PROPOSITION XXVIII. THEOREM.

829. The area of a lune is to the area of the surface of the sphere as the number of degrees in the angle of the lune is to 360.

also

Let ABEC be a lune, BCDF the great circle whose pole is A; let A denote the number of degrees in the angle of the lune, L the area of the lune, and S the area of the surface of the sphere.

Proof. The arc BC measures the A of the lune.

Hence, arc BC: circumference BCDF = A: 360.

§ 779

If BC and BCDF are commensurable, let their common measure be contained m times in BC, and n times in BCDF. Then arc BC: circumference BCDF = m : n.

.. A: 360 m: n.

§ 288

Pass arcs of great through the diameter AE and all the points of the division of BCDF. These arcs will divide the entire surface into n equal lunes, of which the lune ABEC will contain m.

If BC and BCDF are incommensurable, the theorem can be proved by the method of limits as in § 549.

Q.E.D.

830. COR. 1. The number of spherical degrees in a lune is equal to twice the number of angle degrees in the angle of the lune.

If L and S are expressed in spherical degrees,

§ 821

§ 829

831. COR. 2. The area of a lune is equal to one ninetieth of the area of a great circle multiplied by the number of degrees in the angle of the lune.

832. COR. 3. Two lunes on the same sphere or equal spheres have the same ratio as their angles.

833. COR. 4. Two lunes which have equal angles, but are situated on unequal spheres, have the same ratio as the squares of the radii of the spheres on which they are situated.

Ex. 759. Given the radius of a sphere 10 inches.

lune whose angle is 30°.

Find the area of a

Ex. 760. Given the diameter of a sphere 16 inches. Find the area of a lune whose angle is 75°.

PROPOSITION XXIX. THEOREM.

834. The area of a spherical triangle, expressed in spherical degrees, is numerically equal to the spherical excess of the triangle.

Let A, B, C denote the values of the angles of the spherical triangle ABC, and E the spherical excess.

To prove that the number of spherical degrees in AABC=E. Proof. Produce the sides of the AABC to complete circles. These circles divide the surface of the sphere into eight spherical triangles, of which any four having a common vertex, as A, form the surface of a hemisphere.

The AA'BC, AB'C' are symmetrical and equivalent. § 807
And
A ABC +A A'BC lune ABA'C.

Then

Put the ▲ AB'C' for its equivalent, the ▲ A'BC.
A ABC +A AB'C' lune ABA'C.

Also

A ABC+A AB'Clune BAB'C.

And

A ABC +A ABC' lune CAC'B.

Add and observe that in spherical degrees

360,

§ 821

A ABC+A AB'C' + ▲ AB'C+▲ ABC' =
ABA'C + BAB'C + CAC'B = 2 (A+B+ C). § 830

2 A ABC +3602 (A+B+C).

and

Then

835. COR. 1. The area of a spherical triangle is to the area of the surface of the sphere as the number which expresses its spherical excess is to 720.

For the number of spherical degrees in a spherical ▲ ABC is equal to E (§ 834), and the number of spherical degrees in S, the surface of the sphere, is equal to 720.

.. A ABC: S = E: 720.

§ 821

836. COR. 2. The area of a spherical triangle is equal to the area of a great circle multiplied by the number of degrees in E divided by one hundred eighty.

Ex. 761. What part of the surface of a sphere is a triangle whose angles are 120°, 100°, and 95°? What is its area in square inches, if the radius of the sphere is 6 inches ?

Ex. 762. Find the area of a spherical triangle whose angles are 100°, 120°, 140°, if the diameter of the sphere is 16 inches.

Ex. 763. If the radii of two spheres are 6 inches and 4 inches respectively, and the distance between their centres is 5 inches, what is the area of the circle of intersection of these spheres ?

Ex. 764. Find the radius of the circle determined in a sphere of 5 inches diameter by a plane 1 inch from the centre.

Ex. 765. If the radii of two concentric spheres are R and R', and if a plane is drawn tangent to the interior sphere, what is the area of the section made in the other sphere?

Ex. 766. Two points A and B are 8 inches apart. Find the locus in space of a point 5 inches from A and 7 inches from B.

Ex. 767. The radii of two parallel sections of the same sphere are a and b respectively, and the distance between these sections is d. Find the radius of the sphere.

PROPOSITION XXX. THEOREM.

837. If T denotes the number which expresses the sum of the angles of a spherical polygon of n sides, the area of the polygon expressed in spherical degrees is numerically equal to T-(n-2) 180.

Let ABCDE be a polygon of n sides.

To prove that the area of ABCDE expressed in spherical degrees is numerically equal to

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

These diagonals divide the polygon into n - 2 spherical A. The area of each triangle in spherical degrees is numerically equal to the sum of its angles minus 180.

§ 834 Hence, the sum of the areas of all the n 2 triangles expressed in spherical degrees is numerically equal to the sum of all their angles minus (n-2) 180.

Now the sum of the areas of the triangles is the area of the polygon, and the sum of the angles of the triangles is the sum of the angles of the polygon.

Therefore, the area of the polygon expressed in spherical degrees is numerically equal to T (n-2) 180.

Q. E.D.