2 PROP. IV. } TF two spherical triangles have two fides of the one, equal to two fides of the other, and the angle contained by the two fides of the one equal to the correspondent angles of the other, the two triangles will be equal. For, if the two arches containing the angles are equal, their chords or fubtenses are likewise equal, and contain equal angles; a 29. •3 therefore their bases are equal, and remaining angles of the one, equal to the remaining angles of the other, each to each; and the right lined triangles equal; but equal right lines cut off equal b 4 ** circumferences; wherefore the spherical triangles are equal to c 28. 30 one another. COR. I. Hence triangles will be equal and congruous, if two angles of the one be equal to two angles of the other, each to each, and a fide of the one equal to a fide of the other, either the side that lies betwixt the equal angles, or fubtending one of them d. d 29. and 24. 3. and 26. 1. II. Equilateral triangles are likewise equiangular.. III. In ifofceles triangles, the angles at the bases are equal; e 29. and and, if the angles at the bases are equal, the triangles are ifof celes f. 24. 3. and Ι. Ι. f 29 and 24. 3. 5. IV. Any two fides of a triangle are greater than the third; for any two of their chords or fubtenfes, are greater than the third 8. and 6, 1. 8 20, 1. PROP. V. ANY fide of a spherical triangle is less than a femicircle. Let AC, AB, the fides of the triangle ABC, be produced till they meet in D, then the femicircle ACD is greater than the arch AC. PROP. VI. HE three fides of a spherical triangle are less than a whole For For, BD, DC, two fides of the triangle BCD, are greater I cor. 4. than the third BC. Add BA, AC; then DBA, DCA, the two semicircles, are greater than the three fides of the triangle BCD b. Wherefore, &c. 4. 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 side. For, make the angle BAD equal to the 2 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 AC; therefore BC is greater than AC. Wherefore, &c. 2.4. 3 7. 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 opposite angle; and the sum of the internal angles at the base will be greater than two right angles ; if equal, equal, and, if less, less. Let ABC be the spherical triangle; if the two fides, AB, BC, be greater than a femicircle, the internal angle BAC, at the base, will be greater than the external and oppofite angle BCD; if equal, equal, and, if less, less; and the angles A and ACB will likewife 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 femicircle; 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. 1 IF the poles of the fides of any spherical triangle be joined by great circles, they constitute 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 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, ΤΜΝΟ, 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 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 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 fupplement of KB. For the fame 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 fame 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 fides HD, GH, and GD. Wherefore, &c. 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 fides of the triangle XMN, are equal to three femicircles ; but the three fides 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 angle are less than fixright angles. Wherefore, &c. PROP. TF, in any great circle, a point is taken, which is not the pole of it, and from that point feveral 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 obtuse angles with the great circle. b 7.3. 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 least; 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, 22. 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 fame reason, RG is greater than RV; therefore, the arch RA is greater than the arch RG, and RG greater than RV. c 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 obtuse angles. PROP. XII. TF the fides containing the right angles of a spherical triangle be of the fame affection with the opposite angles, that is, if the fides 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, having 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 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 fide AR fubtending the angle AGR, is greater than a right angle ; a 7. and; because AX is less than a quadrant, the angle AGX is less than a right angle. 1 IF the two fides 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 han RV. PROP. XIV. 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 fame reason, if the hypothenuse is greater than a quadrant, then one of the legs is greater, and the other less, than a quadrant. TF the angles at the base of a spherical triangle be both less, or both greater than quadrants, the perpendicular will fall with in 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 cafe, it will fall within the triangle; but, if not, it will fall without; then, in the triangle APB, the side AP, and angle B, are of the fame affection, and like-Fig. 2. wife the fide AP, and angle ACP: Therefore, since 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. |