pendicular to be let fall from one of the angular points of the oblique triangle in such a manner that the latter may be divided into two right-angled spherical triangles, in one of which there shall be data sufficient for the application of the rules; and, as a general method, it may be observed that such perpendicular should fall upon one of the sides from an extremity of a known side and opposite to a known angle. Thus, as an example, let the sides AB and AC, with the included angle A, be given, in the oblique spherical triangle ABC; and let the side BC with the angles at B and C be required. A B Imagine the arc BD to be let fall from the point в upon the side AC, then in the right-angled triangle AD B, the side AB and the angle at A are given besides the right angle at D; and with these data let the segment AD be found. If the rules of Napier be employed, it is to be observed that the angle at a will be the middle part, and that the hypotenuse AB with the side AD are adjacent extremes; therefore by the first of the two rules above (as in the formula (f') art. 62.) r. cos. BAD cotan. AB tan. AD; thus AD may be found, and, subtracting it from AC, the segment DC may also be found. The side BC will be most readily obtained by forming in each of the triangles ADB, BDC, an equation corresponding to (d) in art. 60., and dividing one by the other; each of these equations may be formed by the method given in that article, or by the second of Napier's Rules. Using the latter method, and considering AB and BC in those triangles to be the middle parts, in the first triangle let AD, DB, and in the second let BD, DC be the opposite or disjoined extremes; then by the Rule, r. cos. A B = cos. AD cos. BD and r. cos. BC cos. DC cos. BD, and dividing the former by the latter, we have COS. AB COS. AD = COS. BC COS. DC ; whence cos. AD: cos. DC:: cos. AB: cos. BC, or cos. DC cos. AB = cos. AD cos. BC .... (A), and either from the equation or the proportion the value of BC may be found. The angles ABC and ACB might now be found by Prop. 2., using the proportions sin. BC: sin. A:: sin. AB : sin. C, sin. BC sin. A:: sin. AC: sin. B; or the angle at c may be conveniently found from the triangle BDC, by Napier's first rule, considering the angle at C as the middle part, and the sides BC, DC as the adjacent parts; therefore (as at (f) art. 62.) COS. ACB cotan. BC tan. DC. Thus all the unknown parts in the triangle ABC are determined. 65. The formulæ given in the first and third Propositions, and in the second corollary to the latter, are unfit to be immediately employed in logarithmic computations, since they contain terms which are connected together by the signs of addition and subtraction; and, therefore, before they can be rendered subservient to the determination of numerical values in the problems relating to practical astronomy, they must be transformed into others in which all the terms may be either factors or divisors; for then, by the mere addition or subtraction of the logarithms of those terms, the value of the unknown quantity may be obtained. The transformations which are more immediately necessary are contained in the two following Propositions; and those which may be required in the investigations of particular formulæ for the purposes of astronomy and geodesy will be given with the Propositions in which those formulæ are employed. PROPOSITION IV. 66. To investigate a formula which shall be convenient for logarithmic computation, for finding any one angle of a spherical triangle in terms of its sides. From Prop. I. we have COS. ACB = - COS. AB COS. AC Cos. BC and subtracting the members of this equation from those of sin. AC sin. BC But (Pl. Trigon., art. 36.) 1 cos. ACB, or the versed sine of ACB, is equivalent to 2 sin.2 ACB; and, substituting in the numerator of the second member the equivalent of the difference between the two cosines (Pl. Trigon., art. 41.), observing that AB is greater than ACBC, since two sides of a triangle are greater than the third, the last equation becomes 2 sin. (AB+ AC-BC) sin. (AB - AC + BC) sin. AC sin. BC 2 sin.2 ACB= But if P represent the perimeter of the triangle, (ABAC BC) P-BC, and (ABAC + BC) P— A C, therefore the formula becomes sin.2 ACB (or vers. sin. ACB)= sin. (PBC) sin. († P (PAC) . . . (1). sin. AC sin. BC Again, if to the members of the above equation for cos. ACB (Prop. I.) there be added those of the identical equation sin. AC sin. BC 1= there will be obtained sin. AC sin. BC' sin. AC sin. BC 1+ cos. ACB = sin. But (Pl. Trigon., art. 36.) 1 + cos. ACB is equivalent to 2 cos. 2 ACB; and, substituting in the numerator of the second member the equivalent of the difference between the two cosines (Pl. Trigon., art. 41.), the last equation becomes 2 sin. § (AC+BC+AB) sin. † (AC+BC —AB) sin. AC sin. BC 2 cos. 2 ACB= putting, as before, P for (AC+BC+AB) and P - AB for (AC+ BC — AB), sin. P sin. (PAB) we have cos. 21⁄2 ACB = ... · sin. AC sin. BC (II). Thirdly, dividing the formula (1) by (11), member by mem In the above investigations it has been supposed that the radius is unity if it be represented by r, the numerators of the formulæ (1), (11) and (Î11) must (art. 60.) be multiplied by r2. PROPOSITION V. 67. To investigate formulæ convenient for logarithmic computation in order to determine two of the angles of any spherical triangle when there are given the other angle and the two sides which contain it. In a spherical triangle ABC, let AB and AC be the given sides, and BAC the given angle; it is required to find the angles at B and c. From (a) and (b) respectively in Prop. I. we have cos. AB = cos. ACB sin. AC sin. BC + cos. AC cos. BC.. .... .... (h) (k) cos. BC cos. BAC sin. AC sin. A B+ cos. AC cos. AB Multiplying both members of this last equation by cos. AC we get COS. BC COS. AC= cos. BAC sin. AC cos. AC sin. AB + cos.2 AC Cos. AB, and substituting the second member for its equivalent in (h) there results COS. AB cos. ACB sin. AC sin. BC + cos. BAC sin. AC cos. AC sin. AB +cos.2 AC cos. AB, or cos. AB (1 cos.2 AC) cos. ACB sin. AC sin. BC +cos. BAC sin. AC cos. AC sin. AB. In this last substituting sin.2 AC for 1 - cos.2 AC, and dividing all the terms by sin. AC, we have cos. AB sin. AC = cos. ACB sin. BC + cos. BAC cos. AC sin. AB.... (m). In like manner, from the formula (c) and (b) in Prop. I., and from (k) above, we get cos. AC sin. AB = Cos. ÀBC sin. BC + cos. BAC Cos. AB sin. AC.... (n); and it may be perceived that this last equation can be obtained from that which precedes it by merely substituting B for c and c for B. Adding together the equations (m) and (n), and afterwards subtracting (n) from (m) we have, respectively, putting A, B, and c for the angles BAC, ABC, and ACB, cos. AB sin. AC + sin. AB cos. AC = sin. BC (cos. C + cos. B) + cos. A (cos. AC sin. AB + sin. AC cos. AB), cos. AB sin. AC sin. AB Cos. AC = sin. BC (cos. C + cos. A (cos. AC sin. AB sin. AC Cos. AB). COS. B) also (Pl. Trigon., art. 32.) sin. AC cos. AB± cos. AC sin. AB= sin. (AC± AB): therefore, after transposition and substitution, the last equations become sin. (AC + AB) (1 — cos. ▲): 1 2 sin. BC cos. (B + C) cos. § (B −C), (B + C) sin. 1⁄2 (B — C). 2 sin. a cos.a, A: therefore But again (Pl. Tr., arts. 35, 36.) sin. a 1-cos. A= 2 sin.2A, and 1 + cos. A = 2 cos. the last equations may be put in the form 2 sin. (AC + AB) COS. (AC sin. BC cos. + AB) sin.2 21 A = (B-c)... (p), sin. BC sin. (B+ C) sin. (B-C). . . (2). Now from Prop. II. we have sin. B : sin. C :: sin. aC : sin. AB; whence sin. B + sin. C : sin. B and sin. AC sin. AC + sin. AB sin. C: sin. AC + sin. AB: sin. AB, = (B-C) (AC-AB) CAB = tan.2 14 (B + C), cos. (AC+AB) sin. 1 (AC — AB) Multiplying (r) by (s) we have Cos.2 (ABAC) 1 Cos. cotan.2 AC) cotan. CAB = tan. 1 (B + c) . . . . (1). COS. (AB+ AC) Again, dividing (r) by (s) and extracting the roots, AC) cotan. CAB tan. (B - c).... (11). sin. (AB + AC) Thus from the formulæ (1) and (11) there may be obtained the values of the angles ABC and ACB in terms of the angle CAB and of the sides A B and AC, which contain it. |