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

457 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 58