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 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 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 Triangles.

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.

Given. 1 B, D, and BC, two angles, and a side opposite one of them.

Sought.

C.

2 B, C, and D. BC, two angles, and the side between them.

BC, CD, BD.

3

and B.

4

and B.

CoS, BC: R :: CoT, B : T, BCA. 19. Likewise by 24. CoS, B: S, BCA :: CoS. D: S, DCA; wherefore BCD is the sum or difference of the angles DCA, BCA according as the perpendicular CA falls within or without the triangle BCD; that is (16. of this), according as the angles B, D are of the same or different affection.

CoS, BC: R :: CoT, B : T, BCA.| 19. and also by 24. S, BCA: S, DCA :: CoS, B: CoS, D; and according as the angle BCA is less or greater than BCD, the perpendicular CA falls within or without the triangle BCD; and therefore (16. of this) the angles B, D will be of the same or different affection.

R: COS, B:: T, BC: T, BA. 20 and CoS, BC: CoS, BA :: CoS, DC |: CoS, DA. 25. and BD is the sum or difference of BA, DA.

BC, DB, CD. R: CoS, B :: T, BC: T, BA. 20. and CoS, BA: CoS, BC :: CoS, DA: CoS, DC 25. and according as DA, AC are of the same or different affection, DC will be less or greater than a Iquadrant. 14.

Given.

Sought.

The 12 A, B, Csides. Fig. 7.

See Fig. 7.

In the triangle DEF, DE, EF, FD, are respectively the supplements of the measures of the given angles B, A, C, in the triangle BAC; the sides of the triangle DEF are therefore given, and by the preceding case, the angles D, E, F may be found, and the sides BC, BA, AC, are the supplements of the measures of these angles.

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.

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THE END.

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