Mr. Schroeter's Observations 
32S 
For, EcG = AcB = 90° 
Ec C = AcK 
And hence CcG = KcB. 
To find, therefore, the angle Sr T, calculate for the 12 th 
of March, 1790, at six o'clock in the evening. 
The true heliocentric long, of Venus 5 s 18 0 37' 35" 
Long, of the earth 5 22 21 20, 
Difference 3 43 45, 
Heliocentric lat. of Venus 3 23 o 
Log. sin. 3 0 43' 45" = L. 9>999°795 
+ Log . cos. 3 23 o = L. 9,3992424 
Total = cos. c S T = 9*99832 19 
= L. 5 0 2' o" 
Now in order to find the angle S c T, according to this 
heliocentric distance of Venus from the earth, we have in the 
triangle Sc T, the angle at S, and the two sides c S and T S, 
being the distances of Venus and our earth from the sun. We 
then find the log. c S = 4857022, and the log. TS = 
4997758, and the angle at S being — 5 0 2 / ; the sum of the 
two angles will be = 174 0 58', the half of their sum = 87° 29k 
and then the computation will be as follows : 
