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

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. 67