Fig. 232.

Fig. 233.

the greater; take BO = AC, and join OC. The two sides BO, BC, are equal to the two AC, BC; and the angle OBC contained by the first is equal to the angle ACB contained by the second. Consequently the two triangles have their other parts equal (480), namely, OCB = ABC; but the angle ABC is, by hypothesis, equal to ACB; whence OCB is equal to ACB, which is impossible; AB then cannot be supposed unequal to AC; therefore the sides AB, AC, opposite to the equal angles B, C, are equal.

484. Scholium. It is evident, from the same demonstration, that the angle BAD=DAC, and the angle BDA=ADC. Consequently the two last are right angles; therefore, the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to this base, and divides the angle opposite into two equal parts.

THEOREM.

485. In any spherical triangle ABC (fig. 232), if the angle A is greater than the angle B, the side BC opposite to the angle A will be greater than the side AC opposite to the angle B; conversely, if the side BC is greater than AC, the angle A will be greater than the angle B.

Demonstration. 1. Let the angle A➤B; make the angle BAD = B, and we shall have AD = DB (483); but

AD+DC>AC;

in the place of AD substitute DB, and we shall have DB + DC or BC > AC.

2. If we suppose BC>AC, we say that the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC =AC; and if BAC were less than ABC, it would follow, according to what has just been demonstrated, that BC<AC, which is contrary to the supposition; therefore the angle BAC is greater than ABC.

THEOREM.

486. If the two sides AB, AC (fig. 233), of the spherical triangle ABC are equal to the two sides DE, DF, of the triangle DEF described upon an equal sphere, if at the same time the angle A is greater than the angle D, we say that the third side BC of the first triangle will be greater than the third side EF of the second.

The demonstration of this proposition is entirely similar to that of art. 42.

THEOREM.

487. If two triangles described upon the same sphere or upon equal spheres are equiangular with respect to each other, they will also be equilateral with respect to each other.

Demonstration. Let A, B, be the two given triangles, P, Q, their polar triangles. Since the angles are equal in the triangles A, B, the sides will be equal in the polar triangles P, Q (476); but, since the triangles P, Q, are equilateral with respect to each other, they are also equiangular with respect to each other (482); and, the angles being equal in the triangles P, Q, it follows that the sides are equal in their polar triangles A, B. Therefore the triangles A, B, which are equiangular with respect to each other, are at the same time equilateral with respect to each other.

This proposition may be demonstrated without making use of polar triangles in the following manner.

Let ABC, DEF (fig. 234), be two triangles equiangular with Fig.234. respect to each other, having A=D, BE, C=F; we say

that the sides will be equal, namely, AB DE, AC = DF, BC EF.

=

Produce AB, AC, making AG = DE, AH = DF; join GH, and produce the arcs BC, GH, till they meet in I and K.

The two sides AG, AH, are, by construction, equal to the two DF, DE, the included angle GAH=BAC=EDF, consequently the triangles AGH, DEF, are equal in all their parts (480); therefore the angle AGH= DEF= ABC, and the angle

In the triangles IBG, KBG, the side BG is common, and the angle IGB = GBK; and, since IGB+BGK is equal to two right angles, as also GBK+IBG, it follows that BGK IBG. Consequently the triangles IBG, GBK, are equal (481); therefore IG = BK, and IB = GK.

In like manner, since the angle AHG=ACB, the triangles ICH, HCK, have a side and the two adjacent angles of the one respectively equal to a side and the two adjacent angles of the other; consequently they are equal; therefore IH CK, and HK=IC.

=

Now, if from the equals BK, IG, we take the equal CK, IH, the remainders BC, GH, will be equal. Besides, the angle BCA AHG, and the angle ABC-AGH. Whence the triangles ABC, AHG, have a side and the two adjacent angles of the one respectively equal to a side and the two adjacent angles of the other; consequently they are equal. But the triangle DEF is equal in all its parts to the triangle AHG; therefore it is also equal to the triangle ABC, and we shall have AB= DE, AC = DF, BC=EF; hence, if two spherical triangles are equiangular with respect to each other, the sides opposite to the equal angles will be equal.

488. Scholium. This proposition does not hold true with regard to plane triangles, in which, from the equality of the angle, we can only infer the proportionality of the sides. But it is easy to account for the difference in this respect between plane and spherical triangles. In the present proposition, as well as those of articles 480, 481, 482, 486, which relate to a comparison of triangles, it is said expressly that the triangles are described upon the same sphere or upon equal spheres. Now, similar arcs are proportional to their radii; consequently upon equal spheres two triangles cannot be similar without being equal. It is not therefore surprising that equality of angles should imply equality of sides.

It would be otherwise, if the triangles were described upon unequal spheres; then, the angles being equal, the triangles would be similar, and the homologous sides would be to each other as the radii of the spheres.

THEOREM.

489. The sum of the angles of every spherical triangle is less than six, and greater than two right angles.

Demonstration. 1. Each angle of a spherical triangle is less than two right angles (see the following scholium); therefore the sum of the three angles is less than six right angles.

2. The measure of each angle of a spherical triangle is equal to the semicircumference minus the corresponding side of the polar triangle (476); therefore the sum of the three angles has for its measure three semicircumferences minus the sum of the sides of the polar triangle. Now, this last sum is less than a

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. II. 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 semicircumference, as also Fig. 224. 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.

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 4S, 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 II. 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