PROPOSITION XXIV. THEOREM

751. If the right angle be taken as angular unit, and the tri-rectangular triangle as unit of surface, the area of a spherical triangle is measured by the spherical

To prove.

=

Area ABC A+B+C-2, if the angular unit is the right angle, and the unit of surface the tri-rectangular triangle.

Proof. Complete the circumference ACDE, and produce AB and CB until they meet ACD in D and E respectively. Then, since ▲ ABC+A BED is equivalent to the lune, whose angle is equal to Z ABC,

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By adding these equations, we obtain

3 ABC + BED + ABE + BCD = 2 (A + B + C).

But ABC+BED + ABE + BCD equals a hemisphere or 4.

752. COR. The area of a spherical triangle is to the surface of the entire sphere as its spherical excess is to 8.

Ex. 1207. If the surface of a sphere is 40 sq. ft., find the area of a spherical triangle whose angles are respectively 40°, 60°, and 100°.

Ex. 1208. What fraction of the surface of a sphere is covered by a spherical triangle whose angles are 90°, 100°, and 110°, respectively?

Ex. 1209. How large is the angle of an equiangular spherical triangle whose area is one-fourth of the surface of the sphere?

Ex. 1210. What is the ratio of a lune whose angle is equal to 80°, and the area of an equilateral spherical triangle whose angle is equal to 80° ?

Ex. 1211. The arms of an isosceles spherical triangle whose base angles are 80° are produced so as to form a lune. Find the vertical angle if the triangle is the third part of the lune.

Ex. 1212. The angles of a spherical triangle are 80°, 90°, and 100°. Find the angle of an equivalent lune.

PROPOSITION XXV. THEOREM

753. If the right angle be taken as the angular unit, and the tri-rectangular triangle as unit of surface, the area of any spherical polygon is equal to the sum of its angles diminished by twice the number of its sides less

two:

B

E

Hyp. ABCD... is a spherical polygon of n sides.

To prove ABCD... is measured by (A+B+C+.....) — 2(n-2).

B

C

E

Proof. Draw all diagonals from A, which will divide ABCD... into n -2 triangles.

The area of each triangle is equal to the sum of its angles less 2.

Hence the area of the polygon ABCD..... is equal to

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Q.E.D.

754. SCHOLIUM. Props. XXIV and XXV do not give us the absolute, but the relative size of areas on the sphere.

Ex. 1213. Find the area of an equiangular hexagon whose angle is 160°. (The tri-rectangular triangle taken as unit.)

Ex. 1214. What fraction of the surface of the sphere is covered by the hexagon of the preceding exercise.

Ex. 1215. The angles of a spherical quadrilateral are 90°, 100°, 110°, and 120°. Find the angle of an equivalent equilateral triangle.

Ex. 1216. Find the angle of an equiangular spherical triangle equivalent to the sum of three equilateral triangles whose angle is 70°.

Ex. 1217. The medians of a spherical triangle meet in a point.

Ex. 1218. The intersection of two spheres is a circle whose plane is perpendicular to the line of centers.

Ex. 1219. The line of centers of two spheres is 13. Find the radius of the circle of intersection if the radii of the spheres are 5 and 12 respectively.

Ex. 1220. The volume of a circumscribed polyedron is equivalent to its surface multiplied by one-third of the radius.

PROPOSITION XXVI. THEOREM

755. The area generated by the revolution of a straight line about an axis in its plane is equal to the projection of the line upon the axis multiplied by the circumference of a circle whose radius is the perpendicular erected at the midpoint of the line and terminating in the axis.

Hyp. AB revolving about an axis XY in its plane produces surface ABK, CD is the projection of AB upon XY, and EF is the perpendicular bisector of AB terminating in XY.

Proof. Draw EG to BD and AH || to CF.

The surface ABK is the lateral surface of a cone of revolution.

Hence

But

Whence

or

area ABK = AB × 2 πEG. (1) (668)

ΔΑΒΗ ~AEFG.

AB AH

EF EG

AB × EG: = AH × EF,

AB × EG: = CD x EF.

Multiplying both sides by 2 we have from (1).

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.. area ABK :

= CD x 2 TEF.

Q.E.D.

756. DEF. A zone is a portion of a spherical surface included between two parallel planes.

757. DEF. The circumferences of the sections made by the planes form the bases of the zone, and the distance between the two planes is the altitude of the zone.

Ex. 1221. In the diagram for Prop. XXVI, find the area ABK, if EF is equal to 10 in., AB is equal to 8 in., and angle BH is one-third of a right angle.

758. The surface of a sphere is equivalent to four times the area of a great circle.

Hyp. S is the area of the surface, R the radius of a sphere generated by revolving semicircle ABE about AE.

Proof. Inscribe in the semicircle half of a regular polygon of an even number of sides, as ABCDE.

Draw BG, CO, and DH perpendicular to AE, and denote the common distance of the chords AB, BC, CD, and DE from O by d.

Then

area AB* = AG × 2 πd,

area BC = GO × 2 πd, etc.

* Area AB means area generated by AB.

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