Of KEPLER'S PROBLEM. 22^ 



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



A 



log. tan — = 9.7080866 



log. fin 45 ° = 9.8494850 



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

 therefore, 2 A = 42 ° 19' 46", and p> — 47 ° 40' 14" j this value of 

 p' is greater than BC. 



3. For the third term p'\ we have e zz e x 



1 



i-O-/0 



fin 21° TO' 1 a r fin 2I°Io r 



; , : and tan A rr e x fee 45 ° = s — - X fee 45 . 



arc 2 1 ° 1 o ^ J arc 2 1 ° 1 o ^ J 



log. fecant 45° = 10. 1505 150 



log. fin 21 ° 10' = 9.5576060 



fum — 10= 9.7081210 

 add conft. log. = 3*5362739 



r 3- 2 443949 

 fubtract log. 1270' = 3.1038037 



log. tan A = 1 0.14059 1 2 

 therefore A = 54 6' 58". 



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



log. tan — = 9.7082530 

 log. fin 45 = 9.8494850 



log. fin x = 9.5577380 - log. fin 21 10' 24". 

 Therefore 2X = 42 20' 48", and/' = 47 39' i 2 " ; this value 

 of p" is lefs than BC. 



The 



