· Let D be the centre of the sphere, and let AD, BD, CD, be joined, and let the sines AE, CF, CG of the arches AB, BC, AC be drawn; let the side BC be greater than BA, and let BH be made equal to BC; AH will therefore be the excess of the sides BC, BA; let HK be drawn perpendicular 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 Lemma I. 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 A ED 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 parellels 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. * Cor. Hence, in any spherical triangle, the rectangle contained by the sines of any two sides is to the square of the radius, as the excess of the versed sines of the third side or base, and the arch which is the sum of the sides is to the coversed sine of the angle opposite to the base. LEMMA II. Fig. 23. The rectangle contained by half 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 rect. angle 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. LEMMA III. Fig. 19, 24. The rectangle contained by half 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 will bisect it in E, (4. 1.) BE or AE therefore is the sine of the arch DB or AD, the half of AB: Let BF be perpendicular to AC, and AF will be the versed sine of the arch BA; but, because of the similar 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 sine of the arch AD, as the same AE to AF, the versed sine of the arch AB. Wherefore, (by 16. 6.) the proposition is manifest. * LEMMA IV. If there are three magnitudes, half the sum, and half the difference of one of them, and the excess of the other two, are equal to the two excesses of half the sum of the three magnitudes above each of the first mentioned two. Let AB, AC, CD be three magnitudes, and BC the excess of AC above AB: then will A B В FG D E (CD+ BC,)and (CD-BC) be=f(AB+AC+CD)-AB, and {(AB+AC +CD-AC. For bisect BD in F, and CD in G, and produce AD to E, so that DE may be=AB; then since BD=2FD, and CD= 2CG, the remainder BC=2FG. Now FG being thus= ABC, and CG=CD, the remainder CF is={(CD–BC) and BF is = }(CD+BC). And because AB=DE, the whole A Eis= AB+AC+CD, and BD being bisected in F, AF is therefore =FE=\AE= (AB+AC+CD); and BF which was shown to be = (CD+BC), is thus also= !(AB+AC+CD-AB; and CF, which was shown to be=}(CD-BC,) is also=\(AB+AC+ CD)-AC. In a spherical triangle, the rectangle contained by the sines of any two sides, is to the square of the radius, as the rectangle contained by the sines of the two arches, which are the excesses of half the perimeter above each of those sides, to the square of the sine of half the angle opposite to the base. Let ABC be a spherical triangle, of which the two sides are AB, BC, and base AC, and let the less side B A be produced, so that BD shall be equal to BC: AD therefore is the excess of BC, BA; and it is to be shown, that the rectangle contained by the sines of BC, BA is to the square of the radius, as the rectangle contained by the sines of the two arches wbich are the excesses of half the perimeter above each of the sides BC, BA, to the square of the sine of balf the angle ABC, opposite to the base AC. Since by Prop. 28. the rectangle contained by the sines of the sides BC, BA is to the square of the radius, as the excess of the versed sines of the base AC and AD to the versed sine of the angle Bthat is, (since magnitudes have the same ratio to one another which their equimultiples have,) (15. 5.) as the rectangle contained by half the radius, and that excess, to the rectangle contained by half the radius, and the versed sine of B; therefore (by Lemmata II. and III.) the rectangle contained by the sines of the sides BC, BA is to the square of the radius, as the rectangle contained by the sine of the arch, which is half the sum of AC, AD, and the sine of the arch, which is half the difference of the same AC, AD is to the square of the sine of half the angle ABC; consequently, (by Lemma IV.) the rectangle contained by the sines of the sides BC, BA is to the square of the radius, as the rectangle contained by the sines of the two arches, which are the excesses of half the perimeter above each of the sides BC, BA to the square of the sine of half the angle ABC. Q. E. D. * LEMMA V. Half the supplement of an arch is the same with the complement of half that arch. Let ADB be an arch of a circle; draw the diameter AB, and the arch DB will be the supplement of the arch AD, (def. 3.) Bisect the arches AD, DB in the points C and E; then D E will the arch DE be the complement of the arch CD. For, since the arch AD is double CD, and the arch DB double DE, the whole arch ADB is double CE. But A B the arch ADB is the semi-circumference of the circle, therefore the arch CE is a quadrant; consequently DE (which is the difference between the arch CD and the quadrant CE) is the complement of CD. CoR. Hence the sine and cosine of half the supplement of an arch are the cosine and sine of half the arch. LEMMA VI. The difference of the versed sines of any two arches is to the versed sine of any other arch, as the rectangle contained by the sines of half the sum and half the difference of the two arches is to the square of the sine of half the other arch. G Let BD, BE be two arches, and BG any other arch of the circle AEB; draw the diameter ACB, and EI, DL, GM perpendicular to AB. Join ED, GB, and draw CF, CH perpendicular to them, meeting the circumference in F, H; also draw FK perpendicular to AB: then BL, BI are the versed sines of the arches BD, BE, and IL is the difference of these versed sines. And because CF drawn through the centre is perpendicular to D H For the same reason, the PV arch BG and its chord BG A CI K L M B are bisected in H, P; therefore DF is half the difference of the arches BD, BE, and DV is its sine; also BH is half the arch BG, and BP is its sine, and BM is the versed sine of BG. Also because DE, the difference of the arches BD, BE is bisected in F, BF is half the sum (Lem. III.) of BD, BE, and FK is its sine, Then I say that IL: BM :: the rectangle FK.DV: BP. |