OF THE CIRCULAR, PARTS. IN any right angled spherical triangle ABC, the complement 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 fol. lowing 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, if any three of these five be taken, they either will be all contiguous or adjacent, or guous to either of the other on them will not be conti. in the first case, the part which is between the other two is called the Middle part, and the other two ave 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 hy. pothenuse 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 hypothenuse is not adjacent to either of the sides, viz. on account of the complements of the two angles B and C intervening betseen it and the sides, the complement of the hypothenuse BC will be the middle part, and the sides, BA, AC, opposite ex. tremes. 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 ease. RULE I. dle part, is equal to the rectangle contained by the tangents of RULE II. The rectangle contained by the radius and the sine of the mid dle part, is equal to the rectangle contained by the co-sines of First, Let either of the sides, as BA, be the middle part, and Fig. 16. 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 Řxş, BA=Co-T, BẤT, 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, BCXS, 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, BCXT, BA. Again, Let the complement of the angle B be the middle part, and the complement of the angle C, and the side AC wih. 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, ACXS, 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 Rx Co-S, BC=Co-T, Cx 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 RxCo-S, BC=Co-S, BAXCo-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 Table the solutions are derived from the pre ceding propositions. It is obvious that the same solutions may be derived from Baron Napier's two rules above demonstrat. ed, which, as they are easily remembered, are commonly used in practice. Case Given So't с AC, В. R: Co-S, AC :: $,C: Co-S, B : and B is 2 AC, B C Co-S, AC:R :: Co-S, B: S, C : by 22. 3 B, C AC S, C:Co-S, B::R: Co-S, AC: by 22. and R:Co-S, BA:: Co-S, AC:Co-S, BC. 21. and if both BA, AC be greater or less than a 4 BA, AC BC quadrant, BC will be less than a quadrant. But if they be of different affections, BC Co-S, BA:R:: Co-S, BC : Co-S, AC. 21. 5 BA, BC AC and if BC be greater or less than a qua, drant, BA, AC will be of different or the same affection : by 15. 6 BA, AC B S, BA:R:: T, CA: T, B. 17. and B is of the same affection with AC, 13. ܝܪ Case Given So't? 7 BA, B JAC R: S, BA :: T, B': T, AC. 17. And ac АС AC is of the same affection with B. 13. 8 AC, B BA T, B:R::T, CA: S, BA. 17. : R: CO-S, C::T, BC:T, CA. 20. If BC be BC, CAC less or greater than a quadrant, C and B , c a will be of the same or different affection. 15. 13. Co-S, C:R::T, AC : T, BC. 20. And BC 10 AC, C 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 11 BC, CA c be less or greater than a quadrant, CA and AB, and therefore CA and C, are of the 1 same or different affection. 15. : 12 R: S, BC :: S, B:S, AC. 18. And AC is BC, B AC of the same affection with B. S, BC:R::S, AC: S, B. 18. And B is 14 BC, ACB of the same affection with AC. 15 B, C T, C:R:: Co-T, B:Co-S, BC. 19. And BC according as the angles B and Care of different or the same affection, BC will be greater or less than a quadrant. 14. 16 | BC, C R: C0-S, BC::1,C: Co.T, B. 19. If BC Bbe less or greater than a quadrant, C and B will be of the same or different affection. 15. The second, eighth, and thirteenth cases, which are common. ly called ambiguous, adroit 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, lèt 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 man. ner, 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 inyersion, the radius is to the sine of DC as the sine of the angle D to the sine of AC: therefore ex aequo 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. IŅ 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. |