88 On finding the Latitude. 
This is Dr. Brinkley’s third example (pp. 11 and 12). It is 
an unfavourable instance for his rules, requiring several compu- 
tations and corrections to arrive at a right result. It admits of 
two solutions but without ambiguity, if the latitude by account 
be sufficient to ascertain that the true latitude is less than 5° 46’, 
the complement of p. The other latitude is 1° 31’-6. 
These examples will be sufficient for showing the method of 
calculation. 1 proceed now to consider the correction required 
for the sun’s change of declination in the interval between the 
observations. The true place of the pole will now be at P’, with- 
out the great circle DO, which bisects the are AB, because the 
polar distances P’ A and P’B are unequal. Draw P’ P perpen- 
dicular to that great circle, and complete the isosceles triangle 
APB. The ares AP and PB make equal angles with the circle 
P’P; and hence in the small change of place from P to P’, one 
of the two ares AP’ and BP’ will increase just as much as the 
other decreases ; and each of the arcs A P and P B will be equal 
to half the sum of the polar distances P’ A and P’B. We shall 
therefore obtain the are ZP by the method already explained ; 
and, having drawn Pm perpendicular to ZP’, the correction we 
are seeking is mP’ =PP’ x sin P’ Pm=PP’ x sin ZPO= 
PP’ x sin S. Also, by the lemma, 
, _ cos P’ A —cos P’B 
PP sire ao 
Now, d being the declination at B the greater altitude, and D 
the mean declination as before, we have 
P’A = PA—(D—d), 
P’B = PA+ (D—4d); 
And-hence, PP’ = (D—d) x ss = as, 
Wherefore, 
Pm = (D—d) x @*% 
sin t * 
The corrected latitude will therefore be 
a—(D—d) x eas | 
sin ¢ 
or, independently of S; because sin S = obey. 
cosa ? 
A—(D—d) x —“*_. 
sin ¢ cos A 
Again the arcs P’A and P’B may be considered as making 
equal angles with PP’: consequently the horary circle at the 
middle time, which bisects the angle AP’B, will be perpendicular 
to PP’, Hence the true horary angle of the middle time is equal 
to the complement of ZP’P. But from the triangle ZP P’, we get 
: Sin 
