extremes; and by prop. 18. S, B C is to the radius, as S, BA to S, C; therefore Rx S, BA== S, BC × S, C.

Secondly, Let the complement of one of the angles, as B, be the middle part, and the complement of the hypothenuse, and the side BA will be adjacent extremes: and by Cor. prop. 20. Co-S, B is to Co-T, BC, as T, BA, is to the radius, and therefore R x Co-S, B Co-T, BC × T, BA.

Again, Let the complement of the angle B be the middle part, and the complement of the angle C, and the side AC will be opposite extremes: and by prop. 22, Co-S, AC is to the radius, as Co-S, B is to S. C: and therefore R x Co-S, B⇒Co-S, AC × S, C.

Thirdly, Let the complement of the hypothenuse be the middle part, and the complements of the angles B, C, will be adjacent extremes: but by Cor. 2. prop. 19, Co-S, BC is to Co-T, B, as Co-T, C to the radius: therefore R × Co-S, BC= Co-T, C × Co-T, A.

Again, Let the complement of the hypothenuse be the middle part, and the sides AB, AC, will be opposite extremes: but by prop. 21. Co-S, AC is to the radius, as Co-S, BC, to Co-S, BA; therefore Rx Co-S, BC Co-S, BA x Co-S, AC. Q. E. D.

SOLUTION OF THE SIXTEEN CASES OF RIGHT ANGLED SPHERICAL TRI

ANGLES.

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.

In the following Table the solutions are derived from the preceding propositions. It is obvious that the same solutions may be derived from Baron Napier's two rules above demonstrated, which, as they are easily remembered, are commonly used in practice.

4

5

S, C: Co-S, B:: R: Co-S, AC: by 22. and
AC is of the same species with B. 13.

R: Co-S, BA:: Co-S, AC: Co-S, BC. 21. and if both BA, AC be grearer or less than a quadBA, AC BC rant, BC will be less than a quadrant. But if they be of different affections, BC will be greater than a quadrant. 14.

BA, BC AC

Co-S, BAR:: Co-S, BC Co-S, AC, 21. and if BC be greater or less than a quadrant, BA, AC will be of different or the same affection: by 15.

S, BA : R :: T, CA : T, B, 17, and B is of the same affection with AC, 13.

R: S, BAT, B: T, AC. 17. And AC is of the same affection with B. 13.

8

AC, B BAT, B: R:: T, CA: S, BA. 17.

R: Co-S, C: T, BC: T, CA. 20. If BC be AC less or greater than a quadrant, C and B will be of the same or different affection. 15. 13.

AC, C BC

Co-S, CR: T, AC : T, BC. 20. And BC is less or greater than a quadrant, according as C and AC or C and B are of the same or different affection. 14. 1.

T, BC R:: T, CA: Co-S, C. 20. If BC be less or greater than a quadrant, CA and AB, and therefore CA and C, are of the same or different affection. 15.

R: S, BC: S, B: S, AC. 18. And AC is of the same affection with B.

AC, B BCS, B: S, AC : : R : S, BC. 18.

The second, eighth, and thirteenth cases, which are commonly called ambiguous, admit of two solutions: for in these it is not determined whether the side or measure of the angle sought be greater or less than a quadrant.

< PROP. XXIII. FIG. 16.

IN spherical triangles, whether right angled or oblique angled, the sines of the sides are proportional to the sines of the angles opposite to them.

First, Let ABC be a right angled triangle, having a right angle at A; therefore by prop. 18, the sine of the hypothenuse BC is to the radius (or the sine of the right angle at A) as the sine of the side AC to the sine of the angle B. And in like manner, the sine

of BC is to the sine of the angle A as the sine of AB to the sine of the angle C; wherefore (11. 5.) the sine of the side AC is to the sine of the angle B, as the sine of AB to the sine of the angle C.

Secondly, Let BCD be an oblique angled triangle, the sine of either of the sides BC, will be to the sine of either of the other two CD, as the sine of the angle D opposite to BC is to the sine of the angle B opposite to the side CD. Through the point C, let there be drawn an arch of a great circle CA perpendicular upon BD; and in the right angled triangle ABC (18. of this) the sine of BC is to the radius, as the sine of AC to the sine of the angle B; and in the triangle ADC (by 18. of this): and, by inversion, the radius is to the sine of DC as the sine of the angle D to the sine of AC: therefore, ex æquo perturbate, the sine of BC is to the sine of DC, as the sine of the angle D to the sine of the angle B. Q. E. D.

PROP. XXIV. FIG. 17. 18.

IN oblique angled spherical triangles, having drawn a perpendicular arch from any of the angles upon the opposite side, the co-sines of the angles at the base are proportional to the sines of the vertical angles.

Let BCD be a triangle, and the arch CA perpendicular to the base BD; the co-sine of the angle B will be to the co-sine of the angle D, as the sine of the angle BCA to the sine of the angle DCA.

For, by 22, the co-sine of the angle B is to the sine of the angle BCA as (the co-sine of the side AC is to the radius; that is, by prop. 22, as) the co-sine of the angle D to the sine of the angle DCA; and, by permutation, the co-sine of the angle B is to the co-sine of the angle D, as the sine of the angle BCA to the sine of the angle DCA. Q. E. D.

PROP. XXV. FIG. 17. 18.

THE same things remaining, the co-sines of the sides BC, CD, are proportional to, the co-sines of the bases BA, AD.

For, by 21, the co-sine of BC is to the co-sine of BA, as (the co-sine of AC to the radius; that is, by 21, as) the co-sine of CD is to the co-sine of AD; wherefore, by permutation, the co-sines of the sides BC, CD are proportional to the co-sines of the bases BA, AD. Q. E. D.

PROP. XXVI. FIG. 17. 18.

THE same construction remaining the sines of the bases BA, AD are reciprocally proportional to the tangents of the angles B and D at the base.

For, by 17, the sine of BA is to the radius, as the tangent of AC to the tangent of the angle B; and by 17, and inversion, the radius is to the sine of AD, as the tangent of D to the tangent of AC: therefore, ex æquo perturbate, the sine of BA is to the sine of AD, as the tangent of D to the tangent of B.

PROP. XXVII. FIG. 17. 18.

THE CO-sines of the vertical angles are reciprocally proportional to the tangents of the sides.

For, by prop. 20, the co-sine of the angle BCA is to the radius as the tangent of CA is to the tangent of BC; and by the same prop. 20, and by inversion, the radius is to the co-sine of the angle DCA, as the tangent of DC to the tangent of CA: therefore, ex æquo perturbate, the co-sine of the angle BCA is to the co-sine of the angle DCA, as the tangent of DC is to the tangent of BC. Q. E. D.

LEMMA. FIG. 19. 20.

IN right angled plane triangles, the hypothenuse is to the radius, as the excess of the hypothenuse above either of the sides to the versed sine of the acute angle adjacent to that side, or as the sum of the hypothenuse, and either of the sides to the versed sine of the exterior angle of the triangle.

Let the triangle ABC have a right angle at B; AC will be to the radius as the excess of AC above AB, to the versed sine of the angle A adjacent to AB; or as the sum of AC, AB to the versed sine of the exterior angle CAK.

With any radius DE, let a circle be described, and from D the centre, let DF be drawn to the circumference, making the angle EDF equal to the angle BAC, and from the point F, let FG be drawn perpendicular to DE; let AH, AK be made equal to AC, and DL to DE: DG therefore is the co-sine of the angle EDF or BAC, and GE its versed sine; and because of the equiangular triangles ACB, DFG, AC or AH is to DF or DE, as AB to DG: therefore (19. 5.) AC is to the radius DE as BH to GE, the versed sine of the angle EDF or BAC: and since AH is to DE, as AB to DG (12. 5.), AH or AC will be to the radius DE as KB to LG, the versed sine of the angle LDF or KAC. Q. E. D.

PROP. XXVIII. FIG. 21. 22.

IN any spherical triangle, the rectangle contained by the sines of two sides, is to the square of the radius, as the excess of the versed sines of the third side or base, and the arch, which is the excess of the sides, is to the versed sine of the angle opposite to the base.