THEODOSIUS’S SPHAERICA 
351 
Now, the triangle NLO being right-angled at L, JS r O > NL. 
Measure NT along NO equal to NL, and join TB'. 
Then in the triangles B'NT, B'NL two sides B'N, NT are 
equal to two sides B'N, NL, and the included angles (both 
being right) are equal; therefore the triangles are equal in all 
respects, and Z NLB' = Z NTB'. 
Now 2R:2p = OC'-.C'K 
= ON: NL 
= ON: NT 
[ = tan NTB': tan NOB'] 
> ¿NTB': ¿NOB' 
> ¿NLB'-.¿NOB' 
> ¿COB-.¿NOB' 
> (arc BG): (arc B'C'). 
If a', b', c' are the sides of the spherical triangle A B'C', this 
result is equivalent (since the angle GOB subtended by the arc 
CB is equal to A) to 
1: sin b' = tan A : tan a' 
> a:a', 
where a = BG, the side opposite A in the triangle ABC. 
The proof is based oh the fact (proved in Euclid’s Optics 
and assumed as known by Aristarchus of Samos and Archi 
medes) that, if a., /5 are - angles such that > a > (3, 
tan (x / tan /3 > a//3. 
While, therefore, Theodosius proves the equivalent of the 
formula, applicable in the solution of a spherical triangle 
right-angled at G, that tan a = sin b tan A, he is unable, for 
want of trigonometry, to find the actual value of a/a', and 
can only find a limit for it. He is exactly in the same position 
as Aristarchus, who can only approximate to the values of the 
trigonometrical ratios which he needs, e.g. sin 1°, cos 1°, sin 3°, 
by bringing them within upper and lower limits with the aid 
of the inequalities 
tan a a. sin a 
tan /3 > (3 > sin /3 ’ 
where \ n > a > /3.