2i6 A NEW and UNIVERSAL SOLUTION 



being determined in magnitude only, will correfpond to two 

 arches, the one lefs, and the other greater than 90°. Of thefe 

 two arches, the former will give the pofition of the line FH, 

 having the part GH, infcribed in the angle DCB, equal to CA ; 

 and the latter will give the pofition of the line FKL, having the 

 part KL, infcribed in the angle ACL, equal to CA^ 



The other value of fin 2v, viz. fin 2v — ^ ^ , 



will give no folution when e is not greater that i ; becaufe, in 



that cafe, 



v/2 r^ + 1 + I • 



t^ 



is greater than i : But, for other va^ 



lues of e when ^^ '^ ^ is lefs than i, this value of fin 2v 



will alfo determine the pofition of two ftraight lines, both lying 

 in the angle ACD, that will fatisfy the problem. 



In the particular application we have in view, e being never 

 greater than i, we muft compute fin ax by the formula 

 fm ''y = v/2 e^ + I — I . ^^^ jjU ambiguity will be taken away» 



by the confideration that 2v, or the arch of eccentric anomaly, 

 muft be lefs than the arch of mean anomaly. 



The formula fin 2i 



v/2 e^ + I — I . 



is not very convenient 



in pradice : we fliall obtain another method of computing fin 2v, , 

 greatly preferable in this refped, in the following manner. 

 Resume the formula, 



I — fin 2v 



lin ^ 2'j 



Suppofe fin 2v - jV"^.Qf A " ^^^"' ^^ fubftitution, we ftiall 



find 



cof A^x(i +cofA) _ j!ii:A-_ixtan^-A: 

 cof'A 2cof^A 2 



therefore tan A :=: f X ^/T = <? X fee 45 °. 



Also, 



f ' = - X — 



2 



