therefore the measures of the angles A, B, C are greater than a femicircle; and hence the angles A, B, C are greater than two right angles.

All the external and internal angles of any triangle are equal to fix right angles; therefore all the internal angles are lefs than fix right angles.

IF

PROP. XII. FIG. 8.

F from any point C, which is not the pole of the great circle ABD, there be drawn arches of great circles CA, CD, CE, CF, &c. the greatest of these is CA, which paffes through H the pole of ABD, and CB the remainder of ACB is the leaft, and of any others CD, CE, CF, &c. CD, which is nearer to CA, is greater than CE, which is more remote.

Let the common fection of the planes of the great circles ACB, ADB be AB; and from C, draw CG perpendicular to AB, which will alfo 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 greateft, and GB the leaft; (7. 3.) and GD, which is nearer to GA, is greater than GE, which is more remote. The triangles CGA, CGD are right-angled at G, and they have the common fide CG; therefore the fquares of CG, GA together, that is, the fquare of CA, is greater than the fquares of CG, GD together, that is, the fquare of CD: And CA is greater than CD, and therefore the arch CA is greater than CD. In the fame manner, fince GD is greater than GE, and GE than GF, &c. it is fhown that CD is greater than CE, and CE than CF, &c. and confequently, the arch CD greater than the arch CE, and the arch CE greater than the arch CF, &c. And fince GA is the greatest, and GB the leaft of all the ftraight lines drawn from G to the circumference ADB, it is manifest that CA is the greatest, and CB the leaft of all the ftraight lines drawn from C to the circumference: And therefore the arch CA is the greatest, and CB the leaft of all the circles drawn through C, meeting ADB. Q. E. D.

Ii2

PROP.

Fig. 10.

PROP. XII. FIG. 9.

IN a with the oppofite angles; that

Na right-angled fpherical triangle the fides are of

is, if the fides be greater or lefs than quadrants, the oppofite angles will be greater or less than right angles.

Let ABC be a spherical triangle right-angled at A, any fide AB, will be of the fame affection with the opposite angle ACB. Cafe 1. Let AB be lefs than a quadrant, let AE be a quadrant, and let EC be a great circle paffing 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 lefs than a right angle.. Q. E. D.

Cafe 2. Let AB be greater than a quadrant, make AE a qua-' drant, 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.

Fig.

PROP. XIV.

F the two fides of a right-angled spherical triangle be of the fame affection, the hypothenufe will be lefs than a quadrant; and if they be of different affection, the hypothenuse will be greater than a quadrant.

Let ABC be a right-angled spherical triangle, if the two fides AB, AC be of the fame or of different affection, the hypothenufe BC will be lefs or greater than a quadrant.

Cafe 1. Let AB, AC be each lefs 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, fince CB is farther off from CGD than CE. In the fame manner, it is shown that CB, in the triangle CBD, where the two fides CD, BD are each greater than a quadrant, is lefs than CE, that is, lefs than a quadrant. Q. E. D.

Cafe

Cafe 2. Let AC be lefs, and AB greater than a quadrant; Fig. 10. then the hypothenufe 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 CB is greater than CE by prop. 12. fince AC paffes through the pole of ABD. Q. E. D.

IF

PROP. XV.

F the hypothenufe of a right-angled triangle be greater or less than a quadrant, the fides will be of different or the fame affection.

This is the converse of the preceding, and demonstrated in the fame manner.

PROP. XVI.

IN any spherical triangle ABC, if the perpendicular AD from A upon the bafe BC, fall within the triangle, the angles B and C at the base will be of the fame 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) fince Fig. 11, ADB, ADC are right-angled fpherical triangles, the angles B, C muft each be of the fame affection as AD.

2. Let AD fall without the triangle, then (13. of this) the Fig. 12. angle B is of the fame affection as AD; and by the fame the angle ACD is of the fame 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 fame affection, the perpendicular will fall within the bafe; 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 oppofite 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 fame affection, contrary to the fuppofition.

PROP. XVII. FIG. 13.

[N right-angled fpherical triangles, the fine of either of the fides about the right angle, is to the radius of the sphere, as the tangent of the remaining fide is to the tangent of the angle oppofite to that fide.

Let ABC be a triangle, having the right angle at A; and let AB be either of the fides, the fine of the fide AB will be to the radius, as the tangent of the other fide AC to the tangent of the angle ABC, oppofite 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 fine of the arch AB, and from the point E, let there be drawn in the plane BDC the ftraight 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 paffes through DE is perpendicular to the plane AEF, (18. 11.) and the plane AEF perpendicular to ABD: The plane ACD or AFD is alfo perpendicular to the fame ABD : Therefore the common fection, 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 will be to the radius as AF to the tangent of the angle AEF, ( Pl. Tr.); but AE is the fine of the arch AB, and AF the tangent of the arch AÇ, and the angle AEF is the inclination of the planes CBD, ABD, (6. def. 11.) or the fpherical angle ABC: Therefore the fine of the arch AB is to the radius as the tangent of the arch AC, to the tangent of the oppofite angle ABC.

COR. 1. If therefore of the two fides, and an angle oppofite to one of them, any two be given, the third will also be given.

COR. 2. And fince by this propofition the fine of the fide AB is to the radius, as the tangent of the other fide AC to the tangent

tangent of the angle ABC oppofite to that fide; and as the radius is to the co-tangent of the angle ABC, fo is the tangent of the fame angle ABC to the radius, (Cor. 2. def. Pl. Tr.) by equality, the fine of the fide AB is to the co-tangent of the angle ABC adjacent to it, as the tangent of the other fide AC to the radius.

PROP. XVIII. FIG. 13.

N right-angled fpherical triangles the fine of the byApothenufe is to the radius, as the fine of either fide is to the fine of the angle opposite to that side.

Let the triangle ABC be right-angled at A, and let AC be either of the fides; the fine of the hypothenufe BC will be to the radius as the fine of the arch AC is to the fine 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 fine of the hypothenuse BC; and from the point G let there be drawn in the plane ABD the ftraight line GH perpendicular to DB, and let CH be joined; CH will be at right angles to the plane ABD, as was shown in the preceding propofition of the straight line FA; Wherefore CHD, CHG are right angles, and CH is the fine of the arch AC; and in the triangle CHG, having the right angle CHG, CG is to the radius as CH to the fine of the angle CGH: (1. Pl. Tr.) But fince CG, HG are at right angles to DGB, which is the common fection of the planes CBD, ABD, the angle CGH will be equal to the inclination of these planes; (6. def. 11.) that is, to the fpherical angle ABC. The fine therefore, of the hypothenufe CB is to the radius as the fine of the fide AC is to the fine of the oppofite angle ABC. Q. E. D. COR. Of these three, viz. the hypothenufe, a fide, and the angle oppofite to that fide, any two being given, the third is alfo given by prop. 2.