PROP. XII.

The three angles of a spherical triangle are greater than two, and less than six, right angles.

The measures of the angles A, B, C, in the triangle ABC, together with the three sides of the supplemental triangle DEF, are (11.) equal to three semicircles; but the three sides of the triangle FDE, are (8.) less than two semicircles; therefore the measures of the angles A, B, C are greater than a semicircle; and hence the angles A, B, C are greater than two right angles.

And because the interior angles of any triangle, together with the exterior, are equal to six right angles, the interior alone are less than six right angles. Q. E. D.

PROP. XIII.

If to the circumference of a great circle, from a point, in the surface of the sphere, which is not the pole of that circle, arches of great circles be drawn; the greatest of these arches is that which passes through the pole of the first-mentioned circle, and the supplement of it is the least; and of the other arches, that which is nearer to the greatest is greater than that which is more remote.

Let ADB be the circumference of a great circle, of which the pole is H, and let C be any other point; through C and H let the semicircle ACB be drawn meeting the circle ADB in A and B ; and let the arches CD, CE, CF also be described. From C draw CG perpendicular to AB, and then, because the circle AHCB which passes through H, the pole of the circle ADB, is at right angles to ADB, CG is perpendicular to the plane ADB. Join GD, GE, GF, CA, CD, CE, CF, CB. H

Because AB is the diameter of the circle ADB, and G a point in it, which is not the centre, (for the centre is in the point where the perpendicular from H meets AB), therefore AG, the part of the diameter in which the centre is, is the greatest, (7. 3.), and GB the least of all the straight lines that can be drawn from G to the circumference; and GD, which is nearer to AB, is greater than GE, which is more remote. But the triangles CGA, CGD are right angled at G, and therefore AC2= AGGC2, and DC2=DG2+GC2; but AG2+GC2 7DG3+GC2; because AG DG; therefore AC2¬DC2, and AC DC. And because the chord AC is greater than the chord DC, the arch AC is

D

E

greater than the arch DC. In the same manner, since GD is greater than GE, and GE than GF, it is shewn that CD is greater than CE, and CE than CF. Wherefore also the arch CD is greater than the arch CE, and the arch CE greater than the arch CF, and CF than CB; that is, of all the arches of great circles drawn from C to the circumference of the circle ADB, AC which passes through the pole H, is the greatest, and CB its supplement is the least; and of the others, that which is nearer to AC the greatest, is greater than that which is more remote. Q. E. D.

PROP. XIV.

In a right angled spherical triangle, the sides containing the right angle, are of the same affection with the angles opposite to them, that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles, and conversely.

Let ABC be a spherical triangle, right angled at A, any side AB will be of the same affection with the opposite angle ACB.

C

G

Produce the arches AC, AB, till they meet again in D, and bisect AD in E. Then ACD, ABD are semicircles, and AE an arch of 90o. Also, because CAB is by hypothesis a right angle, the plane of the circle ABD is perpendicular to the plane of the circle ACD, so that the pole of ACD is in ABD, (cor. 1. 4.), and is therefore the point E. Let EC be an arch of a great circle passing through E and C.

Then because E is the pole of the circle ACD, EC is a (2.) quadrant, and the plane of the circle EC (4.) is at right angles to the plane of the circle ACD, that is, the spherical angle ACE is a right angle; and therefore, when AB is less than AE, the angle ACB, being less than ACE, is less than a right angle. But when AB is greater than AE, the angle ACB

is greater than ACE, or than a right angle. In the same way may the converse be demonstrated. Therefore, &c. Q. E. D.

1

PROP. XV.

If the two sides of a right angled spherical triangle about the right angle be of the same affection, the hypotenuse will be less than a quadrant; and if they be of different affection, the hypotenuse will be greater than a quadrant.

Let ABC be a right angled spherical triangle; according as the two sides AB, AC are of the same or of different affection, the hypotenuse BC will be less, or greater than a quadrant.

The construction of the last proposition remaining, bisect the semicircle ACD in G, then AG will be an arch of 90°, and G will be the pole of the circle ABD.

1. Let AB, AC be each less than 90°. Then, because C is a point on the surface of the sphere, which is not the pole of the circle ABD, the arch CGD, which passes through G the pole of ABD is greater than CE, (13.), and CE greater than CB. But CE is a quadrant, as was before shewn, therefore CB is less than a quadrant. Thus also it is proved of the right angled triangle CDB, (right angled at D), in which each of the sides CD, DB is greater than a quadrant, that the hypotenuse BC is less than a quadrant.

2. Let AC be less, and AB greater than 90o. Then because CB falls between CGD and CE, it is greater (13.) than CE, that is than a quadrant. Q. E. D.

1

COR. 1. Hence conversely, if the hypotenuse of a right angled triangle be greater or less than a quadrant, the sides will be of different or the same affection.

COR. 2. Since (14.) the oblique angles of a right angled spherical triangle have the same affection with the opposite sides, therefore, according as the hypotenuse is greater or less than a quadrant, the oblique angles will be of different, or of the same affection.

COR. 3. Because the sides are of the same affection with the opposite angles, therefore when an angle and the side adjacent are of the same affection, the hypotenuse is less than a quadrant; and conversely.

PROP XVI.

In any spherical triangle, if the perpendicular upon the base from the opposite angle fall within the triangle, the angles at the base are of the same affection; and if the perpendicular fall without the triangle, the an gles at the base are of different affection.

Let ABC be a spherical triangle, and let the arch CD be drawn from C perpendicular to the base AB.

1. Let CD fall within the triangle; then, since ADC, BDC are right angled spherical triangles, the angles A, B must each be of the same affection with CD, (14).

B

D

A

B D A

2. Let CD fall without the triangle; then (14.) the angle B is of the same affection with CD; and the angle CAD is of the same affection with CD; therefore the angle CAD and B are of the same affection, and the angle CAB and B are therefore of different affections. Q. E. D.

COR. Hence, if the angles A and B be of the same affection, the perpendicular will fall within the base; for if it did not, A and B would be of different affection. And if the angles A and B be of different affection, the perpendicular will fall without the triangle; for, if it did not, the angles A and B would be of the same affection, contrary to the supposition.

PROP. XVII.

If to the base of a spherical triangle a perpendicular be drawn from the opposite angle, which either falls within the triangle, or is the nearest of the two that fall without; the least of the segments of the base is adjacent to the least of the sides of the triangle, or to the greatest, according as the sum of the sides is less or greater than a semicircle.

E

B

Let ABEF be a great circle of a sphere, H its pole, and GHD any circle passing through H, which therefore is perpendicular to the circle ABEF. Let A and B be two points in the circle ABEF, on opposite sides of the point D, and let D be nearer to A than to B, and let C be any point in the circle GHD between H and D. Through the points A and C, B and C, let the arches AC and BC be drawn, and let them be produced till they meet G the circle ABEF in the points E and F, then the arches ACE, BCF are semicircles. Also ACB, ACF, CFE, ECB are four spherical triangles contained by arches of the same circles, and having the same perpendiculars CD and CG.

F

H

K k

1. Now because CE is nearer to the arch CHG than CB is, CE is greater than CB, and therefore CE and CA are greater than CB and CA, wherefore CB and CA are less than a semicircle; but because AD is by supposition less than DB, AC is also less than CB, (13.), and therefore in this case, viz. when the perpendicular falls within the triangle, and when the sum of the sides is less than a semicircle, the least segment is adjacent to the least side.

2. Again, in the triangle FCA the two sides FC and CA are less than a semicircle; for, since AC is less than CB, AC and CF are less than BC and CF. Also, AC is less than CF, because it is more remote from CHG than CF is; therefore in this case also, viz. when the perpendicular falls without the triangle, and when the sum of the sides is less than a semicircle, the least segment of the base AD is adjacent to the least side.

3. But in the triangle FCE the two sides FC and CE are greater than a semicircle; for, since FC is greater than CA, FC and CE are greater than AC and CE. And because AC is less than CB, EC is greater than CF, and EC is therefore nearer to the perpendicular CHG than CF is, wherefore EG is the least segment of the base, and is adjacent to the greater side.

4. In the triangle ECB the two sides EC, CB are greater than a semicircle; for, since by supposition CB is greater than CA, EC and CB are greater than EC and CA. Also, EC is greater than CB, wherefore in this case, also, the least segment of the base EG is adjacent to the greatest side of the triangle. Therefore, when the sum of the sides is greater than a semicircle, the least segment of the base is adjacent to the greatest side, whether the perpendicular fall within or without the triangle: and it has been shewn, that when the sum of the sides is less than a semicircle, the least segment of the base is adjacent to the least of the sides, whether the perpendicular fall within or without the triangle. Wherefore, &c. Q. E. D.

PROP. XVIII.

In right angled spherical triangles, the sine of either of the sides about the right angle, is to the radius of the sphere, as the tangent of the remaining side is to the tangent of the angle opposite to that side.

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 arch AB, and from the point F, let there be drawn in the plane BDC the straight line FE at right angles to BD, meeting DC in E, and let AE be joined. Since therefore the straight