PARTICULARLY APPLICABLE TO SOME GEODETICAL PROBLEMS. 135 



Thus the problem has two solutions, Avhich may be deduced from 

 Art. 14. and Formula vii. Art. 18. as follows: 



1. Find A, B, C the angles of the triangle whose sides are a, h, c. 



2. Also A', S', C the angles of a triangle whose sides are pa, qb. re. 



3. Thenx = ^^ 



\/{q-— 2qr cos (A ± A') +r'^\ ' 

 There are like expressions for y and s, which are more simply. 



q r 



y = - X, z = - X. 

 P P 



34. The value of x may be found from the formula by subsidiary 

 angles. The radical -^/(^^-S^';- cos a + /-°) being the base of a triangle 

 whose sides are q and r, and the contained angle a = A±A', its value 

 may be found by the formula? of Trigonometry; for example, by one 

 which I proposed in the Transactions of the Royal Society of Edinburgh, 

 Vol. X. viz.. 



Let a, b, c, be the sides, and A, B, C the opposite angles ; 

 « + i : a-b = tan\{A+B) : tanl(^-^), 

 cos \{A-B) : cos \{A + B) = a + b : c; 

 Also sin \{A-B) : sin {{A + B) = a-b : c. 

 Hence, to find x, find E, such that 



Q — r 



cotan E = ^ cot A a. 



q + r ■^ 



Then x 



_ pa sin E 

 q + r' ain i a 



_ ^irt cos E 

 q — r' cos \ a ' 



35. We have seen that the first problem in this paper is convertible 

 into another, in appearance easier. The solution of the latter is usually 



