On finding the Latitude. 93 
. Now let P be the elevated pole; PA and PB the horary cir- 
eles passing through the sun’s centre at the two observations ; 
then, if each of the ares PA and PB be equal to the polar di- 
stance of the sun, or to the complement of his declination at the 
middle of the elapsed time, A and B will represent the two ap- 
parent places of the sun. Conceive two small circles described 
upon the surface of the sphere about the poles A and B, and at 
distances from them respectively equal to the complements of 
the observed altitudes; these circles will intersect in two points 
Zand Z’ situated at equal distances on opposite sides of the great 
cirele passing through A and B, and, generally speaking, either 
of the points Z or Z’ may represent the place of observation. The 
problem therefore admits of two solutions, the latitude sought be- 
ing the complement of either of the polar distances Z P or Z’P. 
Draw the great circle PDOD’ to bisect the vertical angle of 
the isosceles triangle AP B; which circle will therefore intersect 
the base AB at right angles and will bisect it. Draw the ares 
ZD and Z’ D perpendicular to the circle PO; these arcs will be 
equal, because Z and Z’ are similarly situated with regard to both 
the circles AB and PO. Then the angle AP B is known, for it 
is equal to the elapsed time converted into degrees at the rate 
of 15° to 1; wherefore in the right-angled triangle A PO, the 
hypothenuse A P and the angle AP O, equal to half AP B, being 
known, the sides AO and OP may be found by the rules of sphe- 
rical trigonometry. In the triangle AZB, the two sides ZB 
aud’Z A, being the complements of the observed altitudes, are 
known; and as the are AO, half AB, has been found, we may 
compute the ares ZD and DO by the premised |emma. Now 
P D is the difference, and P D’ the sum, of the ares PO and OD; 
and hence the two sides about the right angle are known in each 
of the triangles ZPD, Z’PD': wherefore we may find the polar 
distances ZP and Z’ P, and likewise the angles ZPO and Z’ PO, 
which are the horary angles at the middle time, and the problem 
will be completely solved. 
Although the problem is ambiguous in theory, yet, in most 
cases, it becomes determinate in practice. In the first place there 
is no ambiguity when the arc Z’P is equal to, or greater than, 
90°: for the distance ZP between the place of observation and 
the elevated pole is always less than a quadrant. In order to 
find a criterion for determining this point without actually com- 
puting both latitudes, it is to be.observed that the angle contained 
between the circles ZO and O P is always less than a right angle ; 
and, because in right-angled spherical triangles the sides are of the 
same affection with the angles opposite to them, it follows that the 
are ZD will be less than 90°. Wherefore, Z’ D being less than 90°, 
the polar distance Z’ P will be greater than, or equal to,a quadrant, 
L2 according 
