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

527
[Fig. 126]
Fig. 126
Now refer to fig. 126. Let NS be the ecliptic and
NM the celestial equator. The quadrant NS equal to 90°
is divided into six equal parts at S₁, S₂ etc. The successive
declinations of S₁, S₂ etc. are given by Brahmagupta as
362, 703, 1002, 1238, 1388, 1440 where the radius is given
to be 3438'. In other words 1440 = H Sin ω = H Sin 24°.
These can be easily verified from the formula
H Sin δ = (H sin λ × H sin ω) / R
putting λ = 15°, 30° etc. and R = 3438'.
[Fig. 127]
Fig. 127
Now refer to fig. 127. R M is the lunar orbit, R being
Rāhu. RL is the ecliptic. Using the formula.