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.