Of KEPLER'S PROBLEM. 22^ 



A 



Then, fin x — tan — x fin 45', and p' — 90' — 2X, 



log. tan — = 9.7080866 



log. fin 45 ° - 9.8494850 



log. fin X = 9.5575716 - log. fin 21 " 9' 53", 

 therefore, 2X = 42° 19*46", and / = 47" 40' 14" ; this value of 

 p' is greater than BC. 



fin-^-:=i^' 



3. For the third term />", we have e — b x 



fin 21° 10' , . ^ „ fin 2i°io' 



1 . : and tan A = ^ x fee 45° = — ;- x fee 4c'. 



arc 21° 10 ^^ arc 21° 10 ^^ 



log. feeant 45° = 10.1505150 



log. fin 21 ° 10' = 9.5576060 



fiim — 10= 9.7081210 

 add conft. log. = 3-5362y^g 



13-2443949 

 fubtraft log. 1270' = 3.1038037 



log. tan A = 10.1405912 

 therefore A = 54° 6' 58". 



Now fin X = tan — X fin 45 ", and p" = 90° — 2X ; 



log. tan - = 9.7082530 

 log. fin 450 =r 9.8494850 



log. fin X = 9.5577380 = log. fin 21° 10' 24". 

 Therefore 2X = 42° 2o'48', and/'= 47° 39 12"; this value 

 of />" is kfs than BC. 



The 



