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

190 From M drop a perpendicular ML on r ♎ so that SL will be perpendicular from S on r ♎ by the theorem of three perpendiculars. SL = H sin λ. Also ML will be equal to the perpendicular from S on the diameter of the diurnal circle of S parallel to r ♎ so that ML = SN = H sine of the arc in the diurnal circle corresponding to Kr ∴ SW² = SL² - LN² = H sin² λ - H sin² δ ∴ SN = √(H sin² λ - H sin² δ) ∴ The length of the perpendicular from K on r ♎ = R × √(H sin² λ - H sin² δ) / (H cos δ) since corresponding lines of the diurnal circle and the equator stand in the ratio of H cos δ : R (Vide fig. 20). But this ⊥ᵃʳ is H sin α = R √(H sin² γ - H sin² δ) / (H cos δ) . If α₁, α₂, α₃ be the Right ascensions of the points on the ecliptic whose modern longitudes are 30°, 60° and 90°, expressed in asus, then α₁, α₂ - α₁, α₃ - α₂ will give the rising times of the arcs of the ecliptic which stand for Sāyana Meṣa, Sāyana Vriṣabha and Sāyana Mithuna. The rising times of the next three Rasis will be the same in reverse order since Karkata is symmetric with Mithuna with respect to the Equator and similarly Simha and Kanya symmetric with Vriṣabha and Meṣa. The next three are again symmetric with Meṣa, Vriṣabha and Mithuna and the last three with Mithuna, Vriṣabha and Meṣa. Verse 56. The H cosines of the ends of the Rasis Karkata etc., being multiplied by the radius, and divided by the H cosines of their respective declinations, and the arcs of those H cosines being taken, subtract as before the preceding from the succeeding. Then we have the rising times of the Rasis beginning with Karkata.