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

181 Taking the equinoctial shadow equal to 1 Angula means tan ϕ = 1/12. As charajyā is proportional to tan ϕ, for any equinoctial shadow of s angulas, tan ϕ being equal to s/12 the charajya got above is to be multiplied by S only to obtain the charajyā in any place where the equi- noctial shadow is s angulas. Putting the modern longitu- des equal to 30° & 60°, if the corresponding declinations be δ₁, δ₂ sin δ₁ = sin 30 sin ω, sin δ₂ = sin 60 sin ω. Tak- ing ω = 24° and applying logarithmic tables log sin δ₁ = 9.6990 + 9.6093 = 9.3083 so that δ₁=11°–44′ log sin δ₂ = 9.9375 + 9.6093 = 9.5468 so that δ₂=20°–38′ Now from the formula for charajya cited above viz. H sine (chara) = R tan δ tan ϕ or sine (chara) = tan ϕ tan δ, putting tan ϕ = 1/12 and applying tables, using the values of δ got above, charajya for 30° = (tan 11°–44′) / 12 and charajyā for 60° = (tan 20°–38′) / 12 so that log (sine chara) = 9.3175 – 1.0792 for 30° and for 60° log (sine chara) = 9.5758 – 1.0792 ∴ Chara for 30° or C (30)° = 59′ and C (60°) = 1° – 48′ Converting these arcs into their rising times at the rate of 6′ per Vinadi, we have C (30°) = 10, and C (60°) = 18 Noting δ₃ = ω, sin (chara) for 90° = (tan 24°) / 12 so that log sin (C 90°) = 9.6486 – 1.0792 = 8.5694 so that C (90°) = 2°–8′ = 21⅓ Vinādis. Thus (C 30̄) = 10, C (60) – C (30) = 18 – 10 = 8 C (90°) – C (60°) = 21⅓ – 18 = 3⅓ so that the chara Segments are respectively 10, 8, 3⅓ as given by Bhāskara.