A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

PTOLEMY'S SYNTAXIS 285
not quote those propositions, as he might have done, but proves
them afresh by means of Menelaus's theorem. 1 The application
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.
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. 14, 16.)
By Menelaus's theorem applied to the arcs AE', E'P cut by
the arcs A H', PC which also intersect one another in B.
that is,
crd.2PH' crd.2PB crd. 2CA
crd. 2 H'E f
crd. 2 BC '
sin PH'
sin PB
cvd. 2 AE'
sin CA
sin H'E' ~~ sin BC sin AE'
Now sin PH' = cos H'E', sinPB^cosBC, and smAE'=l;
therefore cot H'E'= cot BC . sin CA
in other words, in the triangle ABC right-angled at C,
cot A — cot a sin b,
or tana = sin b tan A.
1
Syntaxis, vol. i, p. 169 and pp. 126-7 respectively.
2 A, vol. i, pp. 121-2.
286 TRIGONOMETRY
Thus AC is found, and therefore FC-AC or FA.
The lengths of BG, FG 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
PBCP' through the poles, and the great circle PKLP'
passing through the poles of both the ecliptic and the equator.
f
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 OP sin BF
sin KL ~ sirTBG '
sin FK'
But sinPZ=l, sin KL = sin BFG, sinCP=l, sinPZ = l,
and it follows that
sin BG= sin BF sin BFC,
corresponding to the formula for a triangle right-angled at C,
(2) We have
sin a = sin c sin A.
sin PK sin PB sin GF
sin KL sin BG sin FL
'
and sin PK = cos KL = cos BFC, sin PB = cos BG, sin FL = 1
so that
corresponding to the formula
tan BG = sin GF tan BFG,
tan a = sin b 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 Analcmma.
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