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: ABBM :: IL: BM, (1. 6. Euc.), therefore IL: BM::rect. FK.DV: BP2. 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 For, make DFAC, 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. For, 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 half the perimeter and its excess above the base, to the square of the cosine of half the angle opposite to the base, by Prop. 30; and as the rectangle contained by the sines of the two excesses of half the perimeter above each of the sides to the square of the sine of half the angle opposite to the base, by Prop. 29: therefore, the rectangle contained by the sines of half the perimeter and its excess 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 cosine to the square of the sine of half the angle opposite to the base, that is,) as the square of the radius to the square of the tangent of half the angle opposite to the base. Q. E. D. * Solution of the several Cases of Right Angled and Oblique Angled Spherical Triangles. GENERAL PROPOSITION. In a right angled spherical triangle, of the three sides and three angles, any two being given, besides the right angle, the other three may be found. 1. Rules for the sixteen Cases of Right Angled Spherical Trigonometry. See Fig. 16. The two quantities, in the same analogy, marked with asteriscs, are both of the same affection; that is, both at the same time less, or both at the same time greater, than 90°. Cases 7, 8, 9, are doubtful, for two triangles may have the given things, but have the things sought in one of them the supplements of the things sought in the other. In the remaining cases, if the two given quantities which occur in the previous terms of the same analogy are both of the same affection, the last term will be less than 90°; but if they are of different affection, the last term will be greater than 90°. These limitations are founded on the 13th, 14th and 15th Propositions. 2. Rules for the sixteen Cases of Right Angled Spherical |