ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

TIME OF RISING 129
Greek Method
In the same figure¹ let PSC be the triangle and γMQ be the transversal. Then Menelaus's theorem gives
(sin PM / sin MS) × (sin Sγ / sin γC) × (sin CQ / sin QP) = 1
or (1 / sin δ) × (sin l / 1) + (sin ω / 1) = 1
or sin δ = sin l × sin ω.
Indian Method
Again by the Indian method from the same two similar triangles we get
mn : nS = OK : OC
or, mn : R sin l = R cos ω : R
∴ mn = (R sin l × R cos ω) / R
Again MN : mn = OM : Om
i.e., R sin R. A. : mn = R : R cos δ
∴ R sin R.A. = (R sin l × R cos ω) / (R cos δ)
Greek Method
Take PQM for the triangle and γSC for the transversal.
Then,
(sin PC / sin CQ) × (sin Qγ / sin γM) × (sin MS / sin SP) = 1
or (cos ω / sin ω) × (1 / sin R.A) × (sin δ / cos δ) = 1
or sin R.A. = tan δ cot ω,
The Indian form of the equation is different from that of Ptolemy's. It is also better for the purpose of calculation.
Note :—From the same two similar triangles we have
On : ON = R cos δ ; R
∴ On : R cos l = (R cos R.A. × R cos δ) / R ......(3)
Again, tan R.A. = mn / on
= (R sin l × R cos ω) / (R × R cos l) ......(4)
Again, mn : Sm = OK : KC

1. Manitius' Edition of Syntaxis, I, 51-53.