EXAMPLES. 1. Given in the spherical triangle ABC (figs. 4. and 5.), a = 149°, c = 49°, and B = 88°; to find the angles A and C. 2. Given in the spherical triangle ABC (figs. 4. 79. Theorem. Two spherical triangles have all (861) their sides and angles respectively equal in either of the following cases; First. When they have two sides and the included angle respectively equal. Secondly. When they have one side and the two adjacent angles respectively equal. Thirdly. When they have their sides respectively equal. Fourthly. When they have their angles respectively equal. Demonstration. These propositions are deduced at once from the fact, that the solutions given in articles 36, 38, 46, and 65 led but to one triangle, which (862) can solve the problem; either, when two sides and the included angle, when a side and the two adjacent angles, when the three sides or the three angles are given. 80. Theorem. Two spherical right triangles have (863) all their sides and angles equal in the following cases not included in the preceding theorem. First. When they have the hypothenuse and one of the angles respectively equal. Secondly. When they have the hypothenuse and one of the legs respectively equal. Demonstration. This theorem may be proved in (864) the same way as the preceding one by a reference to articles 19 and 22. 81. Scholium. The case in which the hypothe(865) nuse is equal to 90° is, by articles 21 and 24, an exception to the preceding theorem. CHAPTER IV. Surfaces of Spherical Triangles. 82. All spherical surfaces vary with the radius of (866) the sphere to which they belong. Consequently some one spherical surface must be assumed as a standard of measure to which they may be referred. We shall assume the surface of the hemisphere as this standard, and shall suppose it to be divided into 360 equal (867) parts, which we shall call degrees of surface. These degrees may be again subdivided into minutes and (868) seconds. So that 1° of surface is of the surface of the hemisphere, and 35° of surface are of the surface of the hemisphere. (869) 83. Definition. A lunary surface is a part of the surface of a sphere comprehended between two semicircumferences of great circles, which terminate in a common diameter. (870) 84. Theorem. A lunary surface is measured by double the angle included between its sides. Demonstration. Let the lunary Fig.6 surface be AMNA' (fig. 6.), of which the angle is A. Then, if the semicircumference e! M N Α (871) AMA' were to depart from coincidence with ANA', and turn round on the diameter AA' till the angle A (872) had become equal to 180°; it would have passed over the surface of the hemisphere. And as the sphere is symmetrical, the angle A must increase proportionally with the lunary surface. The lunary surface AMNA' (873) is therefore the same part of the hemisphere, which the angle A is of 180°, or which 2 A is of 360°, or the surface AMNA' 2 A surface of hemisphere ; 360° (874) that is, 2 A is equal to the number of degrees of surface in AMNA' or is the measure of its surface, as in (875) (870). 85. Theorem. Two spherical triangles are equal (876) in surface, when their sides and angles are equal. Demonstration. If these triangles cannot be directly applied to each other, they must be situated as are ABC and DEF in (figs. 7. and 8.); and it is only in this case that any demonstration is required. Draw the arcs MP and NP perpendicular to the middle of the sides BC and AC. From P draw PB, PC, and PA. As the right triangles PMB and PMC have their two legs respec(879) tively equal, they must, by (861), give PB equal to PC. and in the same way it may be proved that Again, draw ER and FR, mak(881) ing the angles FER and EFR re spectively equal to PBC and PCB, and join RD. The triangles ERF and BCP have the side BC (882) equal to EF and the two adjacent angles equal, their other sides and angles are therefore, by (861), equal: The triangles APC and DRF have the sides DF and FR equal to AC and PC, and the included angle (883) ACP ACB — PCB = = DFE-EFR = DFR; their other sides and angles are therefore equal, by (861). (884) The triangles ABP and DER having their sides equal, must also have their angles equal by (861). These different triangles can also be respectively (885) applied to each other, because they are, by (836) and (861), isosceles, and they are therefore equal in surface. The given triangle ABC being then, in (fig. 7.), the (886) sum of the three partial triangles ABP, ACP, and BCP, and in (fig. 8.) the difference between ABP and the sum of the other two, must, as in (876), be equal to the triangle DEF, which, in (fig. 7.), is the sum of the three partial triangles DER, DFR, and EFR, and in (fig. 8.) is the difference between DER and the sum of the other two triangles. |