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

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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,