(892) (893) (894) (899) (900) is A = (895) the surface ABC + the surface ABC' 2 C, (896) the surface ABC + the surface A'BC2 A; = the lunary surface CABC' 2 C, and, by (887), (897) = the surface ABC + the surface A'BC' 2 B, for the sides BC and AB are by (892) supplements (898) of BC' and A'B; and the angle ABC is equal to the angle ABC'. The sum of (895), (896), and (897), A Fig.10 3 x the surface ABC + the surface A'BC But the surface of the hemisphere is, by (867), the surface ABC' + the surface A'BC' 360°; which, subtracted from (899), gives 2 x surface ABC = 2 A+ 2 B + 2 C-360° or surface ABC A+B+C 180°, as in (891). - 360° (901) 88. Theorem. The surface of a spherical polygon is equal to the excess of the sum of its angles over as (903) many times two right angles as it has sides minus two. Fig.11 E (902) Demonstration. Let ABCDE (fig. 11.) be the given polygon. Draw from the vertex A the arcs AC, AD, which divide it into as many triangles as it has sides minus two. By the preceding theorem (891), the sum of the surfaces of all these triangles (906) or the surface of the polygon is equal to the sum of all their angles diminished by as many times two right angles as there are triangles; that is, the surface (907) of the polygon is equal to the sum of all its angles diminished by as many times two right angles, as it has sides minus two, which agrees with (903). (904) (905) |