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

322 From (1) and (2) S. S. = 12 / (7/2) × Agrā = 24/7 × 875 = 3000 From (1) and (3) Taddhṛti = k / s × Agrā = 25/2 × 2/7 × 875 = 3125. Note. This is a beautiful example exhibiting Bhās- kara's dexterity in algebra. Verse 96. Given that the sum of H sin δ, S. S. and Taddhṛti – Kujyā = 1440, and the sum of Agrā, S. S. and Taddhṛti = 800, I shall deem him whoever finds s and the longitude of the Sun, as the very Sun illuminating the lotuses of astronomers. Verse 97. Answer to the problem above. The second sum divided by the first and multiplied by 12 gives k from which s could be got. Then the first sum divided by s + 12 + k̄ gives H sin δ from which the longitude of the Sun could be got. Comm. Comparing the third and the fifth latitudinal triangles, we have Agrā / Krāntijyā = S. S. / (Taddhṛti – Kujyā) = Taddhṛti / S. S. (1) (2) (3) = k / 12 = (Agtā + S. S. + Taddhṛti) / (Krāntijyā + Taddhṛti – Kujyā + S. S.) = 1800 / 1440 = 5 / 4 I (4) (5) (6) Equating (4) and (6) k = (12 × 5) / 4 = 15 ∴ k² = 225 = 12² + s² ∴ s = 9. Again comparing the fourth latitudinal triangle, with the fundamental,