769. If two circumferences of great circles intersect on the surface of a hemisphere, the sum of the opposite triangles thus formed is equivalent to a lune whose angle is equal to that included by the semi-circumferences.

B

Let the semi-circumferences BAD and CAE intersect at A on the surface of a hemisphere.

We are to prove ▲ ABC + ADA E equivalent to a lune whose angle is BA C.

The semi-circumferences produced intersect on the opposite hemisphere at A'.

Then each of the arcs AD and A'B is the supplement of A B,

770. COROLLARY. The sum of two spherical pyramids, the sum of whose bases is equivalent to a lune, is equivalent to a wedge whose base is the lune.

PROPOSITION XXXIII. THEOREM.

771. The area of a spherical triangle is equal to its spherical excess multiplied by the area of the tri-rectangular triangle.

B

E

Let ABC be a spherical triangle, and T the area of the tri-rectangular triangle.

We are to prove

AABC=T(A+B+C-2).

Complete the circumference ABDE.

Produce A C and BC to meet this circumference in D and E.
Then AABC + BCD (= lune A)

=

TX2ZA. § 767
A ABC + ACE (= lune B) = TX 2 ≤ B,
Z
$ 767
AABC+DC E (=lune C) (§ 769)=TX2 / C. §767

By adding these equalities,

2▲ A B C + ▲ A B C + BCD+ ACE + DCE

=TX 2 (A + B + C).

=

But AABC + B C D + ACE + DCE 4 T, § 758
(the surface of a hemisphere is equal to 4 tri-rectangular ▲).
... 2 A ABC +4T=TX 2 ( A + B + C);

·

.. AABC=TX(A+B+C – 2). Q. E. D.

772. SCHOLIUM 1. If ▲ A = 140°, ▲ B= 120° and ≤ C= 100°, a right angle being the unit,

140° 120° 100°

+

+

90° 90° 90°

773. SCHO. 2. To find the area of a spherical triangle on a given sphere, the angles of the triangle being given, we may multiply the area of the hemisphere by the ratio of the spherical excess to 360°.

=

140°, Z B

=

120° and

Thus if A the hemisphere is 2 R2, we have ▲ A B C

ZA+B+ ≤ C · 180°

360°

C 100°, since

PROPOSITION XXXIV. THEOREM.

774. The area of a spherical polygon is equal to its spherical excess multiplied by the area of the tri-rectangular triangle.

Let P denote the area of the spherical polygon; S the sum of its angles; n the number of its sides; t, t', t'... the areas of the triangles formed by drawing diagonals from any vertex A; s, s', respectively the sums of the angles of these triangles; and T the tri-rectangular triangle.

الى

...

(the area of a spherical ▲ is equal to its spherical excess multiplied into the

775. COROLLARY. The volume of a spherical pyramid is to the volume of the tri-rectangular pyramid, as the base of the pyramid is to the tri-rectangular triangle. And, since the volume of the tri-rectangular pyramid is the volume of the sphere, and the area of the tri-rectangular triangle is of the surface of the sphere; the volume of a spherical pyramid is to the volume of the sphere as its base is to the surface of the sphere.

776. DEF. A Zone is the part of the surface of a sphere included between two parallel circles of the sphere; as the surface included between the circles A B C and EFG.

777. DEF. The Bases of a zone are the circumferences of the intercepting circles; as circumferences ABC and EFG. If the plane of one base become tangent to the sphere, that base becomes a point, and the zone will have but one base.

778. DEF. The altitude of a zone is the perpendicular distance between the planes of its bases.

779. DEF. A Spherical Segment is a part of the sphere included between two parallel planes.

780. DEF. The Bases of a spherical segment are the bounding circles.

One of the planes may become a tangent plane to the sphere. In this case the segment has but one base.

781. DEF. The Altitude of a spherical segment is the perpendicular distance between the planes of its bases.

782. DEF. A Spherical Sector is a part of a sphere generated by a circular sector of the semicircle which generates the sphere; as AO C K.

783. DEF. The Base of a spherical sector is the zone generated by the arc of the circular sector; as AC K.

The other bounding surfaces of a spherical sector may be one conical surface, or two conical surfaces; or one conical and one plane surface.

Thus, let A B be the diameter around which the semicircle ACB revolves to generate the sphere. The solid generated by the circular sector AOC will be a spherical sector having the zone A C K for its base, and for its other bounding surface the conical surface generated by CO.

The spherical sector generated by COD has for its base the zone generated by CD, and for its other surfaces the concave conical surface generated by D O, and the convex conical surface generated by CO.

The spherical sector generated by EOF has for its base the zone generated by EF, and for one surface the plane surface generated by E O, and for the other surface the concave conical surface generated by FO.