Solution of the Sixteen Cases of Right 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.

In the following Tables 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.

Case. Given. Sought.

R: CoS, AC:: S, C: CoS, B; And 1 AC, C B B is of the same species, with CA,

2

3

by 22. and 13.

AC, B C CoS, AC: R:: CoS, B: S, C: By 22.
S, C: CoS, B:: R: CoS, AC; By
B, CAC 22, and AC is of the same species with
B. 13.

R: CoS, BA :: CoS, AC : CʊS, BC. 21. and if both AB, AC be greater or less than a quadrant, BC will be less 4 BA, AC BC than a quadrant. But if they be of dif ferent affection, BC will be greater than a quadrant. 14,

CoS, BAR:: CoS, BC: CoS, AC. 5 BA,BC AC 21. and if BC be greater or less than a quadrant, BA, AC will be of different or the same affection. By 15.

6 BA,AC B

S, BAR:: T, CA: T, B. 17. and B lis of the same affection with AC. 13.

Case. Given. Sought.

7 BA, BAC RS, BA::T, BT, AC. 17. And AC is of the same affection with B. 13.

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

9 BC, CAC

R: CoS, C:: T, BC : T, CA. 20. If
BC be less or greater than a quadrant,
C and B will be of the same or differ-
ent affections. 15. 13.

CoS. CR: T, AC: T, BC. 20.
And BC is less or greater than a qua-

10 AC, C BCdrant, according as C and A or C and B are of the same or different affections. 14. 15.

T, BC: R::T, CA: CoS. C, 20. If
BC be less or greater than a quadrant,

11 BC,CA C CA and AB, and therefore CA and C are of the same or different affections. 15.

R: S, BC :: S, B: S, AC. 18. And 12 BC, B AC AC is of the same affection with B.

13 AC, B BCS, B: S, AC:: RS, BC. 18.

14 BC,AC B

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

T, C: R:: CoT, B: CoS, BC. 19. And according as the angles B and C 15 B, C BC are of different or the same affection, BC will be greater or less than a quadrant. 14.

Fig. 17. 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 perturbato, 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 cosines of the angles at the base are proportional to the sines of the verticalangles.

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

For by 22. the cosine of the angle B is to the sine of the

angle BCA as (the cosine of the side AC is to the radius; that is, by Prop. 22. as) the cosine of the angle D to the sine of the angle DCA; and, by permutation, the cosine of the angle B is to the cosine 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 cosines of the sides BC, CD, are proportional to the cosines of the bases BA, AD.

For by 21, the cosine of BC is to the cosine of BA, as (the cosine of AC to the radius; that is, by 21. as) the cosine of CD is to the cosine of AD: wherefore, by permutation, the cosines of the sides BC, CD are proportional to the cosines 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 perturbato, 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 Cosines of the vertical angles are reciprocally proportional to the tangents of the sides.

For by Prop. 20. the cosine 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 cosine of the angle DCA, as the tangent of DC to the tangent of CA: Therefore, ex æquo perturbato, the cosine of the angle BCA is to the cosine 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 cosine 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 ra dius, 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.

Let ABC be a spherical triangle, the rectangle contained. by the sines of AB, BC will be to the square of the radius, as the excess of the versed sines of the base AC, and of the arch, which is the excess of AB, BC to the versed sine of the angle ABC opposite to the base.

Let D be the centre of the sphere, and let AD, BD, CD be joined, and let the sines AE, CF, CG of the arches AB, BC, AC be drawn; let the side BC. be greater than BA, and let BH be made equal to BC; AH will therefore be the excess of the sides BC, BA; let HK be drawn per