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 C F G D E (CD+BC,) and (CD-BC) be=4(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= BC, and CG = CD, the remainder CF is = (CD—BC) and BF is (CD+BC). And because AB-DE, the whole AE is AB+ AC+CD, and BD being bisected in F, AF is therefore = FE=4AE= (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=4(AB+AC+ CD)-AC. = PROP. XXIX. FIG. 25. 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 BA 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 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, 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 B; that 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 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 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. WON BY 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. D E 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 ED, the arch ED and its chord ED are bisected in F, V (30. and 3. 3. Euc.) For the same reason, the arch BG and its chord BG are bisected in H, P; P A CIKLM B 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 : BP2. For since EFC, EIC are right angles, E, F, I, C are points in a circle; therefore the angle FEI is equal to VCI; consequently the two triangles EOD, FCK are equiangular; therefore FC: FK:: ED: DO or IL, and the rectangle FC.IL is the rectangle FK.ED=2FK.DV. But because AB is= 2FC, the rectangle AB.IL is=2FC.IL=4FK.DV; and the rectangle AB.BM is= BG2, (Cor. 8. 6. Euc.), that is, (since BG is bisected in P)=4BP2. Wherefore the rectangle AB.IL AB.BM: FK.DV: BP2. But the rectangle AB.IL: AB. BM :: IL: BM, (1. 6. Euc.), therefore IL: BM::rect. FK.DV: BP. Q. E. D. COR. The difference of the versed sines of two arches BD, BE is to the coversed sine of a third arch AG, as the rectangle contained by the sines of half the sum and half the difference of the arches BD, BE is to the square of the cosine of half the arch AG. For the arch BG being the supplement of AG, the coversed sine of AG is BM, the versed sine of BG; and the cosine of half the arch AG is the sine of half the arch BG, (Lemma V.) But by this Lemma, the difference of the versed sines of BD, BE is to the versed sine of BG, (or coversed sine of AG,) as the rectangle contained by the sines of half the sum and half the difference of the arches BD, BE is to the square of the sine of half the arch BG, or square of the cosine of half the arch AG. * LEMMA VII. If there are three magnitudes, half the difference between one of them and the sum of the other two is equal to the difference between half the sum of the three magnitudes, and the one magnitude. Let AB, BC, CD be three magnitudes, and let AC=AB + BC, and AD ट E F Ď =AB+BC+ A B CD; then will (AC/CD)=AD/CD. For, make DF=AC, then CF is= AC CD: bisect AD in E; because AE is = ED, and AC=DF, the remainder EC is equal to the remainder EF; therefore EC is=4CF, that is (ACCD); and because ED is AD, the same EC is AD CD. Q. E. D. * PROP. XXX. FIG. 7. In a spherical triangle, the rectangle contained by the sines of any two of the sides is to the square of the radius, as the rectangle contained by the sines of the two arches which are half the perimeter of the triangle, and the excess of half the perimeter above the base to the square of the cosine of half the angle opposite to the base. Let ABC be a spherical triangle, of which the two sides are AB, BC, and the base AC, the rectangle contained by the sines of AB, BC is to the square of the radius, as the rectangle contained by the sine of half the perimeter of the triangle, and the sine of the excess of half the perimeter above the base AC to the square of the cosine of half the angle ABC. For, the rectangle contained by the sines of AB, BC is to the square of the radius as the difference between the versed sines of AC, and the sum of the sides AB, BC to the coversed sine of the angle ABC, (Cor. Prop. 28.) that is, as the rectangle contained by the sines of half the sum and half the difference of AC and the sides AB, BC taken together, to the square of the cosine of half the angle ABC, (Cor. Lemma VI.) But half the sum of AC and AB, BC taken together is half the perimeter of the triangle; and half the difference between. AC and AB, BC taken together is equal to the excess of half the perimeter above AC, (Lemma VII.) Wherefore the rectangle contained by the sines of AB, BC is to the square of the radius, as the rectangle contained by the sines of half the perimeter and the excess of half the perimeter above AC, to the square of the cosine of half the angle ABC. Q. E. D. * PROP. XXXI. FIG. 7. In a spherical triangle, the rectangle contained by the sines of half the perimeter, and the excess of half the perimeter above the base, is to the rectangle contained by the sines of the two excesses of half the perimeter above each of the sides, as the square of the radius to the square of the tangent of half the angle opposite to the base. |