pendicular to AD, and since AG is the versed sine of the base AC, and AK the versed sine of the arch AH, KG is the excess of the versed sines of the base AC, and of the arch AH, which is the excess of the sides 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 base HF will be equal to the base FC, and the angle HFD equal to the right angle CFD: The straight 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 passes through DF (18. 11.). In like manner, since 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 passes through DG: And it was shown, that the plane CFH or CFL was perpendicular to the same plane BDH: therefore the common section of the planes CFL, CGL, viz. the straight line CL, is perpendicular to the plane BDA (19. 11.) and therefore CLF is a right angle: In the triangle CFL having the right angle CLF, by the Lemma CF, is to the radius as LH, the excess, viz. of CF or FH above FL, is to the versed sine of the angle CFL; but the angle CFL is the inclination of the planes BCD, BAD, since FC, FL are drawn in them at right angles to the common section BF: The spherical angle ABC is therefore the same with the angle CFL; and therefore CF is to the radius as LH to the versed sine of the spherical angle ABC; and since 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 tò 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 same radius; that is (23. 6.) the ratio of the rectangle contained by AE, CF to the square of the radius, is the same with the ratio compounded of the ratio of GK to LH, and the ratio of LH to the versed sine of the angle ABC; that is, the same with the ratio of GK to the versed sine of the angle ABC; therefore, the rectangle contained by AE, CF, the sines of the sides AB, BC, is to the square of the radius as GK, the excess of the versed sines AG, AK, of the base AC, and the arch AH, which is the excess of the sides to the versed sine of the angle ABC opposite to the base AC.! Q. E. D. PROP. XXIX. FIG. 23. THE rectangle contained by half of the radius, and the excess of the versed sines of two arches, is equal to the rectangle contained by the sines of half the sum, and half the difference of the same arches. Let AB, AC, be any two arches, and let AD be made equal to AC the less; the arch DB therefore is the sum, 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 likewise perpendicular to it in G, and let BH be perpendicular to AE, and AH will be the versed sine of the arch AB, and AG the versed sine of AC, and HG the excess of these versed sines: 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 semicircle, and therefore a right angle; therefore BKC is a right angle; and in the triangles DFB, CBK, the angles FDB, BCK, in the same segment 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 sine of half the arch DAB, that is, half the sum of the arches AB, AC; and half the chord of BC is the sine of half the arch BC, which is the difference of AB, AC. Whence the proposition is manifest. PROP. XXX. FIG. 19. 24. THE rectangle contained by half of the radius, and the versed sine of any arch, is equal to the square of the sine of half the same arch.. Let AB be an arch of a circle, C its centre, and AC, CB, BA, being joined: Let AB be bisected in D, and let CD be joined, which will be perpendicular to BA, and bisect it in E, (4. 1.), BE or AE therefore is the sine of the arch PROP. XXIX. FIG. 23. THE rectangle contained by half of the radius, and the excess of the versed sines of two arches, is equal to the rectangle contained by the sines of half the sum, and half the difference of the same arches. Let AB, AC, be any two arches, and let AD be made equal to AC the less; the arch DB therefore is the sum, 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 likewise perpendicular to it in G, and let BH be perpendicular to AE, and AH will be the versed sine of the arch AB, and AG the versed sine of AC, and HG the excess of these versed sines: 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 semicircle, and therefore a right angle; therefore BKC is a right angle; and in the triangles DFB, CBK, the angles FDB, BCK, in the same segment 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 sine of half the arch DAB, that is, half the sum of the arches AB, AC; and half the chord of BC is the sine of half the arch BC, which is the difference of AB, AC. Whence the proposition is manifest, PROP. XXX. FIG. 19. 21. THE rectangle contained by half of the radius, and the versed sine of any arch, is equal to the square. of the sine of half the same arch. a te Let AB be an arch of a circle, C its centre, and AC, CB, BA, being joined: Let AB be bisected in D, and let CD be joined, which will be perpendicular to BA, and bisect it in E, (4. 1.), BE or AE therefore is the sine of the arch |