F

B

E

A

Let ABC be a triangle, having the right angle at A ; and let AB be either of the sides, the sine of the side AB will be to the radius, as the tangent of the other side AC to the tangent of the angle ABC, opposite to AC. Let D be the centre of the sphere; join AD, BD, CD, and let AF be drawn perpendicular to BD, which therefore will be the sine of the arc AB, and from the point F, let there be drawn in the D plane BDC the straight line FE at right angles to BD, meeting DC in E, and let AE be joined. Since therefore the straight line DF is at right angles to both FA and FE, it will also be at right angles to the plane AEF (So. Ge. I. 4); wherefore the plane. ABD, which passes through DF, is perpendicular to the plane AEF (So. Ge. I. 17), and the plane AEF perpendicular to ABD; but the plane ACD or AED is also perpendicular to the same ABD; therefore the common section, namely, the straight line AE, is at right angles to the plane ABD (So. Ge. I. 18), and EAF, EAD, are right angles. Therefore, AE is the tangent of the arc AC; and in the rectilineal triangle AEF, having a right angle at A, AF is to the radius as AE to the tangent of the angle AFE (Pl. Tr. 2); but AF is the sine of the arc AB, and AE the tangent of the arc AC, and the angle AFE is the inclination of the planes CBD, ABD (So. Ge. I. Def. 7), or the sphe rical angle ABC; therefore the sine of the arc AB is to the radius as the tangent of the arc AC to the tangent of the opposite angle ABC.

COR. And since by this proposition the sine of the side AB is to the radius, as the tangent of the other side AC to the tangent of the angle ABC opposite to that side; and as the radius is to the cotangent of the angle ABC, so is the tangent of the same angle ABC to the radius (Pl. Tr. 4, Čor. to Def.); by equality, the sine of the side AB is to the cotangent of the angle ABC adjacent to it, as the tangent of the other side AC to the radius.

PROPOSITION II.

In right-angled spherical triangles the sine of the hypotenuse is to the radius, as the sine of either side is to the sine of the angle opposite to that side.

Let the triangle ABC be right angled at A, and let AC be either of the sides; the sine of the hypotenuse BC will be to the radius as the sine of the arc AC is to the sine of the angle ABC.

Let D be the centre of the sphere, and let CE be drawn perpendicular to DB, which will therefore be the sine of the hypotenuse BC; and from the

D

F

E

B

point E let there be drawn in the plane ABD the straight line EF perpendicular to DB, and let CF be joined; CF will be at right angles to the plane ABD, as was shown in the preceding proposition of the straight line EA; wherefore CFD, CFE, are right angles, and CF is the sine of the arc AC; and in the triangle CFE, having the right angle CFE, CE is to the radius, as CF to the sine of the angle CEF (Pl. Tr. 1). But, since CE, FE, are at right angles to DEB, which is the common section of the planes CBD, ABD, the angle CEF is equal to the inclination of these planes (So. Ge. I. Def. 4), that is, to the spherical angle ABC. The sine, therefore, of the hypotenuse CB is to the radius as the sine of the side AC is to the sine of the opposite angle ABC.

COR. Of these three, namely, the hypotenuse, a side, and the angle opposite to that side, any two being given, the third may be found.

PROPOSITION III.

In right-angled spherical triangles, the cosine of the hypotenuse is to the radius as the cotangent of either of the angles is to the tangent of the remaining angle.

Let ABC be a spherical triangle, having a right angle at A, the cosine of the hypotenuse BC will be to the radius

E

as the cotangent of the angle ABC to the tangent of the angle ACB.

E

Describe the circle DE, of which B is the pole, and let it meet AC in F, and the circle BC in E; and since the circle BD passes through the pole B of the cirele DF, DF will pass through the pole of BD (Sp. Ge. I. Cor. 7). And since AC is perpendicular to BD, AC will also pass through the pole of BD; wherefore, the pole of the circle BD is in the point where the circles AC, DE, intersect, that is, in the point

F. The arcs FA, FD, are therefore quadrants, and likewise the arcs BD, BE. In the triangle CEF, right angled at the point E, CE is the complement of BC, the hypotenuse of the triangle ABC; EF is the complement of the arc ED, which is the measure of the angle ABC; FC, the hypotenuse of the triangle CEF, is the complement of AC; and the arc AD, which is the measure of the angle CFE, is the complement of AB.

But (Sp. Tr. 1) in the triangle CEF, the sine of the side CE is to the radius, as the tangent of the other side EF is to the tangent of the angle ECF opposite to it; that is, in the triangle ABC, the cosine of the hypotenuse BC is to the radius as the cotangent of the angle ABC is to the tangent of the angle ACB.

COR. 1.-Of these three, namely, the hypotenuse and the two angles, any two being given, the third will also be given.

COR. 2.-And since by this proposition the cosine of the hypotenuse BC is to the radius as the cotangent of the angle ABC to the tangent of the angle ACB, and since the radius is to the cotangent of ACB, as the tangent of ACB to the radius (Pl. Tr. 4, Cor. to Def.); therefore, by equality, the cosine of the hypotenuse BC is to the cotangent of the angle ACB, as the cotangent of the angle ABC to the radius.

Schol. The triangle CEF is called the complementary triangle.

PROPOSITION IV,

In right-angled spherical triangles, the cosine of an angle is to the radius, as the tangent of the side adjacent to that angle is to the tangent of the hypotenuse.

F

E

The same construction remaining. In the triangle CEF (Sp. Tr. 1), the sine of the side EF is to the radius, as the tangent of the other side CE is to the tangent of the angle CFE opposite to it; that is, in the triangle ABC, the cosine of the angle ABC is to the radius as the cotangent of the hypotenuse BC to the cotangent of the side AB, adjacent to ABC, or as the tangent of the side AB to the tangent of the hypotenuse, since the tangents of two arcs are reciprocally proportional to their cotangents (Pl. Tr. 4, Cor. to Def.)

B

A

COR. And since by this proposition the cosine of the angle ABC is to the radius, as the tangent of the side AB is to the tangent of the hypotenuse BC; and as the radius is to the cotangent of BC, so is the tangent of BC to the radius; by equality, the cosine of the angle ABC will be to the cotangent of the hypotenuse BC, as the tangent of the side AB, adjacent to the angle ABC, to the radius.

PROPOSITION V.

In right-angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the hypotenuse is to the cosine of the other side.

The same construction remaining. In the triangle CEF, the sine of the hypotenuse CF is to the radius, as the sine of the side CE to the sine of the opposite angle CFE (Sp. Tr. 2); that is, in the triangle ABC, the cosine of the side CA is to the radius as the cosine of the hypotenuse BC to the cosine of the other side BA.

PROPOSITION VI

In right-angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the angle opposite to that side is to the sine of the other angle.

The same construction remaining. In the triangle CEF, the sine of the hypotenuse CF is to the radius as the sine of the side EF is to the sine of the angle ECF opposite to it; that is, in the triangle ABC, the cosine of the side CA is to the radius, as the cosine of the angle ABC, opposite to it, is to the sine of the other angle ACB.

PROPOSITION VII.

In spherical triangles, whether right-angled or obliqueangled, the sines of the sides are proportional to the sines of the angles opposite to them.

First, let ABC be a right-angled triangle, having a right angle at A; therefore (Sp. Tr. 2) the sine of the hypotenuse BC is to the radius (or the sine of the right angle at A) as the sine of the side AC to the sine of the angle B. And in like manner, the sine of BC is to the sine of the angle A, as the sine of AB to the sine of the angle C; wherefore (Pl. Ge. V. 11) the sine of the side AC is to the sine of the angle B, as the sine of AB to the sine of the angle C.

B

A

Secondly, let ABC be an oblique-angled triangle, the sine of any of the sides BC, will be to the sine of any of the other two AC, as the sine of the angle A, opposite to BC, is

to the sine of the angle B, opposite to AC. Through the point C, let there be drawn an arc of a great circle CD perpendicular upon AB; and in the right-angled triangle BCD (Sp. Tr. 2), the sine of BC is to the radius, as the sine of CD to the sine of the angle B; and in the triangle ADC, by inversion, the radius is to the sine of AC as the sine of the angle A to the sine of DC; therefore, by indirect equality, the sine of BC is to the sine of AC, as the sine of the angle A to the sine of the angle B.