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

140 GREEK AND INDIAN METHODS
point A. Then he subtracts 90° from the longitude of A. Thus
having the longitude of N', he next finds the part of the day
elapsed of N' ; from which by the time-altitude equation discussed
above, he finds ZN'. This is of course more accurate than that of
Āryabhaṭa. Bhāskara¹ here follows Brahmagupta.
Greek Method :
Let the ecliptic CN'A cut the lower half of the meridian at
F. Ptolemy takes AK along the ecliptic = 90° and AR along the
horizon = 90°; then the great circle passing through R and K passes
through the nadir Z'. Now take Z'FK for the triangle and ANR
for the transversal, then by Menelaus's theorem.²
(sin FN / sin NZ') × (sin Z'R / sin RK) × (sin KA / sin AF) = 1
∴ sin RK = (sin FN / sin AF) = (cos FZ' / sin AC) = (cos CZ / sin AC) = (sin CH / sin AC)
or