Remark. The three sides of a spherical triangle are together less than the circumference of a great circle. For the plane angles at the point D are together less than four right angles, and consequently the sides of the triangle, which measure the plane angles at D, must together be less than the circumference of a great circle.

PROPOSITION VI.

In a spherical triangle the greater side is opposite the greater angle, and the greater angle opposite the greater side.

Let A B C be a spherical triangle, if the angle B be greater than the angle A, the side A C will be greater

B

Again, if A C be greater than B C, the angle A B C will be greater than the angle A.

For if A B C is not greater than A, it must either be equal to it, or less. If it were equal, then A C would be equal to B C (Spher. Prop. 4.) which it is not. And if A B C were less than A, A would be greater than A B C; and therefore by the former part of this proposition B C would be greater than A C, which it is not. Hence as A B C is neither equal to, nor less than A, it must be greater.

PROPOSITION VII.

Three great circles, whose poles are the angular points of a spherical triangle, will by their intersection form another triangle, whose angular points are the poles of the great circles, which by their intersection form the sides of the first triangle; and the sides of each of the triangles will be the supplements of the arcs which measure the angles of the other. In the annexed figure, if A, B, and C be the poles of D E, E F, and FD respectively, then D, E, and F will be the poles of A C, A B, and B C respectively, and DE, E F, and FD will respectively be the supplements of the arcs which measure the angles B A C, A B C, and AC B, and A C, A B, and B C will respectively be the supple- M ments of the arcs which measure the angles D, E, and F.

D

G

F

L

K

I

B

E

H

For let A B, A C, and B C be produced both ways, till they meet DE, E F, and DF in G, H, I, K, L, and M. Then G H is the measure of the angle B A C, K I the measure of the angle ABC, and M L the measure of the angle A C B. (Spher. Prop. 3.) And as A is the pole of D E, the angle A HD is a right angle, (Spher. Prop. 3. Cor. 3.) and for a like reason, the angle C L D is a right

angle; hence D is the pole of L H. In the same way it may be shewn that E is the pole of G K, and F the pole of MI; and consequently L H is the measure of the angle D, G K the measure of the angle E, and M I the measure of the angle F.

Now as D is the pole of L H, D H is a quadrant; and as E is the pole of G K, EG is a quadrant; therefore E G and D H, or ED and GH together are equal to a semicircle. Hence DE is the supplement of G H, the measure of the angle BA C. And in like manner it may be shewn that E F is the supplement of the measure of A B C, and D F the supplement of the measure of A C B.

Again, because A is the pole of D E, A G is the quadrant; and because B is the pole of E F, B K is a quadrant. Hence A G and B K, or K G and A B together are equal to a semicircle; or A B is the supplement of G K, the measure of the angle E. And in the same way it may be shewn that A C is the supplement of the measure of D, and B C the supplement of the measure of F.

PROPOSITION VIII.

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

For (see last figure) the measures of the angles A, B, C, in the triangle A B C, together with the three sides of the supplemental triangle D E F, are equal to three semicircles. But the three sides of the triangle A B C are less than two semicircles, (Spher. Prop. 5. Remark) therefore the measures of the angles A, B, and C are greater than a semicircle; hence the angles A, B, and C are greater than two right angles.

And as the interior angles of any triangle together with the exterior are equal to six right angles, the interior angles alone must be less than six right angles.

PROPOSITION IX.

If to the circumference of a great circle, from a point in the surface of a sphere, which is not the pole of that circle, arcs of great circles be drawn, the greatest of those arcs is that which passes through the pole of the firstmentioned circle, and of the other arcs that which is nearer to the greatest is greater than the more remote.

D

Let A G F B be the circumference of a great circle, D its pole, and let ADCB be a great circle passing through D. Then the plane of the circle A D C B will be at right angles to the plane of the circle A G F B. (Spher. Prop. 3. Cor. 3.) From C in the plane A C B draw C E at right angles to A B, the line of common section of the planes of the two circles, draw from E any

A

F

B

lines E G, EF, and join A C, GC, and F C, and through G C, and F C, let arcs of great circles be described.

Then (Planes, Def. 2.) C E is at right angles to the A G B, therefore the angles CEA, CE G, and CEF are right angles. Hence C A2 = A E2 + E C2, C G2 = G E2 + E C2, and F C2 = F E2 +E C2.

But (Geo. Theo. 52.) A E is greater than G E, and G E is greater than FE; therefore A E2 is greater than G E2, and G E2 than F E2, consequently A C2 is greater than G C2, and G C2 than F C2, or A C is greater than G C, and G C than G F. Hence the arc ADC is greater than the arc G C, and the arc G C is greater than the arc CF, which is more remote from A D C.

PROPOSITION X,

In right angled spherical triangles the sides containing the right angle are of the same affection as their opposite angles; that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles; and, conversely, if the angles be greater or less than right angles, the opposite sides will be greater or less than quadrants. Let A B C be a spherical triangle, right angled at A; produce A C, A B, till they meet in G, and bisect A B G and A CG in E and F. Then F will be the pole of A B G, and E will be the pole of A F G. Join C E, then C E will be a quadrant, and the angle ECA will be a right angle. Hence when A B is

B

E

G

less than A E, or less than a quadrant, the opposite angle B C A will be less than the angle E C A, or less than a right angle.

Again, let ADC be a spherical triangle right angled at A, having AD greater than a quadrant, then the angle DCA will be greater than the angle E C A, or greater than a right angle.

In a similar manner may the converse be demonstrated.

PROPOSITION XI.

If the sides which contain the right angle of a spherical triangle be of the same affection, the hypothenuse will be less than a quadrant; but if they be of different affections, the hypothenuse will be greater than a quadrant.

For (see the last figure) when A C and A B are each less than a quadrant, C B being farther from C F G than CE, is therefore less than C E, or less than a quadrant. But if A D be greater, and A C

be less than a quadrant, then CD being nearer C F G and C E, is greater than C E, or greater than a quadrant.

Again, in the right angled triangle C G B, where C G and G B are each greater than a quadrant, C B being farther from CFG than CE, is less than C E, or less than a quadrant.

Cor. 1. Hence, conversely, if the hypothenuse of a right angled spherical triangle be greater than a quadrant, the sides will be of different affections, and if the hypothenuse be less than a quadrant, the sides will be of the same affection.

Cor. 2. As the oblique angles of a right angled spherical triangle have the same affection with the opposite sides, those angles will be of the same or different affections, according as the hypothenuse is less or greater than a quadrant.

Cor. 3. When a side and the adjacent angle are of the same affection, the hypothenuse is less than a quadrant, and when a side and the adjacent angle are of different affections, the hypothenuse is greater than a quadrant.

PROPOSITION XII.

In any spherical triangle, if the perpendicular from one of the angles upon the opposite side fall within the triangle, the other angles of the triangle will be of the same affection, but if the perpendicular fall without the triangle, the other angles will be of different affections.

Let A B C be a triangle, and BD a perpendicular drawn from B on A C, or AC produced.

When B D falls within the triangle, the angles A and C of the right angled triangles A D B and B DC being of the same affection with B D, are of the same affection with each other.

But when the perpendicular falls without the triangle, the angles DA B and D C B, being of the same affection with B D, are of the same affection with each other; hence, as D A B is of the same affection with D C B, B A C and C are of different affections.

PROPOSITION XIII.

In any right angled spherical triangle the rectangle of radius and the sine of either of the sides containing the right angle is equal to the rectangle of the tangent of the other side, and the cotangent of the angle opposite to that side.

G
E

A

Let A B C be a spherical triangle, right angled at B; and let D'be the centre of the sphere. From B, in the plane DA B, let BE be drawn perpendicular to DA; and from E in the plane A C D let E F be drawn also perpendicular to D A, meeting DC produced in F, and join F B. Then as DE is perpendicular to E F and E B, it is perpendicular to the plane E B F, therefore the plane E B F is perpendicular to the plane A D B which passes through D E. And as the angle C B A is a right angle, the plane F D B is at right angles to the plane A D B; hence B F, the common section of the planes, E B F and F D B is at right angles to the plane D A B; and consequently the angles FBD and F B E are right angles, and F B is therefore the tangent of B C. Now in the right angled plane triangle E B F, we have E B BF:: rad: tan B E F. But E B is sine of A B, and the angle B E F is the angle made by the planes A D C and AD B, or it is equal to the spherical angle B A C. Hence sin A B tan BC: rad tan B A C. But rad; tan B A C cot B A C rad; therefore sin AB: tan B C

::cot BAC: rad.

Or rad, sin A B = tan B C. cot B A C

And similarly we have rad. sin BC= tan AB; cot B CA.

PROPOSITION XIV.

In any right angled spherical triangle the rectangle of radius and the sine of either of the sides containing the right angle is equal to the rectangle of the sine of the angle opposite that side and the sine of the hypothenuse.

From C (see the last figure) in the plane C D A, let C G be drawn perpendicular to A D; and from G in the plane B D A let GH be drawn also perpendicular to D A, meeting D B in H, and join C H.

Then it may be shewn exactly, as in the last proposition, that C H G is a right angled plane triangle, and that C H is the sine of BC; and the angle C G H is equal to the spherical angle BA C. Now in the plane triangle CG H, we have CG: CH:: rad : sine C G H, or

sin A C sin B C :: rad: sin B A C

:

Hence rad. sin B C = sin A C. sin BAC

And similarly we have rad, sin A B = sin A C. sin B C A

PROPOSITION. XV.

In any right angled spherical triangle A B C (B being the right angle) the following equations obtain.