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 verfed 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 sphere 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 GK, the excefs of the verfed fines AG, AK, of the base AC, and the arch AH, which is the excefs of the fides. to the verfed fine of the angle ABC oppofite to the bafe AC. Q. E. D. T PROP. XXIX. FIG. 23. 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. I et 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 centre 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 excess of these verfed fines: Let BD, BC, BF be joined, and FC also 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 Kk 2 therefore 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. Whence the propofition is manifeft. T PROP. XXX. FIG. 19. 24: 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 centre, 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, becaufe 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 proposition is manifeft. IN PROP. XXXI. FIG. 25. a fpherical triangle, the rectangle contained by the as the rectangle contained by the fine of the arch which is half the fum of the bafe, 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 oppofite to the base. Let Let ABC be a spherical triangle, of which the two fides are AB, BC, and bafe AC, and let the lefs fide BA be produced, fo that BD fhall be equal to BC: AD therefore is the excefs of BC, BA; and it is to be fhown, 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, opposite 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 bafe AC and AD, to the verfed 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. SOLUTION |
