For, BD, DC, two sides of the triangle BCD, are greater than the third BC 2. Add BA, AC; then DBA, DCA, the two femicircles, are greater than the three sides of the triangle BCD b. 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 BAD equal to the a cor. 23. angle B; then AD will be equal to BD ; therefore the fide BDC is equal to AD and DC; but AD, DC, are greater than 2. 4. IN any Spherical triangle, if the sum of two of its fides be great er than a semicircle, then the internal angle at the base will be greater than the external and opposite angle ; and the sum of the internal angles at the base will be greater than two right angles ; if equal, equal, and, if lefs, lefs. Let ABC be the spherical triangle ; if the two sides, AB, BC, be greater than a semicircle, the internal angle BAC, at the base, will be greater than the external and opposite angle BCD; if equal, equal, and, if less, less; and the angles A and ACB will likewise be greater, equal, or less, than two right angles. First, let the semicircles 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 less than a semicircle ; then the angle A will be less 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 less, less. 3. 17. IF. F the poles of the sides of any spherical triangle be joined by great circles, they constitute another triangle, the sides of which are supplements of the arches that measure the angles of the given triangles ; and the arches that are the measures of the angles of the supplementary triangle, are the supplements of the sides 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 supplement 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 supplements of HD, HG, and GD. For, because 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 also a quadrant : Wherefore, M is the pole of the circle GHB. d cor. 1. 3 For the same 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 same reason, XM is the supplement 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 supplement of HD. For the same reason, OC is the supplement of GH, and BA of GD; that is, the measures of the angles X, M, N, are the supplements of the sides HD, GH, and GD. Wherefore, &c. 1 THE 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 sides of the triangle XMN, are equal to three femicircles"; but the three sides of the triangle XMN are less a 9. than two cemicircles b; 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 less than six right angles. Wherefore, &c. PRO P. I F, 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 these arches is that which passes through the pole ; and the remainder of it is the least ; and the arch nearer to that, passing through the pole, is greater than that more remote ; and they make obtufe angles with the great circle. 11. 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 passes through the pole, is the greatest, and RB is the leaft; and the arch RCA is greater than RG, and RG greater than RV. For, because C is the pole of the circle AFB, CD and RS, 2 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 squares of RS, SA, that is, the square of RA, is greater than the squares of RS, SG, that is, than the square of RG, that is, RA is greater than RG. For the lame reason, RG is greater than RV; therefore, the arch RA is greater than the arch RG, and RG greater than RV. 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 obtuse angles. c cor. 3. IF F the sides containing the right angles of a spherical triangle be of the same affection with the opposite angles, that is, if the sides are greater or less than quadrants, the opposite 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 ride AR be greater than a quadrant, the angle AGR will be greater than a right angle; and, if AX be less 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 side AR fubtending the angle AGR, is greater than a right angle “; a 7. and, because AX is less than a quadrant, the angle XGX is less than a right angle. I F the two sides containing the angle of a spherical triangle be both less, 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 hypothenuse will be greater than a quadrant. For, in the triangle ARG, the hypothenuse RG is greater than RF, that is, greater than a quadrant. For the same reason, if the hypothenuse is greater than a quadrant, then one of the legs is greater, and the other less, than a quadrant. IF F the angles at the base of a spherical triangle be both less, 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. wise the fide AP, and angle ACP: Therefore, since the angles ABC, and ACP, are of the same affection, the angles ACB and ABC are of different affections ; þut they are not Where-By Hyp. fore, &c. X Fig. 1. 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 likewise, in the triangle ACP, the angle C, and side AP, are of the fame affecrion; therefore, the angles B and C are of the fame affection: which is inipoisible. Wherefore, &c. a Hyp. IN N right angled spherical triangles, having the same or equal acute angles at the base, the lines of the hypothenuse are proportional to the fines of the perpendicular arches, and the fines of the bases proportional to the tangents of the perpendicular arches. Let the triangles be BAC, BHE, right angled at A and H, and the same acute angle B, at the base BA, the fines of the hypothenuses CP, CE, are to one another as the fines of the per. pendicular arches CD, EF; and the fines of the basis AQ, HK, proportional to IA, GH, the tangents of the perpendicular arches. For, because CD, EF, are perpendicular to the same plain, they are parallel"; as also, FR, DP; therefore the plains of the b 18. 11. triangles EFR, CDP, are parallel b; and CP, ER, the common sections of these plains, with the plains paffing through BE, EO, € 16. 11. will be parallel ; therefore, the triangles CDP, EFR, are equi. angular ; wherefore CP, the fine of the hypothenuse BC, is to CD, the fine of the perpendicular arch CA, as ER, the fine of the hypothenuse BE, is to EF, the fine of the perpendicular d 4. 6. arch EH d. For the same 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 base, is to the fine of the vertical angle, as the coline of the perpendicular is to the radius. ift, Let the triangle be ABC, right angled at A, the coline of B is to the Gne of the angle ACB, as the coline of CA is to radius. For, let the sides AB, BC, CA, be produced, so that BE, BF, CI, CH, be quadrants ; from the poles B, C, draw the great cieples EFDG, IHG; then the angles at E, F, I, H, |