I Cor. 4.

b 4.

For, BD, DC, two fides of the triangle BCD, are greater than the third BC. Add BA, AC; then DBA, DCA, the two femicircles, are greater than the three fides of the triangle BCD. Wherefore, &c.

IN any triangle, the greater angle is fubtended by the greater fide.

Let ABC be the triangle, and A the greater angle, then BC will be the greater fide. For, make the angle BAĎ equal to the a cor. 23. angle B; then AD will be equal to BD 2; therefore the fide BDC is equal to AD and DC; but AD, DC, are greater than AC; therefore BC is greater than AC. Wherefore, &c.

2. 4.

PRO P. VIII.

IN any Spherical triangle, if the fum of two of its fides be greater than a femicircle, then the internal angle at the base will be greater than the external and oppofite angle; and the fum of the internal angles at the bafe will be greater than two right angles ; if equal, equal, and, if lefs, less.

Let ABC be the spherical triangle; if the two fides, AB, BC, be greater than a femicircle, the internal angle BAC, at the bafe, will be greater than the external and oppofite angle BCD; if equal, equal, and, if lefs, lefs; and the angles A and ACB will likewife be greater, equal, or lefs, than two right angles. First, let the femicircles ACD, ABD, be compleated; then, if AB, BC, be equal to ABD, the angle BCD will be equal to BDC, that is, to BAC. 2dly, If AB, BC, be greater than ABD, then BC will be greater than BD, and the angle BDC, that is, the angle BAC, greater than BCD b; if AB, BC, are lefs than a femicircle; then the angle A will be lefs than BCD. And, because the angles BCD, BCA, are equal to two right angles, if the angle A be greater than the angle BCD, then the angles A and ACB will be greater than two right angles, and, if lefs, lefs.

I

PRO P. IX.

F the poles of the fides of any Spherical triangle be joined by great circles, they conftitute another triangle, the fides of which are fupplements of the arches that measure the angles of the given triangles; and the arches that are the measures of the angles of the fupplementary triangle, are the fupplements of the fides of the given triangle.

Let G, H, D, be the angular points of the given triangle GHD; and let the points G, H, D, be the poles of the great circles XCAM, TMNO, XKBN; then XN will be the fupplement of BK, XM of CA, and MN of OT. Likewise, the arches KT, OC, and BA, which are the measures of the angles M, X, N, are the fupplements of HD, HG, and GD.

d

For, becaufe G is the pole of the circle XCAM, GM is a quadrant; and, because H is the pole of the circle TMO, HM a cor. 1. 2. is alfo a quadrant: Wherefore, M is the pole of the circle GH. & cor. 1. 3 For the fame reason, N is the pole of the circle HD, and X the pole of the circle GD. Now, because NK, XB, are each quadrants, XN is the supplement of KB. For the fame reason, XM is the fupplement of AC, and MN of OT; which are the meafures of the angles G, H, D.

Again, because DK, HT, are each quadrants, KT is the fupplement of HD. For the same reason, OC is the fupplement of GH, and BA of GD; that is, the measures of the angles X, M, N, are the fupplements of the fides HD, GH, and GD. Wherefore, &c.

T

PRO P. X.

HE three angles of a spherical triangle are greater than two
right angles, and less than fix.

Let the triangle be GHD; then the three measures of it, with the three fides of the triangle XMN, are equal to three femicircles ; but the three fides of the triangle XMN are lefs a 9. than two cemicircles ; therefore the measure of the three angles b 6. G, H, D, are greater than one; that is, greater than two right angles; but the outward and inward angles of any triangle are together equal to fix right angles; therefore the inward angles are lefs than fix right angles. Wherefore, &c

PRO P.

11.

b 7.3.

PRO P. XI.

IF, in any great circle, a point is taken, which is not the pole of it, and from that point several arches are drawn to its circumference, the greatest of thefe arches is that which passes through the pole; and the remainder of it is the leaft; and the arch nearer to that, paffing through the pole, is greater than that more remote ; and they make obtufe angles with the great circle.

Let AFBE be a great circle, and any point R taken, which is not the pole of it, and from that point the arches RA, RB, RG, RV, of great circles to the circumference of AFBE, the arch RCA, which paffes through the pole, is the greateft, and RB is the leaft; and the arch RCA is greater than RG, and RG greater than RV.

For, becaufe C is the pole of the circle AFB, CD and RS, a 2. and 7. that is parallel to CD, are perpendicular to the plain AFB, from the point S draw SA, SG, SV; then SA is the greatest line, viz. greater than SG, and SG greater than SV b. For, in the right angled plain triangles RSA, RSG, RSV, the fquares of RS, SA, that is, the fquare of RA, is greater than the fquares of RS, SG, that is, than the fquare of RG, that is, RA is greater than RG. For the fame reason, RG is greater than RV; therefore, the arch RA is greater than the arch RG, and RG greater than RV.

cor. 3.

Again, the angle RGA is greater than the angle CGA, which is a right angle ; and the angle RVA greater than CVA, a part of it; therefore the angles RGA, RVA, are obtufe angles.

PROP. XII.

IF F the fides containing the right angles of a spherical triangle_be of the fame affection with the oppofite angles, that is, if the fides are greater or less than quadrants, the oppofite angles will be greater or less than right angles.

Let AGR, AGX, be right angled spherical triangles, ha ving the angles GAR, GAX, right ones; then, if the ide AR be greater than a quadrant, the angle AGR will be greater than a right angle; and, if AX be lefs than a quadrant, the angle AGX is less than a right angle.

For, if AC is a quadrant, C is the pole of the circle AFB, and the angles AGC, AVC, are right ones; therefore the fide AR fubtending the angle AGR, is greater than a right angle a; a 7. and, because AX is less than a quadrant, the angle AGX is lefs than a right angle.

I

F the two fides containing the angle of a spherical triangle be both lefs, or both greater than quadrants, then the hypothenuse is less than a quadrant.

In the triangle ARV, or BRV, let F be the pole of the circle AR; then RF is a quadrant, which is greater than RV.

IF one of the fides is greater, and the other less than a quadrant, then the hypothenufe will be greater than a quadrant.

For, in the triangle ARG, the hypothenufe RG is greater than RF, that is, greater than a quadrant. For the fame reafon, if the hypothenufe is greater than a quadrant, then one of the legs is greater, and the other lefs, than a quadrant.

PRO P. XV.

IF the angles at the base of a spherical triangle be both lefs, or both greater than quadrants, the perpendicular will fall within the triangle; but, if one be greater, and the other less, the perpendicular will fall without the triangle.

Let ABC be the triangle; from the point A let fall the perpendicular AP; in the first case, it will fall within the triangle; but, if not, it will fall without; then, in the triangle APB, the fide AP, and angle B, are of the fame affection, and like-Fig. 2. wife the fide AP, and angle ACP: Therefore, fince the angles ABC, and ACP, are of the fame affection, the angles ACB and ABC are of different affections; but they are not. Where-By Hyp. fore, &c.

In the fecond cafe, if the perpendicular does not fall without, let it fall within. Then, in the triangle ABP, the angle B, and fide AP, are of the fame affection; and likewife, in the triangle ACP, the angle C, and fide AP, are of the fame af fection; therefore, the angles B and C are of the fame affection which is impoffible. Wherefore, &c.

PROP. XVI.

IN right angled fpherical triangles, having the fame or equal acute angles at the base, the fines of the hypothenufe are proportional to the fines of the perpendicular arches, and the fines of the bafes proportional to the tangents of the perpendicular arches.

Let the triangles be BAC, BHE, right angled at A and H, and the fame acute angle B, at the base BA, the fines of the hypothenufes CP, QE, are to one another as the fines of the perpendicular arches CD, EF; and the fines of the bafis AQ, HK, proportional to IA, GH, the tangents of the perpendicular

arches.

For, because CD, EF, are perpendicular to the fame plain, they are parallel; as alfo, FR, DP; therefore the plains of the triangles EFR, CDP, are parallel ; and CP, ER, the common fections of thefe plains, with the plains paffing through BE, EO, will be parallel ; therefore, the triangles CDP, EFR, are equi angular; wherefore CP, the fine of the hypothenufe BC, is to CD, the fine of the perpendicular arch CA, as ER, the fine of the hypothenufe BE, is to EF, the fine of the perpendicular arch EII. For the fame reason, the triangles QAI, KHG, are equiangular. Wherefore QA is to AI, as KH is to HG, the tangents of the perpendicular arches. Wherefore, &c.

IN any right angled Spherical triangle, the cofine of the angle at the bafe, is to the fine of the vertical angle, as the cofine of the perpendicular is to the radius.

ft, Let the triangle be ABC, right angled at A, the cofine of B is to the fine of the angle ACB, as the cofine of CA is to radius. For, let the fides AB, BC, CA, be produced, so that BE, BF, CI, CH, be quadrants; from the poles B, C, draw the great circles EFDG, IHG; then the angles at E, F, I, H,