Solutions to Important Problems. 131 



(0+}) (2z^2£) 



v V COS ft J 



— 2 



(a+b) sin 2 0) 



C08 ft 



a? = (a + J) sin 2 (-^) . sec ft 



The formulas which solve the problem being both adapted to 

 logarithmic calculation. 



By introducing the radius r of the tables, we may write them 

 as follows : — 



tan 2 ft = ifj-fnC sin(A + B + C).r 2 (1) 



(a + by . sin A . sin B 



(a + b) . sin 2 (£\ . sec ft 



3-^ (2) 



r 3 



If the segments be arcs of a geodesic in the reduced surface 

 of the earth, and that a and b are given in circular mea- 

 sure, we know, from modern geometry that by substituting the 

 sines of the arcs a, b, and x instead of the arcs themselves the 

 above formulas will have their equations true also ; and therefore 

 in such case we have 



sin a . sin b . sin C . sin (A + B + C) . r 4 



tan 2 ft = 2/ fl + fe x 2s a —b\ . . . _ ( 3 ) 



Bin y—j- ) . cos \~T~) ■ Bm A ' sm B 



... 2sin(*4>).co S (^).sin*(<L). 6 ec0 

 em a? = (4J 



f»-sp= -n. • - -U • ii, 4. sin C . sin (A + B + C* „, • 



B|2p It is obvious that v — : — =r— - — ' or h is 



sm A . sm B 



constant for all positions of the point P : and this fact is of con" 

 siderable importance in practice, as it enables us (by shifting the 

 position of P,) to test the accuracy of our observations and the 

 quality of the instrument used. Having, from various positions 

 for P, obtained the value of h with precision, we know then that 

 we can calculate x with an accuracy equal to that by which a and 



