125. Napier's rules for the solution of right angled spherical triangles, though applicable to all cases, do not give results of that degree of accuracy, which is sometimes required, when the sought part expressed by its sine is very small, or expressed by its cosine, is very near 90°. The following formulæ may in such cases be used. I, At Art. 86, by the formula which follows, (6), we have, R being 1, 1-cos p2 sin2 p in a similar manner, we might obtain changing p into a, and substituting the value of cos a, given COS (B+c) tan?a= -c) cos (B which is a formula to be employed, when в and c are given and a required. II. With the same data to find b use the formula tan b = √{tan [(B—c)+45°] tan [}(B+c)—45°] } derived from cos b= COS B sin c and formula (6), Art. 74, III. The hypothenuse a and the side c being given to find the adjacent angle в, use the formula derived in a manner similar to that in Case I, Finally, to obtain b when the opposite angle в and the hypothenuse a are given, we have, by Napier's rules, 1-tan x whence, observing that 1+tan x tan (45°-x) (Art. 119) and making tan x= sin a sin в, tan (45°—b) = √ tan 45° a sin B. This last gives the value of b, 2 being calculated by the equation tan x = sin N. B. In the above five formulæ, R is supposed to be 1. To introduce it correctly when logarithms are applied to the formulæ, it is only necessary to observe that the two members of each must be homogeneous. 126. When, in the case considered at Art. 87, the only part required happens to be the side opposite the given angle, the finding of the other two angles then becomes merely a subsidiary operation, and the determination of the required side, by Napier's analogies, seems somewhat lengthy. But a shorter method of solution is deducible from the fundamental formula, cos c= cos a cos b + sin a sin 6 cos c (1). For substituting cos a tan a for its equal sin a it becomes COS C= cos a (cos b + tan a sin b cos c), Assume COS O tan a cos C=cot w= i sin o) then sin w cos b + sin b cos a COS C = COS a sin w cos a sin (a + b). sin a Hence, to find the side c, we must determine a subsidiary angle w from the equation cot w = tan a cos C (2); after which c is found by the equation cos a sin (w +6) (3). COS C sin a EXAMPLE. a 1. In a spherical triangle are given a=38° 30', b = - 70°, and c=31° 34' 26", to find ca a Other formulas for the determination of c might be easily deduced from the same equation (1), but this is as short and as convenient as any. We might also introduce here a distinct formula for the determination of one of the angles a, by help of a subsidiary arc w; but as little or nothing would be gained, in point of brevity, over the process by Napier's analogies, we shall not stop to investigate it. 127. If where two angles and the included side are given, the angle opposite to the given side be the only part required, a more compendious method of solution may be obtained by introducing a subsidiary arc, as in the last case. Thus from the fundamental formula for the cosine of a side in terms of the three angles, might be obtained by aid of the polar triangles, a formula which becomes when cos a tan a is substituted for sin A, cos c = cos a (tan a sin B COS C-COS B); or assuming Hence, having found a subsidiary angle w, by the equation coto = tan A COS C (1); the sought angle is determined by the equation Cos A sin (B—01) sin w COS C EXAMPLE In a spherical triangle are given two angles equal to 39° 23' and 33° 45' 3" and the interjacent side 689 46' 2" to find the third angle. tan a 39° 23' 0' 9.914302 COS A 9.888133 cos c 68 46 2 9.558898 sin w, ar. comp. 0.018392 . cot w 73 26 331 9.473200 sin(B—w)39°41'301" 9.805268 COS B . sin a C As (B --w) is negative, cos c must be negative; hence c is the supplement of this, viz. 120° 59' 47". 128. The part of a spherical triangle determined by the proportion sin a : sin b :: sin a : sin B admits of a double value, since two arcs answer to the same sine ; it becomes necessary, therefore, for us to inquire under what circumstances both these values are admissible, and how we may know which to choose when but one solution exists. Referring to the fundamental formula (Art. 82,) we have cos b COS a COS C i (2) in which expression we may remark that if cos b is numerically greater than either cos a or cos c, the second member must take the sine of cos b, consequently, B and b must be of the same species if sin b<sin a , or sin b< sin c, that is, an angle must be of the same species as its opposite side, if the sine of this side is less than the sine of either of the other sides. But if cos b is numerically less than cos a, then whether the right hand member be + or - will depend upon the magnitude of cos c, or cos c will have two values corresponding to + cos B, and COS B; hence an angle has two values, when the sine of its opposite side is greater than the sine of the other given side. In the proportion sin A : sin B :: sin a : sin b |