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

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Here also assume H sin (Unnata) to be the Taddhṛti
as formerly done. Then as the shadow is 16″,
K̇ = √(16² + 12²) = 20″. Then H cos z = (12 R) / K = 12 / 20 × 3438
Unnatakāla = 8 nāḍīs = 48°.
∴ H sin (48°) = assumed Taddhṛti. Then from the
fourth latitudinal triangle, (H sin 48) / (H cos z) = k / 12
∴ Approximate value of k is (12 H sin 48) / (12/20 × 3438) = (20 H sin 48) / 3438 .
This is a known quantity from which s could be computed
since 12² + s² = k². Again from the fifth latitudinal
triangle s / k = (H sin δ) / (S. S.) .
Here s, k and S. S. (Samamandala-Sanku) are known
∴ H sin δ could be got approximately. Thus we have
found approximately the required quantities s and H sin δ.
From this H sin δ and s we have to compute again H cos δ,
Carajyā, Kujyā, etc. and applying the procedure of verse 54
[H sin (Unnata-cāra) × H cos δ] / R + Kujyā = Iṣṭa Hṛti; this
is nearer value of Iṣṭa Hṛti than the assumed Taddhṛti.
From this again obtain as before s and H sin δ; we could
not apply the proportion “ If by the assumed Taddhṛti we
have the previous H sin δ, what shall we have for this
computed Taddhṛti (Iṣṭa Hṛti) ” for the reason given
below. So Repeat the entire process till an invariable
quantity is got which will be the correct value of H sin δ.
Here repeating the entire calculation is correct and
not taking the proportion because Taddhṛti = (R sin δ) / (Sin ϕ cos ϕ)
where sin δ and sin ϕ are both to be computed.
Formerly we could take the proportion in verse 81
because the latitude of Ujjain being known, in the magni-