322 



On the Problem of the 



[Oct. 



iuation of the point D which 

 if exterior to the triangle as at 

 fig- I. 



K 360o — (P + 7T + C) 

 If interior as at fig. II. 

 R=$ (C-P-tt) 



If the angle of the triangle is 

 meant, as at fig III. 



B = (C - P — at) 



Another method is given by- 

 Puissant in his " Geodesic," I 

 believe first proposed by De- 

 lambre, the solution of which 

 is probably conducted in the 

 same way as made use of by me 

 at page 141 of this journal. In 

 this the half sum of the angles 

 x and y being known, the half 

 difference is found thus— 



Tang § diffr. = tang (0 - 45°). tang \ R 



a. sin 7r 

 Tang 0 = • ;— 



b. sm P 



the value of R being as before. 



This method is far better than the last given, being much more easily 

 applied, and admitting of the greatest accuracy, with being operose. 



In testing the correctness of the quantity found, the value of the 

 line D C may be computed from each thus :— 



D C = — - — sin x. =-h — sin tf. 

 sin P. sm it. 



which will not agree if any error has been committed. 



There is another still more convenient mode of computing the values 



required, by first getting a rough approximation to the value of the 



angle x by a protraction on paper, and then trying the value of D C : 



thus, make 



L=log (a. sin x. cosec P) and 



\=log (a. sin (R — x). cosec w) 

 should the value of x have been taken correctly, 4 will be equal to \ j 

 otherwise take, 



L 4- D. 0' = \ X 0' where 

 ~ L ~ X 



$ *= D + B 



/''if . 3. 



f 



\rrfe/ 



xi/ 



