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 2A X 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, 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 each be a semi-circumference.

B

C

A

E

Now, if we take away AD from both, we shall have D F equal to B A. 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 D E F, DA E is equivalent to the lune ADFE, whose angle is DA 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 ABC 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 ABC till they

BGFBID=2BX T, and CIH+CFE = 2CXT. 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 − 4 T; 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

D

A

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.

BOOK X.

THE THREE ROUND BODIES.

DEFINITIONS.

567. A CYLINDER is a solid, which may be described by the revolution of a rectangle turning about one of its sides, which remains immovable; as the solid described by the rectangle ABCD revolving about its side A B.

B

D

The BASES of the cylinder are the circles described by the sides, AC, BD, of the revolving rectangle, which are adjacent to the immovable side, A B.

The AXIS of the cylinder is the straight line joining the centres of its two bases; as the immovable line A B.

The CONVEX SURFACE of the cylinder is described by the side CD of the rectangle, opposite to the axis A B.

568. A CONE is a solid which may be described by the revolution of a rightangled triangle turning about one of its perpendicular sides, which remains immovable; as the solid described by the right-angled triangle ABC revolving about its perpendicular side A B.

The BASE of the cone is the circle described by the revolution of the side BC, which is perpendicular to the immovable side.

B

A