Area of a Spherical Triangle. are equilateral or equiangular with respect to each other, are equivalent. 468. Lemma. If two spherical 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 supplements 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. Proof. Let the triangles be ABC and DEF (fig. 189), in which A and D are equal; and AB and AC are respectively supplements of DE and DF. Produce AB and AC till they meet in A'. ABA' and ACA' are, by § 428, semicircumferences. In the triangles 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 AC and DF are equal, being supplements of AC. It follows, therefore, from § 467, that they are equal in surface. But A'BC and ABC compose the lunary surface ABCA' which is measured by 2 A. Therefore the sum of ABC and DEF is also measured by 2 A. 469. Theorem. The surface of a spherical triangle is measured by the excess of the sum of its three angles over two right angles, or 180°. Proof. Let ABC (fig. 190) be the given triangle. Produce AC to form the circumference ACA'C', also produce AB and BC to form the semicircumferences ABA' and CBC'. Then, by 465, the lunary surface CABC' =2 C, the lunary surface ABCA' =2 A, or Area of a Spherical Polygon. the surface ABC + the surface ABC' the surface ABC + the surface A'BC and, by § 468, the surface ABC + the surface ABC' for the sides BC and AB are supplements of BC' and A'B; and the angle ABC is equal to the angle ABC'. The sum of these three equations is 3 × the surface ABC + the surface A'BC = 360°; But the surface of the hemisphere is, by § 463, or the surface ABC: 470. Theorem. The surface of a spherical polygon is equal to the excess of the sum of its angles over as many times two right angles, as it has sides minus two. Proof. Let ABCDEK (fig. 191) be the given polygon. Draw from the vertex A the arcs AC, AD, &c., which divide it into as many triangles as it has sides minus two. By the preceding theorem, the sum of the surfaces of all these triangles, 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 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. Surface described by the revolution of a regular portion of a Polygon. 471. Theorem. If a portion ABCD (fig. 192) of a regular polygon, situated entirely upon the same side of a line FG drawn through the centre O of the polygon, revolve about FG as an axis, the surface generated by ABCD has for its measure the product of the circumference inscribed in the polygon by MQ, which is the altitude of this surface, or the part of the axis comprehended between the extreme perpendiculars AM, DQ. Proof. Let I be the middle of AB, Ol is the radius of the inscribed circle. Draw IK, BN, CP, perpendicular to FG, and AX perpendicular to BN. The measure of the surface described by AB is, by § 398, AB × circumference of which KI is radius, which circumference we will denote by circumf. KI. The triangles OIK, ABX are similar, since their sides are perpendicular to each other; whence, by § 178 and 234, AB: AX=OI: IK= circumf. OI: circumf. IK, or, since AX= MN, AB: MN= circumf. 01: circumf. IK; and, multiplying extremes and means, AB circumf. IK=MN × circumf. OI. Whence the area of the surface described by AB is the product of the circumference of the inscribed circle by the altitude MN. In like manner the area of the surface described by BC is the product of the circumference of the inscribed circle by the altitude NP; and that described by CD is the product of this circumference by PQ. Hence the area of the entire surface described by ABCD is the product of the circumference of the in Area of the Surface of the Sphere. scribed circle by the sum of the altitudes MN, NP, PQ ; that is, by the entire altitude MQ. 472. Corollary. If the axis FG passes through the opposite vertices F, G, the area of the surface described by the semipolygon FACG is the product of the circumference of the inscribed circle by the axis FG. 473. Corollary. If the sides of the polygon are infinitely small, the polygon becomes a circle, the entire surface generated is that of a sphere, of which the generating circle is a great circle; and the surface generated by the circular segment ABCD is a zone. Hence the area of the surface of a sphere is the product of its diameter by the circumference of a great circle. And, the area of a zone is the product of its altitude by the circumference of a great circle. 474. Corollary. Since the area of the great circle is, by 279, half the product of its radius by its circumference; or one fourth of the product of its diameter by its circumference, it is one fourth of the surface of the sphere; that is The surface of a sphere is equivalent to four great circles. 475. Corollary. If we denote by R the radius of the sphere, by C the circumference of a great circle, by S the surface of the sphere, and by the ratio of the circumference to the diameter, as in § 237; we have C 2 π Χ R S ១ лX RX 2 R = 4л × R2. 476. Corollary. If we denote in the same way, by R' and S' the radius and surface of a second sphere, we have Solidity of the Sphere. whence S: S'4 × R2 : 4 π × R22 = R2 : R12, that is, the surfaces of spheres are to each other as the squares of their radii. 477. Corollary. Zones upon the same sphere are to each other as their altitudes; and a zone is to the surface of its sphere as its altitude is to the diameter of the sphere. 478. Theorem. The solidity of a sphere is one third of the product of its surface by its radius. Proof. For the surface of the sphere may be considered as composed of infinitely small planes; and each of these planes may be considered to be the base of a pyramid, which has its vertex at the centre of the sphere, and, consequently, an altitude equal to the radius of the sphere. The sum of the solidities of these pyramids is, then, one third of the product of the sum of their bases by their common altitude, that is, the solidity of the sphere is one third of the product of its surface by its radius. 479. Corollary. In the same way, the base of a spherical pyramid or sector may be considered as composed of planes, and, therefore, the solidity of a spherical pyramid or sector is one third of the product of the polygon or zone, which serves as its base, by its radius. 480. Corollary. Spherical pyramids or sectors of the same sphere are to each other as their bases; and a spherical pyramid or sector is to the sphere of which it is a part, as its base to the surface of the sphere. 481. Corollary. Hence, by § 477, spherical sec |