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

388 Fig. 84 will be equal to (b × B) / 225 × (H sin ω) / R where B is the Bhogya- khanda of λ. To obtain the value of the above for a circle of radius R from a circle of radius b, we have to multiply by R/b. So, the result is (b × B) / 225 × (H sin ω) / R × R / b = (B H sin ω) / 225 . But the value of B is got as follows. 'If for H cos λ equal to R we have the first Bhogyakhanda equal to 225, what shall we have for H cos λ?' The result is (225 × H cos λ) / R . Substituting for B, we have (H sin ω) / 225 × (225 × H cos λ) / R = (H sin ω × H cos λ) / R Now, on account of declination, the Sun's disc is inclined like an umbrella. So LM of fig. 84 will take a position like L'M as shown in fig. 85 where the triangle MLL' is similar to SMO, S being the centre of the Sun's disc, O the centre of the sphere. Hence (L'M) / LM = R / (H cos δ) ∴ L'M = R / (H cos δ) × (H sin ω H cos λ) / R = (H sin ω × H cos λ) / (H cos δ) as got before '.