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.