Fig. 236.

Besides, if we exclude from the definition triangles, the sides and angles of which are so great, it is because the resolution of them, or the determination of their parts, reduces itself always to that of triangles contained in the definition. Indeed, it will be readily perceived, that if we know the angles and sides of the triangle ABC, we shall know immediately the angles and sides of the triangle of the same name, which is the remainder of the surface of the hemisphere.

493. The lunary surface AMBNA (fig. 236) is to the surface of the sphere as the angle MAN of this surface is to four right angles, or as the arc MN, which measures this angle, is to the circumference.

Demonstration. Let us suppose in the first place, that the arc MN is to the circumference MNPQ in the ratio of two entire numbers, as 5 to 48, for example. The circumference MNPQ may be divided into 48 equal parts, of which MN will contain 5; then joining the pole A and the points of division by as many quadrants, we shall have 48 triangles in the surface of the hemisphere AMNPQ, which will be equal among themselves, since they have all their parts equal. The entire sphere then will contain 96 of these partial triangles, and the lunary surface AMBNA will contain 10 of them; therefore the lunary surface is to that of the sphere as 10 is to 96, or as 5 is to 48, that is, as the arc MN is to the circumference.

If the arc MN is not commensurable with the circumference, it may be shown by a course of reasoning, of which we have already had many examples, that the lunary surface is always to that of the sphere as the arc MN is to the circumference.

494. Corollary 1. Two lunary surfaces are to each other as their respective angles.

495. Corollary 11. We have already seen that the entire surface of the sphere is equal to eight triangles having each three right angles (491); consequently, if the area of one of these triangles be taken for unity, the surface of the sphere will be represented by eight. This being supposed, the lunary surface, of which the angle is A, will be expressed by 2A, the angle A being estimated by taking the right angle for unity; for we have 2A: 8:: A: 4. Here are then two kinds of units; one for

angles, this is the right angle; the other for surfaces, this is the spherical triangle, of which all the angles are right angles and the sides quadrants.

496. Scholium. The spherical wedge comprehended by the planes AMB, ANB, is to the entire sphere, as the angle A is to four right angles. For the lunary surfaces being equal, the spherical wedges will also be equal; therefore two spherical wedges are to each other as the angles formed by the planes which comprehend them.

THEOREM.

21

497. Two symmetrical spherical triangles are equal in sarface. Demonstration. Let ABC, DEF (fig. 237), be two symmetri- Fig. 237. cal triangles, that is, two triangles which have their sides equal, namely, AB = DE, AC = DF, CB = EF, and which at the same time do not admit of being applied the one to the other; we say that the surface ABC is equal to the surface DEF.

Let P be the pole of the small circle which passes through the three points A, B, C*; from this point draw the equal arcs PA, PB, PC (464); at the point F make the angle DFQ = ACP, the arc FQ CP, and join DQ, EQ.

The sides DF, FQ, are equal to the sides AC, CP; the angle DFQ= ACP; consequently the two triangles DFQ, ACP, are equal in all their parts (480); therefore the side DQ = AP, the angle DQF=APC.

and

In the proposed triangles DFE, ABC, the angles DFE, ACB, opposite to the equal sides DE, AB, being equal (481), if we subtract from them the angles DFQ, ACP, equal, by construction, there will remain the angle QFE equal to PCB. Moreover the sides QF, FE, are equal to the sides PC, CB; consequently the two triangles FQE, CPB, are equal in all their parts; therefore the side QE PB, and the angle FQE = CPB.

=

If we observe now that the triangles DFQ, ACP, which have the sides equal each to each, are at the same time isosceles, we

*The circle, which passes through the three points A, B, C, or which is circumscribed about the triangle ABC, can only be a small circle of the sphere; for, if it were a great one, the three sides AB, BC, AC, would be situated in the same plane, and the triangle ABC would be reduced to one of its sides.

Fig. 238.

shall perceive that they may be applied the one to the other; for, having placed PA upon its equal QF, the side PC will fall upon its equal QD, and thus the two triangles will coincide; consequently they are equal, and the surface DQF = APC. For a similar reason the surface FCE CPB, and the surface DQE APB; we have a cordingly

=

=

DQF+FQE-DQE = APC + CPB — APB, or DEF = ABC; therefore the two symmetrical triangles ABC, DEF, are equal

in surface.

498. Scholium. The poles P and Q may be situated within the triangles ABC, DEF; then it would be necessary to add the three triangles DQF, FQE, DQE, in order to obtain the triangle DEF, and also the three triangles APC, CPB, APB, in order to obtain the triangle ABC. In other respects the demonstration would always be the same and the conclusion the same.

499. If two great circles AOB, COD (fig. 238), cut each other in any manner in the surface of a hemisphere AOCBD), the sum of the opposite triangles AOC, BOD, will be equal to the lunary surface of which the angle is BOD.

Demonstration. By producing the arcs OB, OD, into the surface of the other hemisphere till they meet in N, OBN will be a semicircumference as well as AOB; taking from each OB, we shall have BN AO. For a similar reason DN CO, and BD=AC; consequently the two triangles AOC, BDN, have the three sides of the one equal respectively to the three sides of the other; moreover, their position is such that they are symmetrical; therefore they are equal in surface (496), and the sum of the triangles AOC, BOD, is equivalent to the lunary surface OBNDO, of which the angle is BOD.

500. Scholium. It is evident also that the two spherical pyramids, which have for their bases the triangles AOC, BOD, taken together, are equal to the spherical wedge of which the angle is BOD.

3

THEOREM.

501. The surface of a spherical triangle has for its measure the excess of the sum of the three angles over two right angles.

Demonstration. Let ABC (fig. 239) be the triangle proposed; Fig. 239. produce the sides till they meet the great circle DEFG drawn at pleasure without the triangle. By the preceding theorem the two triangles ADE, AGH, taken together, are equal to the lunary surface of which the angle is A, and which has for its measure 2A (495); thus we shall have ADE + AGH = 2A ; for a similar reason BGF + BID = 2B, CIH + CFE = 2C. But the sum of these six triangles exceeds the surface of a hemisphere by twice the triangle ABC; moreover the surface of a hemisphere is represented by 4; consequently the double of the triangle ABC is equal to 2A + 2B + 2C — 4, and consequently ABC=A+B+C—2; therefore every spherical triangle has for its measure the sum of its angles minus two right angles.

502. Corollary 1. The proposed triangle will contain as many triangles of three right angles, or eighths of the sphere (494), as there are right angles in the measure of this triangle. If the angles, for example, are each equal to of a right angle, then the three angles will be equal to four right angles, and the proposed triangle will be represented by 4-2 or 2; therefore it will be equal to two triangles of three right angles, or to a fourth of the surface of the sphere.

503. Corollary II. The spherical triangle ABC is equivalent A+B+C to a lunary surface, the angle of which is 2

1; like

wise the spherical pyramid, the base of which is ABC, is equal

to the spherical wedge, the angle of which is

A+B+C
2

1.

504. Scholium. At the same time that we compare the spherical triangle ABC with the triangle of three right angles, the spherical pyramid, which has for its base ABC, is compared with the pyramid which has a triangle of three right angles for its base, and we obtain the same proportion in each case. The solid angle at the vertex of a pyramid is compared in like manner with the solid angle at the vertex of the pyramid having a triangle of three right angles for its base. Indeed the comparison is established by the coincidence of the parts. Now, if the bases of pyramids coincide, it is evident that the pyramids themselves will coincide, as also the solid angles at the vertex. Whence we derive several consequences;

Fig. 240.

1. Two spherical triangular pyramids are to each other as their bases; and, since a polygonal pyramid may be divided into several triangular pyramids, it follows that any two spherical pyramids are to each other as the polygons which constitute their bases.

2. The solid angles at the vertex of these same pyramids are likewise proportional to the bases; therefore, in order to compare any two solid angles, the vertices are to be placed at the centres of two equal spheres, and these solid angles will be to each other as the spherical polygons intercepted between their planes or faces.

The angle at the vertex of the pyramid, whose base is a triangle of three right angles, is formed by three planes perpendicular to each other; this angle, which may be called a solid right angle, is very proper to be used as the unit of measure for other solid angles. This being supposed, the same number, which gives the area of a spherical polygon, will give the measure of the corresponding solid angle. If, for example, the area of a spherical polygon is, that is, if it is of a triangle of three right angles, the corresponding solid angle will also be of a solid right angle.

505. The surface of a spherical polygon has for its measure the sum of its angles minus the product of two right angles by the number of sides in the polygon minus two.

Demonstration. From the same vertex A (fig. 240) let there be drawn to the other vertices the diagonals AC, AD; the polygon ABCDE will be divided into as many triangles minus two as it has sides. But the surface of each triangle has for its measure the sum of its angles minus two right angles, and it is evident that the sum of all the angles of the triangles is equal to the sum of the angles of the polygon; therefore the surface of the polygon is equal to the sum of its angles diminished by as many times two right angles as there are sides minus two.

506. Scholium. Let s be the sum of the angles of a spherical polygon, n the number of its sides; the right angle being supposed unity, the surface of the polygon will have for its measure - 2n + 4.

$ 2n

2) or s