the pole of AC; for the arcs CA and ED meet at right angles (Geom. Prop. 12); and as EF and EA are arcs at right angles to AC, E is a pole of AC (Geom. Prop. 15). Since then from E the vertex of an oblique angle in the right-angled triangle BDE, as a pole, the arc of a great circle AC is drawn, meeting the sides DB and EB produced, the triangle ABC thus formed, having one of its vertices at B, is complemental to BDE.

4. Of two triangles complemental to each other, an oblique angle in one is equal to an oblique angle in the other; the opposite side in each is the complement of the measure of the second oblique angle in the other; and of the sides containing the equal angles, the hypotenuse in either triangle is the complement of the other side in the other triangle.

Thus, of the complemental triangles ABC, DBE, in the preceding figure, the oblique angles at B are equal; and since CF is a quadrant, AC is the complement of AF, which is evidently the measure of the angle E in BDE; and ED is the complement of DF, which is the measure of the angle C in ABC. And of the sides containing the equal angles at B, the hypotenuse BC is the complement of BD, and the hypotenuse BE the complement of AB.

5. In a right-angled spherical triangle, the sine of the hypotenuse is to the radius, as the sine of another side is to the sine of the opposite angle.

Let ABC be a spherical triangle, right-angled at A; then the sine of BC the hypotenuse, is to the radius, as the sine of the side AC is to the sine of the opposite angle ABC.

at right angles to the common section AD, CE must be at right angles to the plane ABD, and to the line EF which it meets in that plane. Now since CEF is a right angle, CE2+EF2=CF2; and since CFD is a right angle, CF2+FD2=CD2; therefore CE2+EF2+FD2= CD2: again, since CED is a right angle, CE2+ED2= CD2: hence, CE2+EF2+FD2=CE2+ED2; and therefore EF2+FD2=ED2: consequently EFD is a right angle (Euc. 1, 48). Since then CF and EF are drawn in the planes BCD, BAD, at right angles to their common section BD, the angle EFC is the angle of inclination between the planes, and is accordingly equal to the spherical angle ABC. Now in the plane triangle CEF right-angled at E, CF is to the radius as CE is to the sine of EFC; but CF is the sine of the arc CB, and CE the sine of CA, and the angle EFC is equal to ABC; therefore the sine of BC is to the radius as the sine of AC is to the sine of the angle ABC. In like manner it may be shown that the sine of BC is to the radius, as the sine of AB is to the sine of ACB.

Cor. Let the angles of the triangle ABC right-angled at A, be denoted by A, B, C, and the opposite sides (or the angles measured by them) by a, b, c, re

spectively; and let the corresponding parts of a triangle complemental to ABC, be denoted by A', B', C', a', b', c', the angle A' being a right angle. Then (Art. 4), if B'=B; C'=comp. b, b'=comp. C, a'=comp. c, c'= comp. a: and if C'=C; B'=comp. c, c'=comp. B, a'= comp. b, b=comp. a.

Now by the preceding proposition,

sin a: R:: sin b: sin B; hence if unity be put for R, (1) sin b sin a sin B.

Likewise, since the same equation holds good in the complemental triangles,

sin b' sin a' sin B';

from which equation are derived the two following; one for the case in which B'=B, the other for that in which C'=C;

sin comp. C-sin comp. c sin B,

sin comp. a=sin comp. b sin comp. c;

or (2) cos C=cos c sin B,

(3) cos a=cos b cos c.

6. In a right-angled spherical triangle, the sine of either of the sides containing the right angle is to the radius as the tangent of the other of these sides is to the tangent of the opposite angle.

Let ABC be a triangle right-angled at A, the sine of AC is to the radius as the tangent of AB is to the tangent of the opposite angle ACB.

Let D be the center of the sphere; join DA, DB, DC,

E

draw AE perpendicular to CD; and AF a tangent to AB, meeting DB produced in F; also draw EF; this will be perpendicular to CD.

For since AF is a tangent to AB, it is perpendicular to AD; and being drawn in the plane BAD at right angles to AD the intersection of this plane with the plane CAD, it must be perpendicular to the latter plane, and perpendicular to AE in that plane. Then in the rightangled triangle FAD, FD2=AF2+AD2; and in the triangle AED, AD2=AE2+ED2; therefore FD2=AF2 +AE2+ED2; but AF2+AE2=FE2, because FAE is a right angle; hence FD2=FE2+ED2; and consequently FED is a right angle (Euc. 1, 48). Since then from E a point in CD the common section of the planes ACD, BCD, the lines EA, EF, are drawn in the two planes at right angles to CD, AEF is the angle of inclination between the planes, and is equal to the spherical angle ACB. Now in the plane triangle AEF right-angled at A, AE is to the radius as AF is to the tangent of the angle AEF; or since AE is the sine of AC, and AF the tangent of AB, the sine of AC is to the radius as the tangent of AB is to the tangent of ACB. In like manner it may be shown that the sine of AB is to the radius as the tangent of AC is to the tangent of ABC.

Cor. Hence (if the quantities concerned be denoted as in the corollary to the last proposition), are derived the following equations:

7. If of the sides and angles in a right-angled triangle, any two besides the right angle be given, the remaining three may in general be determined by means of the equations numbered (1), (2), (3), in the two preceding corollaries. And as in each computation there are three parts concerned, two given and one sought, the following six are all the cases that can occur: namely, those in which the three parts concerned are (1) The hypotenuse and the two oblique angles, (2) The three sides,

(3) The hypotenuse, another side, and the included angle,

(4) The hypotenuse, another side, and the opposite angle,

(5) The two perpendicular sides and one of the oblique angles,

(6) The two oblique angles and a side opposite one of them.

The equations belonging to these cases are, when placed in the corresponding order, as follows:

(1) cot B=cos a tan C.

(2) cos a=cos b cos c
(3) cot a=cos C cot b

(4) sin b=sin a sin B

(5) tan c=sin b tan C.

(6) cos C=cos c sin B

To these must be added four others; for it is evident that there are two equations belonging to each of the last four cases, one of which is to be deduced from the other by changing B into C, 6 into c, and vice versa. These additional equations are the following,