of KEPLER'S PROBLEM. 



213 



fonings that have led us to this conclufion are quite general, and 

 hold good in every ftate 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 

 f, from the equation fm (« — c) = <? x fin v x cof ►-, fuppofing n 

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

 about. 



The given equation, 

 fin (« — v) zz: e X Gn II X cofv, 

 is eafily transformed into 

 this, (Art. 6.) 



fin « cof H , 



fin II cof v 

 Let A D B be a femicirck, 

 and let the diameter DC be 

 drawn at right angles to AB : 

 Take the arch AM rz /;, 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 AB and CD, in the angle DCB, may be equal to 

 CA, the radius of the circle : and, laftlyj 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 



fin AN' 

 cof AM 

 cof AN • 



Now, FH — FG = HG = CA, therefore 



CA z= FC X /filL^ _ S2[AM] 

 Vfm AN cof AN/' 



Vol. v.— P, II. 



Ee 



confequently, 



