be drawn; let the fide BC be greater than BA, and let BH be inade equal to BC: AH will therefore be the excefs of the fides BC, BA; let HK be drawn perpendicular to AD, and fince AG is the verfed fine of the base AC, and AK the verfed fine of the arch AH, KG is the excefs of the verfed fines of the bafe AC, and of the arch AH, which is the excefs of the fides BC, BA: let GL likewise be drawn parallel to KH, and let it meet FH in L, let CL, DH be joined, and let AD, FH meet each other in M.

Since therefore in the triangles CDF, HDF, DC, DH are equal, DF is common, and the angle FDC equal to the angle FDH, because of the equal arches BC, BH, the bafe HF will be equal to the base FC, and the angle HFD equal to the right angle CFD the ftraight line DF therefore (4. 11.) is at right angles to the plane CFH: wherefore the plane CFH is at right angles to the plane BDH, which paffes through DF. (18. 11.) In like manner, fince DG is at right angles to both GC and GL, DG will be perpendicular to the plane CGL; therefore the plane CGL is at right angles to the plane BDH, which paffes through' DG: and it was fhewn, that the plane CFH or CFL, was perpendicular to the fame plane BDH; therefore the common fection of the planes CFL, CGL, viz. the ftraight line GL, is perpendicular to the plane BDA, (19. 11.) and therefore CLF is a right angle in the triangle CFL having the right angles CLF, by the Femma CF is to the radius as LH, the excefs, viz. of CF or FH above FL, is to the verfed fine of the angle CFL; but the angle CFL is the inclination of the planes BCD, BAD, fince FC, FL. are drawn in them at right angles to the common fection BF: the spherical angle ABC is therefore the fame with the angle CFL; and therefore, CF is to the radius as LH to the versed fine of the spherical angle ABC; and fince the triangle AED is equiangular (to the triangle MFD, and therefore) to the triangle MGL, AE will be to the radius of the fphere AD, (as MG to ML; that is, because of the parallels as) GK to LH: the ratio therefore which is compounded of the ratios of AE to the radius, and of CF to the fame radius; that is, (23. 6.) the ratio of the rectangle contained by AE, CF to the fquare of the radius, is the fame with the ratio compounded of the ratio of GK to LH, and the ratio of LH to the verfed fine of the angle ABC; that is, the fame with the ratio of GK to the verfed fine of the angle ABC; therefore, the rectangle contained by AE, CF, the fines of the fides AB, BC, is to the fquare of the radius as CK, the excefs

€

of the verfed fines AG, AK, of the bafe AC, and the arch AH, which is the excefs of the fides to the verfed fine of the angle ABC oppofite to the base AC. QE. D.

TH

HE rectangle contained by half of the radius, and the excess of the verfed fines of two arches, is equal to the rectangle contained by the fines of half the fum, and half the difference of the fame arches.

Let AB, AC be any two arches, and let AD be made equal to AC the lefs; the arch DB therefore is the fum, and the arch CB the difference of AC, AB: through E the center of the circle, let there be drawn a diameter DEF, and AE joined, and CD likewife perpendicular to it in G; and let BH be perpendicular to AE, and AH will be the verfed fine of the arch AB, and AG the verfed fine of AC, and HG the excefs of thefe verfed fines: let BD, BC, BF be joined, and FC alfo meeting BH in K.

Since therefore BH, CG are parallel, the alternate angles BKC, KCG will be equal; but KCG is in a femicircle, and therefore a right angle; therefore BKC is a right angle; and in the triangles DFB, CBK, the angles FDB, BCK in the fame fegment are equal, and FBD, BKC are right angles; the triangles DFB, CBK are therefore equiangular; wherefore DF is to DB, as BC to CK, or HG; and therefore the rectangle contained by the diameter DF, and HG is equal to that contained by DB, BC; wherefore the rectangle contained by a fourth part of the diameter, and HG, is equal to that contained by the halves of DB, BC: but half the chord DB is the fine of half the arch DAB, that is, half the fum of the arches AB, AC; and half the chord of BC is the fine of half the arch BC, which is the difference of AB, AC. propofition is manifeft..

THE

PROP. XXX. FIG. 19. 24.

Whence the

HE rectangle contained by half of the radius, and the verfed fine of any arch, is equal to the fquare of the fine of half the fame arch.

Let AB be an arch of a circle, C its center, and AC, CB, BA being joined: let AB be bifected in D, and let CD be joined, which will be perpendicular to BA, and bifect it in E. (4. 1.) BE or AE therefore is the fine of the arch DB or AD, the half of AB: let BF be perpendicular to AC, and AF will be the verfed fine of the arch BA; but, because of the fimilar triangles CAE, BAF, CA is to AE as AB, that is, twice AE to AF; and by halving the antecedents, half of the radius CA is to AE the fine of the arch AD, as the fame AE to AF the verfed fine of the arch AB, Wherefore by 16. 6. the propofition is manifeft,

IN

PROP. XXXI. FIG. 25,

Na fpherical triangle, the rectangle contained by the fines of the two fides, is to the square of the radius, as the rectangle contained by the fine of the arch which is half the sum of the base, and the excefs of the fides, and the fine of the arch, which is half the difference of the fame to the fquare of the fine of half the angle opposite to the bafe.

Let ABC be a spherical triangle, of which the two fides are AB, BC, and base AC, and let the less fide BA be produced, fo that BD fhall be equal to BC: AD therefore is the excess of BC, BA; and it is to be fhewn, that the rectangle contained by the fines of BC, BA is to the fquare of the radius, as the rectangle contained by the fine of half the fum of AC, AD, and the fine of half the difference of the fame AC, AD to the fquare of the fine of half the angle ABC oppofite to the bafe AC.

Since by prop. 28. the rectangle contained by the fines of the fides BC, BA is to the fquare of the radius, as the excess of the verfed fines of the base AC and AD, to the versed fine of the angle B; that is, (1. 6.) as the rectangle contained by half the radius, and that excefs, to the rectangle contained by half the radius, and the verfed fine of B; therefore (29. 30. of this,) the rectangle contained by the fines of the fides BC, BA is to the fquare of the radius, as the rectangle contained by the fine of the arch, which is half the fum of AC, AD, and the fine of the arch which is half the difference of the fame AC, AD is to the fquare of the fine of half the angle ABC. Q. E. D.

Fig. 26, 17.

SOLUTION OF THE TWELVE CASES OF OBLIQUE-
ANGLED SPHERICAL TRIANGLES.

IN

GENERAL PROPOSITION.

N any oblique-angled fpherical triangle, of the three fides and three angles, any three being given, the other three may be found.

R: COS, B:: T, BC: T, BA. 20. and BC, CD BD. CoS, BC: CoS, BA:: CoS, DC: CoS, DA. 25. and BD is the fum or difference of BA, DA.

3

and B.

R: COS, BT, BC: T, BA. 20. and BC, DB CD. COS, BA: CoS, BC:: CoS, DA: Cob,

and B.

DC. 25. and according as DA, AC are of the fame or different affection, DC will be lefs or greater than a quadrant. 14.

BC.

R: COS, B:: T, BC: T, BA. 20. and ¡T, D: T, B:: S, BA : S, DA. 26. and BD is the fum or difference of BA, DA.

R: COS, B:: T, BC: T, BA. 20. and S, DA: S, BA; : T, B: T, D; and according as BD is greater or lefs than BA, the angles B, D are of the fame or different affection. 16.

COS, BC: R:: Co-T, B : T, BCA. 19. and T, DC: T, BC :: CoS, BCA: COS, DCA. 27. the fum or difference of the angles BCA, DCA is equal to the angle BCD.

COS, BC: R:: Co-T, B: T, BCA. 19. alfo by 27. CoS, DCA: CoS, BCA:: T, BC: T, DC. 27. if DCA and B be of

B, C and DC. the fame affection; that is, (13.) if AD and CA be fimilar, DC will be lefs than a quadrant, 14. and if AD, CA be not of the fame affection, DC is greater than a quadrant. 14.