86. Lemma. If two triangles have an angle of the one equal to an angle of the other; and the sides which include the angle in one triangle are supple- (887) ments of those which include it in the other triangle; the sum of the surfaces of the two triangles is measured by double the included angle, ABC and DEF, the angles A' and D are equal, being both equal to A; A'B and DE are equal, being supplements of AB; and A'C and DF are (889) equal, being supplements of AC. It follows therefore from (861) and (876), that they are equal in surface. But A'BC and ABC compose the lunary surface ABCA' which is measured by 2 A. Therefore the (890) sum of ABC and DEF is also measured by 2 A. 87. Theorem. The surface of a spherical triangle is measured by the excess of the sum of its three (891) angles over two right angles or 180°. (892) (893) (894) (895) = A'Fig.10 the surface ABC + the surface ABC 2 C, (896) the surface ABC+ the surface A'BC= 2 A; (897) and, by (887), 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'. (899) (900) is The sum of (895), (896), and (897), 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 the surface ABC' + the surface A'BC' = 360°; which, subtracted from (899), gives or 2x surface ABC 2 A+2B+2 C-360° (901) surface ABC A+B+C-180°, as in (891). = (902) 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 (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). 1 |