290 
TRIGONOMETRY 
both in the plane LSHG at right angles to the meridian; 
therefore arc SQ — arc L'Z'. 
Hence all four arcs $F, VC, QG, QS are represented in the 
auxiliary figure in one plane. 
So far the procedure amounts to a method of graphically 
constructing the arcs required as parts of an auxiliary circle 
in one plane. But Ptolemy makes it clear that practical 
calculation followed on the basis of the figure. 1 The lines 
used in the construction are SF= sini (where the radius =1), 
FT = OF sin cf), FG = OF sin (90° — 0), and this was fully 
realized by Ptolemy. Thus he shows how to calculate the 
arc SZ, the zenith distance (= cl, say) or its complement $F, 
the height of the sun (= h, say), in the following way. He 
says in effect; Since G is known, and Z F'O'G' = 90° — 0, the 
ratios O'F': F'T and Q'F': O'T' are known. 
[In fact 
O'F' _ D 
Ö'T ~ crd. (180° — 20)’ 
where D is the diameter 
of the sphere.] 
Next, since the arc MS or M'S' is known [ = £], and there 
fore the arc P'S' [= 90° — £], the ratio of O'F' to D is known 
[in fact O'F'/D = {crd. (180 — 2t)}/2D. 
It follows from these two results that 
0'T' = 
crd. (180°- 2t) 
2D 
. crd. (180° —2 0)]. 
Lastly, the arc $F (= h) being equal to B'F, the angle h is 
equal to the angle O'I'T' in the triangle I'O'T'. And in this 
triangle O'F, the radius, is known, while O'T' has been found; 
and we have therefore 
'T' crd. {2h) crd. (180° — 2t) crd. (180°—2 0) „ , 
yp = —j } = ' —jp , from above. 
[In other words, sin h — cos t cos 0 ; or, if u — SC = 90° — t, 
sink = sinu cos 0, the formula for finding sin h in the right- 
angled spherical triangle $F6'.] 
For the azimuth co (arc BV — arc B'X'), the figure gives 
tan co = 
XG' S'F' 
B'F' 
O'F' 
O'F' 
-- tan t 
1 
G'O' ~ F'T' ~ O'F' F'T' sin 0 
See Zeuthen in Bibliotheca mathematica, i 3 , 1900, pp. 23-7.