518 ASTRONOMICAL PROBLEMS. 
In the triangle CP’a we now have 
the angle C=111° 35’ 28’-7 Ca= 89° 46’ 333 
CP'= 29° 25' 47” 
log cos C= 9°56582828 
log tan P’'C= 9°75139824 
log tan Ist arc=19°31722652 
sin Ca 
sin P’a 
log sin C= 9:9684044 
log sin Ca= 9'9999967 
19:9684011 
—log sin Pa= 9:9930562 
sin P’=sin C 
Ist arc= 11° 43’ 40’:77* 
2nd arc=101° 30’ 14:07 
log cos 2nd are= 92998009 
log cos CP’= 9:9399977 
19:2397986 
—log cos Ist are= 9:9908376 
log cos P'a= 9:2489610 
P’a=180° — 79° 46’ 533 
N.P.D. in 1820=100° 13’ 6”-7 
log sin P’= 9-9753449 Decl.= 10° 13' 6-7 8. 
*, P’=70° 52’ 33” From Naut. Alm.= 10° 13/ 3-3 
= 4h 43m 308-2 Dit, =) Of) 07 3/-4 
add, 8-719 x 67 48s 
R.A. =182— 4b 43m 188-2 
= 13 16™ 418-8 by calculation 
135 16™ 438-46 from Naut. Alm. 
Diff= 0b Om 18-66. 
Problem IX. Calculate the R. A. and N. P. D. of Pollux 
B 
(8 Geminorum) for 1. 1. 95 from data given in the Nautical Almanack 
on 1.1.87, 
* The angle C being greater than 90° the 1st arc must be added. 
