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.