84 On finding the Latitude. 
according as the known are P D', or PO + OD, is greater than, 
or equal to, a quadrant; in all which cases there is only one so- 
lution, by means of the triangle ZPD having the side PD equal 
to the difference of POand OD. But when the are PD’, or 
DO + OP, is less than 90°, the same pole will be elevated above 
the horizons of both the zeniths, and recourse must be had to 
other considerations to distinguish the true solution from the 
false one. Now, in this case, the zenith Z’ will be always be- 
tween the great circle AOB and the equator, having a latitude 
less than the complement of the are PO: wherefore, if it be 
known that the latitude of the place of observation is greater 
than the complement of PO, the ambiguity will be removed, and 
the true solution will be obtained by means of the triangle ZPD 
as in the former cases. On the other hand, if the latitude of the 
ship be less than the complement of PO, both latitudes must be 
computed ; if they be on different sides of the complement of 
PO, the case will be determined; but if they be both less than 
that arc, the solution will remain ambiguous unless the latitude 
by account be known so nearly as to enable the calculator to 
make a choice. That both the latitudes may be less than the 
complement of PO, which is the greatest distance between the 
great circle passing through A and B and the equator, will be 
obvious if it be considered that the two zeniths may be as near 
the great circle A B as we please, and may even coincide in one 
point in its cireumference. This ambiguous case can happen but 
rarely; and when it does occur, the problem will have no preten- 
sion to much precision ; because the difference between the ares 
ZB and ZA, will be so nearly equal to the are AB, that very 
small errors in the observed altitudes will occasion a great varia- 
tion in the position of the points Z and Z’. By means of these 
observations the ambiguity of the solution is mostly, but not en- 
tirely, taken away. 
I shall now reduce the foregoing solution into algebraic for- 
mulz of calculation, which wiil be shorter than giving a rule in 
words at length. Let 2 and h’ denote the two altitudes, the let- 
ter without the accent standing for the greater; D the sun’s de- 
clination at the mean time between the two observations ; and 
# the angle found by converting half the elapsed time into de- 
grees at the rate of 15° to 1>: these are the data of the problem. 
Put also J for half the base, and p for the perpendicular, of the 
isosceles triangle AP B; y for the are ZD; ax for the are DO; 
and further, for the sake of abridging, let 
sinh + sin h! 
A =-—— 
> 
sin h — sin h! 
B= Sierig hiits 
Then, 
