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

311 Comm. The data are H sin δ, K and b; since K and b are given a is known. Thus from the triangle PZS (fig. 61) we have sin δ = sin ϕ cos z + cos ϕ sin z sin a, all quantities except ϕ are known. Solving this trigonometrical equation which is of the form a cos ϕ

• b sin ϕ = c, we can have ϕ. Fig. 61 We shall now see how it is solved by Bhāskara. Let s be the equinoctial shadow. Then a = b + s = KA / R = (KH sin δ) / (H cos ϕ) I But 12 / K = 12 / √(s² + 12²) = (H cos ϕ) / R so that H cos ϕ = 12 R / √(s² + 12²). Substituting in I b + s = (KH sin δ √(12² + s²)) / 12 R which reduces to 12² R² (b + s)² = K² H sin² δ (12² + s²) ie. s² (12² R² − K² H sin² δ) + 2b 12² R² s = 12² {(H sin² δ) K² − b² R²} ie. s² (12² − (K² H sin² δ) / R²) + 2. 12² b. s) = 12² ((K² H sin² δ) / R² − b²) Here (K² H sin² δ) / R² is symbolized as L 12² ((K² H sin² δ) / R² − b²) put as Ādya and 12² b is put as para.