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

531 ∴ log tan rA = log cos 10 + log tan 4½° = 9·9934 + 8·8960 = 8·8894 ∴ rA = 4°-26ʹ. From the spherical triangle rAG, cos 90—ω = tan rA / tau Gr log tan Gr = log tan rA — log sin ω = 8·8894 — 9·6093 = 9·2801. Again from the spherical triangle rQG, cos ω = tan rQ / tan Gr ∴ log tan rQ = log tan Gr + log cos ω = 9·2801 + 9·9697 = 9·2498 ∴ rQ = 10°-5ʹ whereas Bhāskara got 10°-22ʹ-28″. This shows how Bhāskara was correct mathematically. The small difference there is, due to his taking Hsines for arcs of 15° instead of for smaller arcs. Note. In the commentary called Sikhā of one Sri Kedāra Datta Joṣī (page 357) we find a mistake committed namely that he subtracted 3 R-10° the longitude of the pāta which is in Rāśis and degrees from the Sphuṭakrānti 349ʹ-45″. What Bhāskara meant was, that since the longitude 100° exceeds 90°, the cosine will be negative which therefore entails △β to be subtracted from △δ. Verse 7. How to know the occurrence of a pāta. If the Sphuṭakrānti of the Moon, when it is maximum be less than that of the Sun, then there could be no occasion for their declinations to become equal in the near future. Comm. The situation in which the maximum decli- nation of the Moon falls short of that of the Sun, arises