equal to its symmetrical, and has equal angles opposite the equal sides (594).

773. Corollary.-The radius being the same, two spherical triangles are equal,

1st. When they have two sides and the included angle of the one respectively equal to those parts of the other, and similarly arranged;

2d. When they have one side and the adjacent angles of the one respectively equal to those parts of the other, and similarly arranged;

3d. When the three sides are respectively equal, and similarly arranged;

4th. When the three angles are respectively equal, and similarly arranged.

774. Corollary.-In each of the four cases just given, when the arrangement of the parts is reversed, the triangles are symmetrical.

POLAR TRIANGLES.

775. If at the vertex of a triedral, a perpendicular be erected to each face, these lines form the edges of a supplementary triedral (590). If the given vertex is at the center of a sphere, then there are two spherical triangles corresponding to these two triedrals, and they have all those relations which have been demonstrated concerning supplementary triedrals.

Since each edge of one triedral is perpendicular to the opposite face of the other, it follows that the vertex of each angle of one triangle is the pole of the opposite side of the other. Hence, such triangles are called polar triangles, though sometimes supplementary.

776. Theorem.—If with the several vertices of a spherical triangle as poles, arcs of great circles be made, then a

second triangle is formed whose vertices are also poles of the first.

777. Theorem. Each angle of a spherical triangle is the supplement of the opposite side of its polar triangle. Let ABC be the given triangle, and EF, DF, and DE be arcs of great circles, whose poles are respectively A, B, and C. Then ABC and DEF are polar or supplementary triangles.

These two theorems are corollaries of Article 589, but they can be demonstrated by c the student, with the aid of

E

B

the above diagram, without reference to the triedrals. 778. The student will derive much assistance from drawing the diagrams on a globe. Draw the polar triangle of each of the following: a birectangular triangle, a trirectangular triangle, and a triangle with one side longer than a quadrant and the adjacent angles very

acute.

INSCRIBED AND CIRCUMSCRIBED.

779. A sphere is said to be inscribed in a polyedron when the faces are tangent to the curved surface, in which case the polyedron is circumscribed about the sphere. A sphere is circumscribed about a polyedron when the vertices all lie in the curved surface, in which case the polyedron is inscribed in the sphere.

780. Problem. Any tetraedron may have a sphere inscribed in it; also, one circumscribed about it.

For within any tetraedron, there is a point equally distant from all the faces (625), which may be the cen

ter of the inscribed sphere, the radius being the perpendicular distance from this center to either face. There is also a point equally distant from all the vertices of any tetraedron (623), which may be the center of the circumscribed sphere, the radius being the distance from this point to either vertex.

781. Corollary.—A spherical surface may be made to pass through any four points not in the same plane.

EXERCISES.

782.-1. In a spherical triangle, the greater side is opposite the greater angle; and conversely.

2. If a plane be tangent to a sphere, at a point on the circumference of a section made by a second plane, then the intersection of these planes is a tangent to that circumference.

3. When two spherical surfaces intersect each other, the line of intersection is a circumference of a circle; and the straight line which joins the centers of the spheres is the axis of that circle.

SPHERICAL AREAS.

783. Let AHF be a right angled triangle and BFD a semicircle, the hypotenuse AF being a secant, and the vertex F in the circumference. From E, the point where AF cuts the arc, let a perpendicular EG fall upon AD.

Suppose the whole of this figure to revolve about AD as an axis. The triangle AFH describes a cone, the trapezoid EGHF describes the frustum of a cone, and the semicircle describes a sphere.

E

B

H

The points E and F describe the circumferences of

the bases of the frustum; and these circumferences lie in the surface of the sphere.

A frustum of a cone is said to be inscribed in a sphere, when the circumferences of its bases lie in the surface of the sphere.

784. Theorem.-The area of the curved surface of an inscribed frustum of a cone, is equal to the product of the altitude of the frustum by the circumference of a circle whose radius is the perpendicular let fall from the center of the sphere upon the slant hight of the frustum.

Let AEFD be the semicircle which describes the given sphere, and EBHF the trapezoid which describes the frustum. Let IC be the perpendicular let fall from the center of the sphere upon the slant hight EF.

Then the circumference of a circle of this radius would be π times twice CI, or 2πCI; and it is to be proved that the area of the curved surface of the frustum is equal to the product of BH by 2πCI.

E

H

C

The chord EF is bisected at the point I (187). From this point, let a perpendicular IG fall upon the axis AD. The point I in its revolution describes the circumference of the section midway between the two bases of the frustum. GI is the radius of this circumference, which is therefore 2GI. The area of the curved surface of the frustum is equal to the product of the slant hight by this circumference (727); that is, EF by 2πGI.

Now from E, let fall the perpendicular EK upon FH. The triangles EFK and IGC, having their sides respectively perpendicular to each other, are similar. Therefore, EF EK :: CI: GI. Substituting for the second term

:

EK its equal BH, and for the second ratio its equimultiple 2 CI: 2πGI, we have

EF: BH :: 2лСI : 2πGI.

By multiplying the means and the extremes,

EFX 2лIG=BH × 2πIC.

But the first member of this equation has been shown to be equal to the area of the curved surface of the frustum. Therefore, the second is equal to the same

area.

785. Corollary.-If the vertex of the cone were at the point A, the cone itself would be inscribed in the sphere; and there would be the same similarity of triangles, and the same reasoning as above. It may be shown that the curved surface of an inscribed cone is equal to the product of its altitude by the circumference of a circle whose radius is a perpendicular let fall from the center of the sphere upon the slant hight.

786. Theorem.-The area of the surface of a sphere is equal to the product of the diameter by the circumference of a great circle.

Let ADEFGB be the semicircle by which the sphere is described, having inscribed in it half of a regular polygon which may be supposed to revolve with it about the common diameter AB.

Then, the surface described by the side AD is equal to 2πCI by AH. The surface described by DE is equal to 2πCI by HK, for the perpendicular let fall upon DE is equal to CI; and so on. If one of the

E

F

D

I

H

K

B

sides, as EF, is parallel to the axis, the measure is the same, for the surface is cylindrical. Adding these sev