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

276 ∴ b = KA / R - s. But (H sin a) / b = R / S where S is the shadow so that S = bR / (H sin a) = (KA / R - s) R / (H sin a) = √(K² - 12²) ∴ (KA - sR) = H sin a √(K² - 12²) ie. K² A² + s² R² - 2 AsRK = H sin² a (K² - 12²) ∴ K² (A² - H sin² a) - 2 AsRK = - s² R² - 12² H sin² a ∴ K² (H sin² a - A²) + 2 AsRK = 12² H sin² a + s² R² I This quadratic is of the form ax² + 2bx = c ie. x² + 2 (b/a) x = c/a II Here 'c'/a is called Adya and b/a Anya. The solution of the above equation is given by x = - b/a ± √(b²/a² + c/a) ie. - Anya ± √(Anya² + Adya). III When b = KA / R + s, putting - s in the place of s in I, K = Anya ± √(Anya² + Adya) IV Out of the four solutions given by III and IV we have taking the positive solutions K = √(Anya² + Adya) ± Anya as stated. Verse 51. If H sin a < A, then in the northern hemisphere ie. where δ is north, ± √(Anya² - Adya) + Anya = K. Comm. We have initially put Anya = H sin² a - A². If H sin a < A, them to avoid a negative value for the