OF THE CIRCULAR PARTS. Fig. 15. In any right angled spherical triangle ABC, the comple ment of the hypothenuse, the complements of the angles, and the two sides, are called The circular parts of the triangle, as if it were following each other in a circular order, from whatever part we begin : Thus, if we begin at the complement of the hypothenuse, and proceed towards the side BA, the parts following in order will be the complement of the hypothenuse, the complement of the angle B, the side BA, the side AC (for the right angle at A is not reckoned among the parts), and lastly, the complement of the angle C. And thus at whatever part we begin, any three of these five be taken, they either will be all contiguous or adjacent, or one of them will not be contiguous to either of the other two : In the first case, the part which is between the other two is called the Middle part, and the other two are called Adjacent extremes. In the second case, the part which is not contiguous to either of the other two is called the Middle part, and the other two Opposite extremes. For example, if the three parts be the complement of the hypothenuse BC, the complement of the angle B, and the side BA; since these three are contiguous to each other, the complement of the angle B will be the middle part, and the complement of the hypothenuse BC and the side BA will be adjacent extremes: But if the complement of the hypothenuse BC and the sides BA, AC be taken: since the complement of the hypothepuse is not adjacent to either of the sides, viz. on account of the complements of the two angles B and C intervening between it and the sides, the complement of the hypothenuse BC will be the middle part, and the sides BA, AC opposite extremes. The most acute and ingenious Baron Napier, the inventor of Logarithms, contrived the two following rules concerning these parts, by means of which all the cases of right angled spherical triangles are resolved with the greatest case. RULE I. The rectangle contained by the radius and the sine of the middle part, is equal to the rectangle contained by the tangents of the adjacent parts. RULE II. middle part, is equal to the rectangle contained by the cosines of the opposite parts. These rules are demonstrated in the following manner : First, Let either of the sides, as BA, be the middle Fig. 16. part, and therefore the complement of the angle B, and the side AC, will be adjacent extremes. And by cor. 2. prop. 17. of this, S, BA, is to the Co-T, B, as T, AC is to the radius, and therefore RxS, BA=Co-T, BRT, AC. The same side BA being the middle part, the complement of the hypothenuse, and the complement of the angle C, are opposite extremes; and by Prop. 18. S, BC is to the radius, as S, BA to S, C; therefore RxS, BA=S, BC XS, 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 Rx Co-S, B=Co-T, BCRT, 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 Rx Co-S, B=CO-S, AC XS, 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, B to the radius ; Therefore R x Co-S, BC=Co-T, Cx Co-T, C. 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 R x Co-S, BC=Co-S, BAX CO-S, AC. Q. E. D. 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, by 22. and 13. 2 AC, BC CoS, AC:R:: CoS, B:S, C: By 22. S, C: CoS, B:: R: CoS, AC; By 3 B, CAC 22. and AC is of the same species with B. 13. 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, BA :R:: CoS, BÇ : 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. S, BA:R:: T, CA: T, B. 17. and B 6 BA,AC B lis of the samne affection with AC. 13. 1 draot, C and B will be of the same or Case. Given. Sought. 7 BA, B AC R: S, BA':: T, B : T, AC. 17 And AC is of the same affection with B. 13. 8 AC, B BA T, B:R::T, CA: S, BA, 17. R: COS, C:: T, BC : T, CA. 20. It BC be less or greater than a quadrant, 9 |BC, C| AC C and B will be of the saine or differ- CoS. C:R:: T, AC: T, BC. 20. And BC is less or greater than a qua10 AC, CBCdrant, according as C and A or Cand B are of the same or different affections. 14. 15. T, BC:R::T, CA: COS. C. 20. I BC be less or greater than a quadrant, 1 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, BBC S, B : S, AC :: R:S, BC. 18. 14 BC,AC B S, BC:R::S, AC:S, B. 18. And 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 qua- R: COS, BC :: T, C, : CoT, B. 19. If BC be less or greater than a qua16 BC, CB different affection. 15. 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 siues 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. Fig. 17. 18. 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 side 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 vertical angles. 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 |