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

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is the centre of the sphere and the Hsine of KP is KN so
that it is the Āyanavalanajyā. Hence
ON² = OK² — KN² ∴ ON = Yaṣṭi = √R² — Āyanavalanajyā²
= Āyanavalanakoṭijyā. From the similarity of the
triangles OKN and GRF,
OK / ON = GR / GF ∴ GF = (GR × ON) / OK = (H sin MR × Yaṣṭi) / R
GR could be taken equal to MR, the latter being
small and GF could be taken to be equal to GR′ ie.
H sin MR′ and so equal to MR′.
∴ MR′ = (MR × Yaṣṭi) / R
Adding MR′ to the declination of M, we get the
modern declination of R′ ie. the Sphuṭakrānti of R.
Note (5) In fact the spherical triangle MKP is
just like the spherical triangle r E ♋ (fig.). In the place
of the paramakrānti E ♋, we have the Āyanavalana KP
(fig. 108), and in the place of Dyujyā ♋ L (fig. 19) we have
Yaṣṭi.
Note (6) An alternative is given in the verse for
this namely MR′ = (MR × H cos δ′) / R where δ′ is the decli-
nation of a point whose longitude is 90+λ. In other
words we have to prove that H cos v = H cos δ′ or v = δ′
(of a point whose longitude is 90+λ. Since declination
is given by the formula H sin δ = (H sin λ × H sin ω) / R
Putting 90+λ for λ, H sin δ′ = (H cos λ × H sin ω) / R I
But from triangle MKP, sin v / sin ω = sin (90 — λ) / sin (90 — δ) = cos λ / cos δ
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