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

321
Verse 95. Answer to the question above.
Divide 12 × Second sum by the first sum, that will
be s. Again (12 × Second sum) / (12 + s + k) = H sin δ. From H sin δ, λ
could be had as before.
Comm. Comparing the third and fifth latitudinal
triangles
Kujyā / Krāntijyā = Krāntijyā / (Taddhṛti — Kujyā) = Agrā / S. S. = s / 12
(Kujyā + Krāntijyā + Agrā) / (Krāntijyā + Taddhṛti + S. S. — Kujyā) = 1960 / 6720 = 7 / 24
∴ s = 7/24 × 12 = 7/2 = 3½″.
Again comparing the third and the first latitudinal
triangles
s / Kujyā = 12 / Krāntijyā = k / Agrā = (s + 12 + k) / (Kujyā + Agrā + Krāntijyā)
(1) (2) (3) (4)
= (s + 12 + k) / 1960
(5)
Equating (2) and (5) Krāntijyā = (12 × 1960) / (12 + s + k)
= (12 × 1960) / (7/2 + 24/2 + 25/2) = (12 × 1960) / 28 = 840 since when s = 7/2
k = 25/2 which is the hypotenuse of the triangle formed
by the equinoctial shadow with the gnomon.
Equating (1) and (5) Kujyā = 245 ; equating (3) and
(5) Agrā = 875.
Now from the fourth latitudinal triangle compared
with the first Agrā / s = S. S. / 12 = Taddhṛti / k
(1) (2) (3)
41