Of KEPLER'S PROBLEM, 



21 



failings chat have led us to this conclufion are quite general, and 

 hold good in every flate of the data of the problem. To have 

 a complete and univerfal folution of this famous problem, it 

 only remains, that we inveftigate a rule for computing the arch 

 »>, from the equation fin [n — v) = e x fin. » X cof v, fuppofing n 

 and e to be given quantities. This is what we are now to fet 

 about. 



Th e given equation, 

 fin [n — v) =r e X fin v X cof v, 

 is eafily transformed into 

 this, (Art. 6.) 



fin n cof n , 

 — fin v cof v 



LetADB be a femicircle, 

 and let the diameter DC be 

 drawn at right angles to AB : 

 Take the arch AM = n, and 

 in CM produced take CF = 



B 



- X CA : From the point F 



draw the ftraight line FGH, 

 fo that the part GH, inter- 

 cepted between \B and CD, in the angle DCB, may be equal to 

 CA, the radius of the circle : and, laftly* through C draw CN, 

 parallel to FH : I fay, that the arch AN is equal to v. 

 From the two triangles FCH, FCG, we have 



fin AM 



fin FHA : fin FCA : : FC : FH = FC X 



fin FGD : fin FCD : : FC : FG =: FC X 

 Now, FH — FG = HG - CA, therefore 



fin AN' 



cof AM 



cof AN ' 



CArzFC 

 Vol. V.— P. II. 



< 



fin AM cof AM \ 

 fin AN cof AN / ; 

 E e 



confequently, 



