Theorem. A lune is to the surface of a sphere as the angle of the lune is to four right angles.

Data: A lune, as ACFD, whose angle is CAD, on the sphere whose center is 0.

Denote the lune by L and the surface

Suppose arc CD and BCEH have a common unit of measure, as CJ, contained in CD m times and in BCEH n times.

Beginning at C, divide BCEH into parts, each equal to the unit of measure CJ, and through the points of division and the poles, 4 and F, of this circumference pass great circles.

By §§ 689, 721, these circles divide the whole surface of the sphere into n equal lunes of which the given lune contains m.

Then,
Hence,

L:S=m: n.

L:S CAD: 4 rt. .

By the method of limits exemplified in § 223, the same may be proved, when arc CD and BCEH are incommensurable.

Therefore, etc.

:Q.E.D.

733. Cor. I. Let A denote the degrees in the angle of a lune. Then, L:SA: 360°.

Since, § 722, s contains 720 spherical degrees,

whence,

L : 720 = 4:360;

L

24;

that is, the numerical measure of a lune expressed in spherical degrees is twice the numerical measure of its angle expressed in angular degrees.

734. Cor. II. Lunes on the same sphere, or on equal spheres, are to each other as their angles.

Proposition XXV

735. On a sphere draw a spherical triangle and complete the great circles whose arcs are its sides. How many triangles having a common vertex with the given triangle occupy the surface of a hemisphere? Since the given triangle plus any one of the others is equivalent to a lune whose angle is equal to one of the angles of the given triangle, or to twice as many spherical degrees as that angle contains angular degrees, how does three times the given triangle plus the other three compare with twice the number of spherical degrees that there are angular degrees in the angles of the given triangle? How many spherical degrees are there in the four triangles occupying the surface of the hemisphere? Then, discover how the number of spherical degrees in the given triangle compares with the sum of the angular degrees in its angles less 180°, that is, with its spherical excess.

Theorem. A spherical triangle is equivalent to as many spherical degrees as there are angular degrees in its spherical excess.

Data: A spherical triangle, as ABC, whose spherical excess is E degrees.

To prove ▲ ABC E spherical degrees.

Proof. Complete the great circles whose B arcs are sides of ▲ ABC.

These circles divide the surface of the

sphere into eight spherical triangles, any

B

four of which having a common vertex, as 4, form the surface of

a hemisphere, whose measure is 360 spherical degrees.

§ 731, ▲ ABC + ▲ AB'C' ≈ a lune whose angle equals angle 4; .. § 733, ▲ ABC +▲ AB'C' 2 A spherical degrees. In like manner, ▲ ABC + ▲ AB'C ≈ 2 B spherical degrees, and ▲ ABC + ▲ ABC'≈2c spherical degrees. (3)

Adding (1), (2), and (3),

(1) (2)

3 ▲ ABC +▲ AB'C' + ▲ AB'C + ▲ ABC' ≈ 2 (A + B + C) sph. deg. But ▲ ABC÷▲ AB'C' +▲ AB'C +▲ ABC' = 360 spherical degrees; .. 2 AABC+360 spherical degrees≈2(A+B+C) spherical degrees; hence, ▲ ABC ≈ (A + B + C – 180) spherical degrees; that is, § 706, AABC E spherical degrees.

Q.E.D.

Proposition XXVI

736. Draw any spherical polygon and divide it into spherical tri angles by diagonals from any vertex. To how many spherical degrees is each triangle equivalent? How does the number of spherical degrees in the sum of the triangles compare with the number of angular degrees in the sum of the spherical excesses of the triangles? To how many spherical degrees, then, is any spherical polygon equivalent?

Theorem. Any spherical polygon is equivalent to as many spherical degrees as there are angular degrees in its spherical excess.

Data: Any spherical polygon, as ABCDF, whose spherical excess is E degrees.

Proof. Divide the polygon into spherical triangles by diagonals from any vertex, as A.

By § 735, each triangle is equivalent to as many spherical degrees as there are angular degrees in its spherical excess.

Hence, the polygon is equivalent to as many spherical degrees as there are angular degrees in the sum of the spherical excesses of the triangles; that is, § 707, in the spherical excess of the polygon. Hence, Therefore, etc.

ABCDFE spherical degrees.

Q.E.D.

737. Cor. The area of any spherical polygon is to the area of the surface of the sphere as the number which expresses its spherical excess is to 720.

Ex. 877. The angle of a lune is 40°. What part of the surface of the sphere is the lune ?

Ex. 878. What is the area of a spherical triangle whose angles are 85°, 120°, and 110°, if the radius of the sphere is 10dm?

Ex. 879. The area of the surface of a sphere is 160 sq. in.; the angles of a spherical triangle on this sphere are 93°, 117°, and 132°. What is the area of the triangle?

Ex. 880. Two spherical triangles on the same sphere, or on equal spheres, are equivalent, if the perimeters of their polar triangles are equal.

738. A solid bounded by a spherical polygon and the planes of its sides is called a Spherical Pyramid.

The center of the sphere is the vertex of the pyramid, and the spherical polygon is its base.

O-ABCDE is a spherical pyramid whose vertex is O and base ABCDE.

739. The portion of a sphere contained between two parallel planes is called a Spherical Segment.

E

A

B

The sections made by the parallel planes are the bases of the spherical segment; the perpendicular distance between its bases is the altitude of the segment.

If one of the parallel planes is tangent to the sphere, the segment is called a segment of one base.

740. The portion of a sphere bounded by a lune and the planes of its sides is called a Spherical Wedge, or Ungula.

741. The portion of a sphere generated by the revolution of a circular sector about a diameter of the circle is called a Spherical Sector.

The zone generated by the arc of the circular sector is the base of the spherical sector.

742. MNEFAB is a semicircle, AD and BC are lines from the semicircumference perpendicular to the diameter MN, and OE and

OF are radii. Then, if the semicircle is revolved about MN as an axis, it generates a sphere.

The arc AB generates a zone whose altitude is DC, and whose bases are the circumferences generated by the points 4 and B.

The figure ABCD generates a spherical segment whose altitude is DC and whose bases are the circles generated by AD and BC.

M

B

C

A

D

E

N

The arc BM generates a zone of one base, and the figure BCM a spherical segment of one base.

The circular sector OEF generates a spherical sector whose bounding surfaces are its base, the zone generated by the arc EF, and the conical surfaces generated by the radii OE and OF.

Proposition XXVII

743. Represent a polyhedron circumscribed about a sphere. If pyramids are formed having the faces of the polyhedron as bases and the center of the sphere as a common vertex, how will the altitudes of these pyramids compare with each other and with the radius of the sphere? What is the volume of each pyramid? What, then, is the volume of the sum of the pyramids? If the number of faces of the polyhedron is indefinitely increased, how will its volume compare with the volume of the sphere? To what, then, is the volume of a sphere equal?

Theorem. The volume of a sphere is equal to the product of its surface by one third of its radius.

Data: A sphere whose center is 0, surface s, and radius R.
Denote its volume by V.

Proof. Circumscribe about the sphere any polyhedron, as D-ABC, and denote its surface by s' and its volume by v'.

Form pyramids, as O-ABC, etc., having the faces of the polyhedron as bases and the center of the sphere as a common vertex. Then these pyramids will have a common altitude equal to R, and, § 560, the volume of each pyramid = its base R.

Now, if the number of pyramids is indefinitely increased by passing planes tangent to the sphere at the points where the edges of the pyramids cut the surface of the sphere,

v

s' approaches s as its limit; .. v' approaches V as its limit.