286 
TRIGONOMETRY 
Thus AC is found, and therefore FC—AC or FA. 
The lengths of EG, FC are found in I. 14, 16 by the same 
method, the four intersecting great circles used in the figure 
being in that case the equator EE', the ecliptic ZZ', the great 
circle PBGP' through the poles, and the great circle PKLP' 
passing through the poles of both the ecliptic and the equator. 
In this case the two arcs PL, AE' are cut by the intersecting 
great circles PC, FK, and Menelaus’s theorem gives (1) 
sin PL _ sin CP sin BF 
sin KL sin BG sin FK « 
But sin PLj = 1, sin KL., = sin BFG, sin CP — 1, sin FK = 1, 
and it follows that 
sin BG = sin BF sin BFG, 
corresponding to the formula for a triangle right-angled at G, 
sin a = sin c sin A. 
(2) We have. 
sin PK _ sin PB sin GF 
sin KLj sin BG sin FL * 
and sin PK = cos KL = cos BFG, sin PB = cos BG, sin FL = 1, 
so that tan BG — sin GF tan BFG, 
corresponding to the formula 
tan a = sin h tan A. 
While, therefore, Ptolemy’s method implicitly gives the 
formulae for the solution of right-angled triangles above 
quoted, he does not speak of right-angled triangles at all, but 
only of arcs of intersecting great circles. The advantage 
from his point of view is that he works in sines and cosines 
only, avoiding tangents as such, and therefore he requires 
tables of only one trigonometrical ratio, namely the sine (or, 
as he has it, the chord of the double arc). 
The AnaJvmma. 
Two other works of Ptolemy should be mentioned here. 
The first is the Analemma. The object of this is to explain 
a method of representing on one plane the different points