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

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∴ x² (9 R² + 16 Carajyā²) = 16 R² Carajyā²
∴ x² = (16 R² Carajyā²) / (9 R² + 16 Carajyā²) ∴ x = (4 R Carajyā) / √(9 R² + 16 Carajyā²)
= (12 Carajyā) / √(81 + (12² Carajyā² / R²)) = (12 Carajyā) / √(9² + (12 Carajyā / R)²)
Here Carajyā being known, H sin δ could be computed.
Verse 100. If you studied what is known as Madh-
yamāharaṇa, then compute λ the longitude of the Sun
given that H sin δ + H cos δ + H sin λ = 5000.
Verse 101. Answer to the problem above.
Let the given sum multiplied by 4 and divided by 15
be Ādya ; then H sin δ =
Ādya − √(910678 − (2 square of the given sum / 337)) .
Comm. Let H sin δ = x ; then H cos δ = √(R² − x²)
and since H sin δ = (H sin ω H sin λ) / R
∴ H sin λ = (x R) / (H sin ω) = (x R) / 1397
∴ The given sum = x + √(R² − x²) + (x R / 1397) = 5000
∴ √(R² − x²) = 5000 − x (1 + R / 1397) = 5000 − (4835 / 1397) x
∴ R² − x² = 5000² + x² (4835 / 1397)² − (2 × 5000 × 4835 / 1397) x
∴ x² {1 + 4835² / 1397²} − (2 × 5000 × 4835 / 1397) = R² − 5000²