Hence, the mutually equiangular triangles A and B are at the same time, mutually equilateral.

Scholium. This proposition is not applicable to rectilineal triangles; in which equality among the angles indicates only proportionality among the sides. Nor is it difficult to account for the difference, in this respect, between spherical and rectilineal triangles. In the proposition now before us, as well as in the preceding ones, which treat of the comparison of triangles, it is expressly required that the arcs be traced on the same sphere, or on equal spheres. Now, similar arcs are to each other as their radii; hence, on equal spheres, two triangles cannot be similar without being equal. Therefore, it is not strange that equality among the angles should produce equality among the sides.

The case would be different, if the triangles were drawn upon unequal spheres; there, the angles being equal, the triangles would be similar, and the homologous sides would be to each other as the radii of their spheres.

PROPOSITION XIV. THEOREM.

The sum of all the angles, in any spherical triangle, is less than six right angles and greater than two.

For, in the first place, every angle of a spherical triangle is less than two right angles: hence, the sum of the three is less than six right angles.

Secondly, the measure of each angle of a spherical triangle is equal to the semicircumference minus the corresponding side of the polar triangle (P. 6); hence, the sum of the three, is measured by the three semicircumferences, minus the sum of the sides of the polar triangle. Now, this latter sum is less than a circumference (P. 2, C.); therefore, taking it away from three semicircumferences, the remainder is greater than one semicircumference, which is the measure of two right angles; hence, the sum of the three angles of a spherical triangle is greater than two right angles.

Cor. 1. The sum of the three angles of a spherical triangle is not constant, like that of the angles of a rectilineal triangle, but varies between two right angles and six, without ever reaching either of these limits. Two given angles therefore do not serve to determine the third.

Cor. 2. A spherical triangle may have two, or even three of its angles right angles; also two, or even three of its angles obtuse.

Cor. 3. If the triangle ABC is bi-rectangular, in other words, has two right angles B and C, the vertex A is the pole of the base BC; and the sides AB, AC, are quadrants (P. 3, c. 2).

B

A

If the angle A is also a right angle, the triangle ABC is tri-rectangular; each of its angles is a right angle, and its sides are quadrants. Two tri-rectangular triangles make half a hemisphere, four make a hemisphere, and eight the entire surface of a sphere.

PROPOSITION XV. THEOREM.

The surface of a lune is to the surface of the sphere, as the angle of the lune, to four right angles; or, as the arc which measures that angle, to the circumference.

Let AMBN be a lune, and NCM the angle included between its two great circles: then will its surface be to the surface of the sphere as the angle NCM to four right angles, or as the arc NM to the circumference of a great circle. For, suppose the arc MN to be

N

P

to the circumference MNPQ, as some one integer number to another, as 5 to 48, for example. Divide the circumference MNPQ, into 48 equal M parts, MN will contain 5 of them; and if the pole A were joined with the several points of division, by as many quadrants, we should in the hemisphere AMNPQ, have 48 triangles, all equal, because all the corresponding parts are equal. The whole sphere

B

would contain 96 of these triangles, and the lune AMBNA, 10 of them; hence, the lune is to the sphere as 10 is to 96, or as 5 to 48; in other words, as the arc MN is to the circumference.

If the arc MN is not commensurable with the circumference, it may still be shown, that the lune is to the sphere as MN to the circumference (B. III., P. 17).

Cor. 1. Two lunes on the same or on equal spheres, are to each other as their respective angles.

Cor. 2. It was shown above, that the whole surface of the sphere is equal to eight tri-rectangular triangles (P. 14, c. 3); hence, if the area of one such triangle be represented by 7, the surface of the whole sphere will be expressed by 87. This granted, if the right angle be assumed equal to 1, the surface of the lune whose angle is A, will be expressed by 2AXT. For,

in which expression, A represents such a part of unity, as the angle of the lune is of one right angle.

Scholium. The spherical ungula, bounded by the planes AMB, ANB, is to the whole solid sphere, as the angle A is to four right angles. For, the lunes being equal, the spherical ungulas are also equal; hence, two spherical ungulas are to each other, as the angles formed by the planes which bound them.

PROPOSITION XVI. THEOREM.

Two symmetrical spherical triangles are equivalent.

Let ABC, DEF, be two symmetrical triangles, that is to say, two triangles having their sides AB-DE, AC=DF, CB=EF, and yet incapable of superposition: we are to show that the surface ABC is equal to the surface DEF.

Let P be the pole of the small circle passing through the three points A, B, C;* from this point draw the equal

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

all their parts (P. 8); consequently, the side_DQ=AP, and the angle DQF=APC.

In the triangles DFE, ABC, the angles DFE, ACB, opposite to the equal sides DE, AB, are equal (P. 10). If the angles DFQ, ACP, which are equal by construction, be taken away from them, there will remain the angle QFE, equal to PCB. The sides QF, FE, are equal to the sides PC, CB; hence, the two triangles FQE, CPB, are equal in all their parts (P. 8); hence, the side QE=PB, and the angle FQE=CPB.

Now, the triangles DFQ, ACP, which have their sides respectively equal, are at the same time isosceles, and capable of coinciding, when applied the one to the other. For, having placed AC on its equal. DF, the equal sides will fall the one on the other, and thus the two triangles will exactly coincide: hence, they are equal; and the surface DQF APC. For a like reason, the surface FQE= CPB, and the surface DQE=APB; hence we have,

or,

DQF+FQE—DQE~~APC+CPB—APB,
DFE ABC;

hence, the two symmetrical triangles ABC, DEF, are equal in surface.

Scholium. The poles P and Q might lie within triangles ABC, DEF: in which case it would be requisite to add the three triangles DQF, FQE, DQE, together, in order to make up the triangle DEF; and in like manner, to add the three triangles APC, CPB, APB, together, in order to make up the triangle ABC: in all other respects, the demonstration and the result would be the same.

If the circumferences of two great circles intersect each other on the surface of a hemisphere, the sum of the opposite triangles thus formed, is equivalent to the surface of a lune whose angle is equal to the angle formed by the circles.

Let the circumferences A OB, COD, intersect on the surface of a hemisphere; then will the opposite triangles AOC, BOD, be equivalent to the lune whose angle is BOD. For, produce the arcs OB, OD, on the other hemisphere, till they meet in N. Now, since AOB and OBN are semicircumferences, if we take away the common part OB, we shall have BNAO. For a like reason, we have DN=CO, and BD=AC. Hence, the two triangles AOC, BDN,

D

B

have their three sides respectively equal: they are therefore symmetrical; hence, they are equal in surface (P. 16). But the sum of the triangles BDN, BOD, is equivalent to the lune OBNDO, whose angle is BOD: hence, AOC+BOD is equivalent to the lune whose angle is BOD.

Scholium. It is likewise evident, that the two spherical pyramids, which have the triangles AOC, BOD, for bases, are together equivalent to the spherical ungula whose angle is BOD.

The surface of a spherical triangle is equal to the excess of the sum of its three angles above two right angles, multiplied by the tri-rectangular triangle.

Let ABC be any spherical triangle: then will its surface be equal to

(A+B+C−2)× T.

For, produce its sides till they meet the great circle DEFG, drawn at pleasure, without the triangle. By the last theorem, the two triangles ADE, AGH, are together