A history of Greek mathematics - Wilbourhall.org

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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
sin PH'
crd. 2 BC '
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.