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 archi AD, as the same AE to AF the versed sine of the arch AB, Wherefore, by 16.6. the proposition is manifest. PROP. XXXI. FIG. 25. In a spherical triangle, the rectangle contained by the sines of the two sides, is to the square of the radius, as the rectangle, contained by the sine of the arch which is half the sum of the base and the excess of the sides, and the sine of the arch, which is half the difference of the same 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 sides of BC, BA is to the square of the radius, as the rectangle contained by the sine of half the sum of AC, AD, and the sine of half the difference of the same AC, AD 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 (1.6.), as the rectangle contained by half the radius, and that excess, to the recta angle contained by half the radius, and the versed sine of B; therefore (29. 30. of this), 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. Q. E. D. 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. PROP. XXXI. FIG. 25. In a spherical triangle, the rectangle contained by the sines of the two sides, is to the square of the radius, as the rectangle contained by the sine of the arch which is half the sum of the base and the excess of the sides, and the sine of the arch, which is half the difference of the same 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 sides of BC, BA is to the square of the radius, as the rectangle contained by the sine of half the sum of AC, AD, and the sine of half the difference of the same AC, AD 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 (1.6.), as the rectangle contained by half the radius, and that excess, to the recta angle contained by half the radius, and the versed sine of B; therefore (29. 30. of this), 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. Q.E.D. |