ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)
Brahmasphuta Siddhanta of Brahmagupta with Commentary
आचार्य ब्रह्मगुप्त द्वारा
पृष्ठ 180, कुल 737 में से
संदर्भ में पढ़ें140 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 sin MN' = sin CH / sin AC.
Here Ptolemy's equation is simpler than that of Āryabhaṭa;
hence they must be independent of each other.
[Figure 12] Problem VII:—
To find the Angle made by the
Vertical through any Point of the
Ecliptic with the Latter
This problem is considered
by Ptolemy but it is not consider-
ed separately in Indian Astronomy,
but from the rule for parallax in
longitude, the rule for its calcula-
Fig. 12 tion can be deduced.
Indian Method :
In Fig. 12 S represents the true position of the
Sun and S' the Sun's position as depressed by parallax. N'SA is
the ecliptic. If from S', S'Q be drawn perpendicular to the ecli-
ptic, then, if P is the horizontal parallax,
- Grahaganita; XII, 3-4.
- Manitius, ibid, pp. 110-111.