Let AB produced meet DE, EF in G, M, and AC meet FD, FE in K, L, and let BC meet FD, DE in N, H. Since A is the pole of FE, and the circle AC passes through A, EF will pass through the pole of AC, (13. 15. I. Th.); and since AC passes through C, the pole of FD, FD will pass through the pole of AC; therefore the pole of AC is in the point F, in which the arches DF, EF intersect each other. In the same manner, D is the pole of BC, and E the pole of AB. And since F, E are the poles of AL, AM, the arches FL and EM (2. of this) are quadrants, and FL, EM together, that is, FÈ and ML together, are equal to a semicircle. But since A is the pole of ML, ML is the measure of the angle BAC, (3. of this,) consequently FE is the supplement of the measure of the angle BAC. In the same manner, ED, DF are the supplements of the measures of the angles ABC, BCA. Since likewise CN, BH are quadrants, CN and BH together, that is, NH and BC together, are equal to a semicircle; and since D is the pole of NH, NH is the measure of the angle FDE, therefore the measure of the angle FDE is the supplement of the side BC. In the same manner, it is shown, that the measures of the angles DEF, EFD are the supplements of the sides AB, AC, in the triangle ABC. Q. E. D. PROP. XI. FIG. 7. The three angles of a spherical triangle are greater than two right angles, 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 (10. of this) equal to three semicircles; but the three sides of the triangle FDE, are (7. of this) 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 since all the external and internal angles of any triangle are together equal to six right angles; therefore all the internal angles alone are less than six right angles. Q. E. D. PROP. XII. FIG. 8. If from any point C, which is not the pole of the great circle ADB, there be drawn arches of great circles CA, CD, CE, CF, &c. the greatest of these is CA, which passes through H the pole of ADB, and CB the remainder of ACB is the least, and of any other CD, CE, CF, &c. CD, which is nearer to CA, is greater than CE, which is more remote. Let the common section of the planes of the great circles ACB, ADB be AB; and from C, draw CG perpendicular to AB, which will also be perpendicular to the plane ADB, (4. def. 11); join GD, GE, GF; CD, CE, CF, CA, CB. Of all the straight lines drawn from G to the circumference ADB, GA is the greatest, and GB the least, (7. 3.); and GD, which is nearer to GA, is greater than GE, which is more remote. But the triangles CGA, CGD are right angled at G, and they have the common side CG; therefore the squares of CG, GA together, that is, the square of CA, is greater than the squares of CG, GD together, that is, the square of CD: And CA is greater than CD, and therefore the arch CA is greater than CD. In the same manner, since GD is greater than GE, and GE than GF, &c. it may be shown that CD is greater than CE, and CE than CF, &c. and consequently, the arch CD greater than the arch CE, and the arch CE greater than the arch CF, and CF than CB. Wherefore, of all the arches of great circles drawn from C to the circumference of the circle ADB, CA, which passes through the pole H, is the greatest, and CB the least; and of the others, that which is nearer to CA the greatest, is greater than that which is more remote. Q. E. Ď. PROP. XIII. FIG. 9, 10. In a right angled spherical triangle, the sides are of the same affection with the opposite angles; that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles. Let ABC be a spherical triangle, right angled at A, any side AB will be of the same affection with the opposite angle ACB. Fig. 10. Fig. 9. Fig. 10. Case 1. Let AB be less than a quadrant, let AE be a quadrant, and let EC be a great circle passing through E, C. Since A is a right angle, and AE a quadrant, E is the pole of the great circle AC, and ECA a right angle; but ECA is greater than BCA, therefore BCA is less than a right angle. Q. E. D. Case 2. Let AB be greater than a quadrant, make AE a quadrant, and let a great circle pass through C, E, ECA is a right angle as before, and BCA is greater than ECA, that is, greater than a right angle. Q. E. D. PROP. XIV. FIG. 9, 10. 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. Case 1. Let AB, AC be each less than a quadrant. Let AE, AG be quadrants; G will be the pole of AB, and E the pole of AC, and EC a quadrant; but, by Prop. 12, CE is greater than CB, since CB is farther off from CGD than CE. In the same manner, it is shown of the right angled triangle CDB, in which each of the sides CD, BD is greater than a quadrant, that the hypotenuse CB is less than CE, that is, less than a quadrant. Q. E. D. Case 2. Let AC be less, and AB greater than a quadrant; then the hypotenuse BC will be greater than a quadrant; for, let AE be a quadrant, then E is the pole of AC, and EC will be a quadrant. But because CB falls between CGD and CE, it is greater than CE by Prop. 12, that is, it is greater than a quadrant. Q. E. D. PROP. XV. 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. This is the converse of the preceding, and demonstrated in the same manner. PROP. XVI. In any spherical triangle ABC, if the perpendicular AD from A on the base BC fall within the triangle, the angles B and C at the base will be of the same affection; and if the perpendicular fall without the triangle, the angles B and C will be of different affection. 1. Let AD fall within the triangle; then (13. of this) since Fig. 11. ADB, ADC are right angled spherical triangles, the angles B, C must each be of the same affection with AD. 2. Let AD fall without the triangle; then (13. of this) the Fig. 12. angle B is of the same affection as AD; and the angle ACD of the same affection as AD; therefore the angle ACB and AD are of different affection, and the angles B and ACB of different affection. COR. Hence if the angles B and C be of the same affection, the perpendicular will fall within the base; for if it did not, (16. of this,) B and C would be of different affection. And if the angles B and C be of different affection, the perpendicular will fall without the triangle; for, if it did not, (16. of this,) the angles B and C would be of the same affection, contrary to the supposition. PROP. XVII. FIG. 13. 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 AE be drawn perpendicular to BD, which therefore will be the sine of the arch AB, and from the point E let there be drawn in the plane BDC the straight line EF at right angles to BD, meeting DC in F, and let AF be joined. Since therefore the straight line DE is at right angles to both EA and EF, it will also be at right angles to the plane AEF, (4. 11,) wherefore the plane ABD, which passes through DE, is perpendicular to the plane AEF, (18. 11,) and the plane AEF perpendicular to ABD. The plane ACD or AFD is also perpendicular to the same ABD; because the spherical angle BAC is a right angle: therefore the common section of the planes AFD, AFE, viz. the straight line AF, is at right angles to the plane ABD, (19. 11.): And FAE, FAD are right angles, (3. def. 11.): therefore AF is the tangent of the arch AC; and in the rectilineal triangle AEF having a right angle at A, AE is to the radius as AF to the tangent of the angle AEF, (1. Pl. Tr.); but AE is the sine of the arch AB, and AF the tangent of the arch AC, and the angle AEF is the inclination of the planes CBD, ABD, (6. def. 11,) or is the same with the spherical angle ABC: Therefore the sine of the arch AB is to the radius as the tangent of the arch AC, to the tangent of the opposite angle ABC. Q. E. D. COR. 1. If therefore of the two sides, and an angle opposite to one of them, any two be given, the third will also be given. COR. 2. 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, (Cor. 2. def. Pl. Tr.) 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. PROP. XVIII. FIG. 13. 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 arch AC is to the sine of the angle ABC. Let D be the centre of the sphere, and let CG be drawn perpendicular to DB, which will therefore be the sine of the |