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Problem I. 
To find in the ecliptic the point of longed afccnfon. 
Analysis. 
Let (Fig. 4.) ABC be the equator, ADC the 
-ecliptic, BD the fituation of the horizon, when D 
is the point of longed afcenfion. Let EFG be an- 
other fituation of the horizon. Then the ratio of 
the fine of EB to the fine of F D is compounded of 
the ratio of the fine of BG to the fine of GD, and 
of the ratio of the fine of AE to the fine of AF i 
but the angles B and E being equal, the arcs EG, 
GB together make a fern i circle ; and, by the ap- 
proach of EG towards GB, the ultimate magnitude 
of BG will be a quadrant, and the ultimate ratio of 
-EB to FD will be compounded of the ratio of the 
radius to the fine of DG (that is, the cofine of BD) 
and of the ratio of the fine of A B to the fine of AD. 
dDraw the arc DH perpendicular to AB. Then, in the 
triangle B DH, the radius is to the cofine of B D, as 
the tangent of the angle BDH tothecotangentof HBD. 
Alfo, in the triangle BDA, the fine of AB is to the 
fine of AD as the fine of the angle B D A (or BDC) 
to the fine of A B D ; therefore, the ultimate ratio 
of B E to DF is compounded of the ratio of the 
tangent of BDH to the cotangent of ABD, and 
of the ratio of the fine of B D C to the fine of 
ABD; which two ratios compound that of the 
redangle under the tangent of BDH and the fine of 
BDC to the rectangle under the cotangent and the 
ifine of the given angle ABD, 
4 
But 
