: 5, it will be as 360° : }s :: A+B+c-180°: its area. Therefore, putting P for the sum of all the angles of the polygon, n for their number, and v for the sum of all the vertical angles of its constituent triangles, it will be, by composition, as 360°: }s :: P + v — 180°. n : surface of the polygon. But v is manifestly equal to 360° or 180' x 2. Therefore,
the polygon. Q. E. D. Cor. 1. If or and d represent the same quantities as in theor. 5, cor. 2, then the surface of the polygon will be ex