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

266 H sin SR = (H sin ST × H sin T̂) / R II and H sin φ = (H sin ZT × H sin T̂) / R III so that (H sin SR) / (H sin φ) = (H sin ST) / (H sin ZT) ∴ H sin ST = (H sin SR / H sin φ) × H sin ZT. Noting that SR = δ and putting ST = D H sin D = (H sin δ / H sin ϕ) × H sin ZT. [Diagrams: Fig. 46 and Fig. 47 showing spherical triangles with vertices P, Q, N, R, S, T, D, δ] Fig. 46 Fig. 47 The same formulae are derivable from fig. 47 also; only in fig. 46 while there is a decrement in the shadow of the day as compared with the shadow on the equinoctial day, in fig. 47, there is an increment. This is seen from the decrease and increase of ST in the zenith-distance ZT of the equinoctial day in the given direction. Now corre- lating fig. 45 with figures 46 and 47, the shadow gL pertains to the zenith-distance ZT on the equinoctial day whereas the shadows gN pertains to the zenith-distance on the day concerned in the same direction. We have, S / √(12² + S²) = (H sin z) / R so that RS / √(12² + S²) = H sin z where S is the shadow at any instant when the zenith-distance is z. The process indicated by saying ‘Obtain H sin φ