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 sines 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 of the radius, and that excess, to the rectangle 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.

SOLUTION OF THE TWELVE CASES OF OBLIQUE ANGLED SPHERICAL TRI

ANGLES.

GENERAL PROPOSITION.

In an oblique angled spherical triangle, of the three sides and three angles, any three being given, the other three may be found.

The 3d, 5th, 7th, 9th, 10th cases, which are commonly called ambiguous, admit of two solutions, either of which will answer the conditions required; for, in these cases, the measure of the angle or side sought, may be either greater or less than a quadrant, and the two solutions will be supplements to each other. (Cor. to def. 4. 6. Pl. Tr.)

If from any of the angles of an oblique angled spherical triangle, a perpendicular arch be drawn upon the opposite side, most of the cases of oblique angled triangles may be resolved by means of Napier's rules.

FINIS.