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

316 (Sūtra × H cos δ) / R = Kalā = (H cos h . H cos δ) / R = ∴ (H cos h . √(R² — x²)) / R But Kalā is the Koti of the fifth latitudinal triangle of which H sin δ is Bhuja. Hence (Kalā × H sin ϕ) / (H cos ϕ) = H sin δ = x ie. (H cos h √(R² — x²)) / R × (H sin ϕ) / (H cos ϕ) = x; but (H sin ϕ) / (H cos ϕ) = s / 12 ∴ Squaring both sides [(R² — H sin² h) (R² — x²)] / R² × s² / 12² = x² ∴ 12² R² x² = s² R² (R² — H sin² h) — s² x² (R² — H sin² h) ie. x² {(12² R² + s² (R² — H sin² h)} = s² R² (R² — H sin² h) ∴ x² = [s² R² (R² — H sin² h)] / [12² R² + s² (R² — H sin² h)] = R² / [(12² R²) / (s² (R² — H sin² h)) + 1] ∴ x = R / √[(12² R²) / (s² (R² — H sin² h)) + 1] = H sin δ as given. From H sin δ, the method of obtaining λ is clear from the formula (H sin λ H sin ω) / R = H sin δ. In the given numerical example h = 5 nādīs = 360° / 12 = 30° since 60 nadis of time correspond to 360°. Thus H sin h = R / 2 ; the remaining work follows. Verse 86. Another question. When the Sun is on the prime-vertical the gnomonic shadow is noted to be 16 inches. The Unnatakāla is 8 nādis. If you could give the H sin δ and s, I shall con- sider you nothing short of one who is an adept in solving the totality of the diurnal problems. Verses 87 and 88. Answer to the question.