14 ° 
Dr. Robertson's demonstrations of the late 
obliquity of the ecliptic. Then in the first and second qua- 
drants of right ascension, for a star whose declination is north, 
A ~ O = SAF = B, the second auxiliary angle, but in these 
quadrants for a star whose declination is south A + O = 
SAF = B. 
In the third and fourth quadrants of right ascension, for a 
star whose declination is north, A -j- O = SAF = B, but in 
these quadrants for a star whose declination is south, A ~ O 
= SAF = B. 
If S be on Pp, as represented in Fig. 3 and 6 , then 90 0 — O 
= SAF == B. 
To find the longitude , 
We have the following proportions cos. SAH : R : ; tan. AH : 
tan, SA, and R : cos. SAF : : tan. SA : tan. AF. 
, TT A y~, cos. SAF tan. AH 
Hence cos. SAH : cos. SAF : : tan. AH : tan. Ah =— i^7sAH“‘ 
. . , cos. B tan. M 
That is tan. longitude = — • 
.a.i . R sill. A . . r 
Or, as tan. A : R : : sin. A: cos. A = tan> A > this being put foi 
cos. A in the preceding expression, we have also tan. longitude 
tan. A cos. B tan. M 
“ ~ R sin. A 
If S be on P p, then R : cos. SAF : : tan. SA : tan. AF 
c > n . . c a , . cos. (QO°— O) tan. declin. 
cosrSAFjai^SA^ That is tan. longitude == — - - g " * 
R 
To find the latitude . 
By trigonometry, sin. AF : R :: tan. SF : tan. SAF , and therefore 
tan. SF= sln,AFt p an ' S — , that is tan. latitude = “ : ' Io " s -g * 
-# v 1 . •»—< tan. AFcos.AF 
But tan . AF : R : : sin. AF : cos. AF, and sin. Ah = g ’ 
and this being put in the preceding expression for sin. AF, we 
tan. AF cos. AF tan. B ___ tan.long.cos.long.tan.B _ 
have also tan. latitude = g*— — — R a 
