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सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

Siddhanta Shiromani Ganitadhyaya of Bhaskaracharya with Commentary

भास्कराचार्य द्वितीय द्वारा

DevanagariHindipublished573 पृष्ठ

पृष्ठ 549, कुल 573 में से

संदर्भ में पढ़ें
पृष्ठ 549

529
when his longitude is equal to zero?" The result is
(Ap × 15) / (Δδ ± Δβ) = II
where rQ is looked upon as the longitude of the Moon at G
called the Goḷa Sandhi of the Moon.
But AP = the Sphuṭa Vikṣepa of the Moon when his
longitude is equal to zero where rA is the Asphuṭa Vikṣepa
at that point. Using his method of calculating AP from
Ar, Ap = (Ar × H cos δ) / R where
H sin δ = (H sin ω × H sin (90 + λ)) / R .
Here AP pertains to the longitude λ equal to zero, so
that H sin δ = (H sin 90° × H sin ω) / R = H sin ω.
∴ H cos δ = H cos ω = H cos 24° = 110 when R = 120'
Hence Ap = (Ar × 110) / 120 = (11 / 12) Ar.
But Ar is the celestial latitude of the Moon to be
calculated from RA taken to be λ
∴ rA = (H sin λ × 270) / 120 = (9 / 4) H sin λ (where 270' = 4½° = i)
taking R = 120 ∴ Ap = (9 / 4) H sin λ × (11 / 12). Hence
rQ from I = (9 / 4) H sin λ × (11 / 12) × 15 / (Δδ ± Δβ)
But (H sin λ / 4) × (135 × 11) / 12 = (123¾ / 4) H sin λ. This has to
be divided by Δδ ± Δβ. In the problem given. λ = 100°
and Δδ is taken as 362. To obtain Δβ Bhāskara adduces
the argument "If at λ = 0, the first Śarakhaṇḍa of 70
corresponds to H cos λ = 120' what amount of Śara-
khaṇḍa corresponds to an arbitray H cos λ?" The result
is (H cos λ × 70) / 120 = (7 / 12) H cos λ. This he calls Kotiphala
because it is based upon H cos λ which is the Kotijyā.
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