EXERCISES.

1. If, from the extremities of one side of a spherical triangle, two arcs of great circles be drawn to a point within the triangle, the sum of these arcs is less than the sum of the other two sides of the triangle.

2. On the same sphere, or on equal spheres, if two spherical triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first will be greater than the third side of the second.

3. To draw an arc perpendicular to a given spherical arc, from a given point without it.

4. At a given point in a given arc, to construct a spherical angle equal to a given spherical angle.

5. To inscribe a circle in a given spherical triangle.

6. Given a spherical triangle whose sides are 60°, 80°, and 100°; find the angles of its polar triangle.

7. The volume of a pyramid is 200 cubic feet; find the volume of a similar pyramid which is three times as high.

8. Find the centre of a sphere whose surface shall pass through three given points, and shall touch a given plane.

9. Find the centre of a sphere whose surface shall pass through three given points, and shall also touch the surface of a given sphere.

10. Find the centre of a sphere whose surface shall touch two given planes, and also pass through two given points which lie between the planes.

761. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle.

Let ABCDE be the circumference of a great circle, and AD the diameter, and OA the radius of a sphere.

We are to prove

=

surface of sphere A D X 2 O A.

П

Let the semicircle and any regular inscribed semi-polygon revolve together about the diameter A D.

The semi-circumference will generate the surface of the sphere,

and the semi-perimeter a surface equal to the sum of the surfaces generated by the sides A B, BC, CD, etc.

Draw from the centre 0, Is OH, OI and OK to the chords AB, BC, CD, etc.

These bisect the chords and are equal;

§ 185

§ 759

Adding, and observing that OH, OI and O K are equal,

=

area A B C D (A P+PR+ RD) X 2 OH.

.. area A B C D = ADX 2π OH.

π

Now, if the number of sides of the regular inscribed semipolygon be indefinitely increased, the surface generated by the semi-perimeter will approach the surface of the sphere as its limit, and OH will approach OA as its limit.

Hence the

762. COROLLARY 1. If R denote the radius of the sphere, then AD will equal 2 R, and OA will equal R. surface of a sphere equals 2 R X 2 T R = 4 π R2.

763. COR. 2. Since the area of a great circle of a sphere is equal to π R2 (§ 381), and the area of the surface of a sphere is equal to 4 π R2, the surface of a sphere is equal to four great circles.

764. COR. 3. If we denote the surfaces of two spheres by S and S', and their radii by R and R', we have S: S':: 4 π R2 : 4 π R2, or S: S':: R2: R2; that is, the surfaces of two spheres have the same ratio as the squares on their radii.

sphere is equivalent to a circle whose radius is equal to the diameter

of the sphere.

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

A

E

Let L denote the lune ABEC whose angle is A; S, the surface of the sphere; and BCDF, a great circle whose pole is A.

Now the arc BC measures the ZA of the lune;
and the circumference B C D F measures 4 rt. s.

§ 714

CASE I. · If BC and B C D F be commensurable.

Find a common measure of BC and B C D F.

Suppose this common measure to be contained in BC 3 times, and in BC D F 25 times.

Pass arcs of great through A and these points of division. The entire surface will be divided into 25 equal lunes, of which lune L will contain 3.

CASE II. If BC and BCDF be incommensurable,

the proposition can be proved by the method of limits, as employed in § 201.

Q. E. D.

767. COROLLARY. If we denote the surface of the tri-rectangular triangle by T, the surface of the whole sphere will be 8 T (§ 758), and if we denote the surface of the lune by L, and its angle by A, the unit of the angle being a right angle, we shall

And if we take the tri-rectangular triangle as the unit of surface in comparing surfaces on the same sphere, we shall have L = 2 A. That is, if a right angle be the unit of angles and the tri-rectangular triangle be the unit of spherical surfaces, the area of a lune is expressed by twice its angle.

768. SCHOLIUM. We may also obtain the area of a lune whose angle is known, on a given sphere, by finding the area of the sphere, and multiplying this area by the ratio of the angle of the lune, expressed in degrees, to 360°. Thus, if the angle of the lune be 60°, the area of the lune will be

of the area of the sphere.

Ex. 1. Given the radius of a sphere is 10 feet; find the area of a lune whose angle is 30°.

2. Given the diameter of a sphere is 16 feet; find the area of a lune whose angle is 75°.

3. Given the diameter of a sphere is 20 inches; find the entire surface of its circumscribed cylinder; and of its circumscribed cone, the vertical angle of the cone being 60°.