Dr. Maskelyne' 's formula for finding the longitude , &c. 143 
its right ascension is in the third quadrant, but F its longitude 
is in the second, and SAC, which is equal to B, is greater than 
EAC or 90°. 
Hence it follows, that if B be greater than 90 °, the star 
must be situated between e A and p A, or between EA and PA, 
and that the consequences with respect to its longitude, must 
be as stated in the third Rule. 
Dr. Maskelyne says, p. 59, Problem XIII. “ Longitude 
will be of the same kind, or in the same quadrant of the circle 
as iH is, unless B exceeds 90°, which can only happen when jR 
is in second semicircle .* Then if M be in third quadrant or 
from 6 s to 9% longitude will be in second quadrant or from 
3 s to 6 s , and the operation will give L. cot. excess of long, 
above 3 s . Or if Jit be in fourth quadrant, or from 9* to 12 s , 
longitude will be in first quadrant ; and the operation will 
give, L.t, long, under 3% or in first quadrant." 
Problem II. 
“ The longitude and latitude of a celestial object, with the 
obliquity of the ecliptic, being given, to find its right ascension 
and declination." 
Using the same figures as in the last article, by trigono- 
metry, sin. AF : R :: tan. SF : tan. SAF 
north or south as the latitude is. 
Let this first auxiliary angle be called A. Then when the 
longitude is in the first or second quadrant A+ 0 =:SAH=B, 
the second auxiliary angle, if the latitude is north, but A ^ O 
= SAH — B, if the latitude is south. 
* The words printed in italics contain a mistake, which would affect the longitude 
of any celestial object situated between Ap, Ae, and pe, both in Fig. 2 and 5. 
R tan.SF R tan. latitude 
sin. AF sin. longitude 5 
