PTOLEMY’S SYNTAX IS 
285 
not quote those propositions, as he might have done, but proves 
them afresh by means of Menelaus’s theorem. 1 The appli 
cation of the theorem in other cases gives in effect the 
following different formulae belonging to the solution of 
a spherical triangle ABC right-angled at C, viz. 
sin a = sin c sin A, 
tan a = sin b tan A, 
cos c = cos a cos b, 
• tan b = tan c cos A. 
One illustration of Ptolemy’s procedure will be sufficient. 2 
Let HAH' be the horizon, PEZH the meridian circle, EE' 
the equator, ZZ' the ecliptic, F an 
equinoctial point. Let EE', ZZ' 
cut the horizon in A, B. Let P be 
the pole, and let the great circle 
through P, B cut the equator at C. h 
Now let it be required to find the 
time which the arc FB of the ecliptic 
takes to rise; this time will be 
measured by the arc FA of the 
equator. (Ptolemy“ has previously found the length of the 
arcs BC, the declination, and FC, the right ascension, of B, 
I. H, 16.) 
By Menelaus’s theorem applied to the arcs AE', E'P cut by 
the arcs AH', PC which also intersect one another in B, 
crd. 2 PH' _ crd. 2 PB crd. 2 CA 
crd.2 H'E' = crd. 2 BC ’ crd. 2 AE 7 ' 
sin PH' sin PB sin G A 
sin~WË' = sin BC ‘ sin A E' ' 
Now sin PH' = cos H'E', sin PB = cos BC, and 
therefore cot H'E' = cot BC. sin GA, 
in other words, in the triangle ABC right-angled at G, 
cot A = cot a sin b, 
or tan a = sin b tan A. 
1 Syntaxes, vol. i, p. 169 and pp. 126-7 respectively. 
2 lb., vol. i, pp. 121-2.