AOC COB-AOB DPF+FPE-DPE, AOCCOB or, АВС Hence the two triangles A B C, D E F are equivalent. 555. Cor. 1. If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they are equivalent. For in that case the triangles will be mutually equilateral. 556. Cor. 2. Hence, also, if two triangles on the same sphere, or on equal spheres, have two sides and the included angle, or have two angles and the included side, in the one equal to those in the other, the two triangles are equivalent. 557. Scholium. The poles O and P might lie within the triangles A B C, DEF; in which case it would be requisite to add the three triangles DPF, FPE, DPE together, to form the triangle DEF; and in like manner to add the three triangles A O C, COB, AOB together, to form the triangle ABC; in all other respects the demonstration would be the same. PROPOSITION XVIII.-THEOREM. 558. The area of a lune is to the surface of the sphere as the angle of the lune is to four right angles, or as the arc which measures that angle is to the circumference. Let A CBD be a lune upon a sphere whose diameter is AB; then will the area of the lune be to the surface of the sphere as the angle DOC to four right angles, or as the arc DC to the circumference of a great circle. For, suppose the arc CD to be to the circumference CDEF C A E D B in the ratio of two whole numbers, as 5 to 48, for example. Then, if the circumference A E D B Hence, the whole sphere must contain 96 of these triangles, and the lune A CBD 10 of them; consequently, the lune is to the sphere as 10 is to 96, or as 5 to 48; that is, as the arc CD is to the circumference. If the arc CD is not commensurable with the circumference, it may still be shown, by a mode of reasoning exemplified in Prop. XVI. Bk. III., that the lune is to the sphere as CD is to the circumference. 559. Cor. 1. Two lunes on the same sphere, or on equal spheres, are to each other as the angles included between their planes. 560. Cor. 2. It has been shown that the whole surface of the sphere is equal to eight quadrantal triangles (Prop. X. Sch.). Hence, if the area of a quadrantal triangle be represented by T, the surface of the sphere will be represented by 8T. Now, if the right angle be assumed as unity, and the angle of the lune be represented by A, we have, Area of the lune: 8T:: A : 4, which gives the area of lune equal to 2 A × T. 561. Cor. 3. The spherical ungula included by the planes AC B, A D B, is to the whole sphere as the angle DOC is to four right angles. For, the lunes being equal, the spherical ungulas will also be equal; hence, two spherical ungulas on the same sphere, or on equal spheres, are to each other as the angles included between their planes. PROPOSITION XIX.-THEOREM. 562. If two great circles intersect each other on the surface of a hemisphere, the sum of the opposite triangles thus formed is equivalent to a lune, whose angle is equal to the angle formed by the circles. Let the great circles BAD, CAE intersect on the surface of a hemisphere, ABCDE; then will the sum of the opposite triangles, BAC, DAE, B be equal to a lune whose angle is DAE. For, produce the arcs AD, AE till they meet in F; and the arcs BAD, ADF will F A E each be a semi-circumference. Now, if we take away AD from both, we shall have DF equal to BA. For a like reason, we have EF equal to CA. DE is equal to BC. Hence, the two triangles BAC, DEF are mutually equilateral; therefore they are equivalent (Prop. XVII.). But the sum of the triangles DEF, DA E is equivalent to the lune A D F E, whose angle is D A E. PROPOSITION XX.-THEOREM. 563. The area of a spherical triangle is equal to the excess of the sum of its three angles above two right angles, multiplied by the quadrantal triangle. Let A B C be a spherical triangle; its area is equal to the excess of the sum of its angles, A, B, C, above two right angles multiplied by the quadrantal triangle. For produce the sides of the triangle A B C till they BGF+BID=2BX T, and CIH+CFE=2CX T. But the sum of these six triangles exceeds the hemisphere by twice the triangle ABC; and the hemisphere is represented by 4T; consequently, twice the triangle ABC is equivalent to 2A XT2B × T+2C × T − 4T; therefore, once the triangle A B C is equivalent to (A+B+C-2) × T. Hence the area of a spherical triangle is equal to the excess of the sum of its three angles above two right angles multiplied by the quadrantal triangle. 564. Cor. If the sum of the three angles of a spherical triangle is equal to three right angles, its area is equal to the quadrantal triangle, or to an eighth part of the surface of the sphere; if the sum is equal to four right angles, the area of the triangle is equal to two quadrantal triangles, or to a fourth part of the surface of the sphere, &c. PROPOSITION XXI.THEOREM. 565. The area of a spherical polygon is equal to the excess of the sum of all its angles above two right angles taken as many times as the polygon has sides, less two, multiplied by the quadrantal triangle. E A C B Let ABCDE be any spherical polygon. From one of the vertices, A, draw the arcs AC, AD to the opposite vertices; the polygon will be divided into as many spherical triangles as it has sides less two. But the area of each of these triangles is equal to the excess of the sum of its three angles above two right angles multiplied by the quadrantal triangle (Prop. XX.); and the sum of the angles in all the triangles is evidently the same as that of all the angles in the polygon; hence the area of the polygon A B C D E is equal to the excess of the sum of all its angles above two right angles taken as many times as the polygon has sides, less two, multiplied by the quadrantal triangle. 566. Cor. If the sum of all the angles of a spherical polygon be denoted by S, the number of sides by n, the quadrantal triangle by T, and the right angle be regarded as unity, the area of the polygon will be expressed by S — 2 (n−2) × T = (S — 2 n + 4) × T. |