circumference (461); consequently, by subtracting it from three semicircumferences the remainder will be greater than a semicircumference, which is the measure of two right angles; therefore the sum of the three angles of a spherical triangle is greater than two right angles.

490. Corollary 1. The sum of the angles of a spherical triangle is not constant like that of a plane triangle; it varies from two right angles to six, without the possibility of being equal to either limit. Thus, two angles being given, we cannot thence determine the third.

491. Corollary 11. A spherical triangle may have two or three right angles, also two or three obtuse angles.

If the triangle ABC (fig. 235) has two right angles B and C, Fig. 235. the vertex A will be the pole of the base BC (467); and the sides AB, AC, will be quadrants.

If the angle A also is a right angle, the triangle ABC will have all its angles right angles, and all its sides quadrants. The triangle having three right angles is contained eight times in the surface of the sphere; this is evident from figure 236, if we suppose the arc MN equal to a quadrant.

492. Scholium. We have supposed in all that precedes, conformably to the definition art. 442, that spherical triangles always have their sides less each than a semicircumference; then it follows that the angles are always less than two right angles. For the side AB (fig. 224) is less than a semicircumfe- Fig. 224. rence as also AC; these arcs must both be produced in order to meet in D. Now the two angles ABC, CBD, taken together, are equal to two right angles; therefore the angle ABC is by itself less than two right angles.

We will remark, however, that there are spherical triangles of which certain sides are greater than a semicircumference, and certain angles greater than two right angles. For, if we produce the side AC till it becomes an entire circumference ACE, what remains, after taking from the surface of the hemisphere the triangle ABC, is a new triangle, which may also be designated by ABC, and the sides of which are AB, BC, AEDC. We see then, that the side AEDC is greater than the semicircumference AED; but, at the same time, the opposite angle B exceeds two right angles by the quantity CBD.

Geom.

22

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.

Fig. 236.

THEOREM.

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.

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, and the angle DQF = APC.

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.

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 FQE CPB, and the surface DQE = APB; we have accordingly

=

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.

Fig. 238.

THEOREM.

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.

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.

A+B+C
2

1; like

503. Corollary II. The spherical triangle ABC is equivalent to a lunary surface, the angle of which is 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;