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

233 We shall now prove it in modern terms from the spherical triangle PZS. We have the formula sin δ = sin ϕ cos z + cos ϕ sin z sin a where PZS = 90 — a, a being the Hindu azimuth measured from the East point. Multiply the above equation by K and divide throughout by cos ϕ where K is called the Chāyākarṇa equal to √(12² + S²), S being the gnomonic shadow at the moment. Hence K sin δ / cos ϕ = K cos z tan ϕ + K sin z sin a (1) Fig. 31 Fig. 32 But from Fig. 31, K cos z = 12, K sin z = S, (2) and from Fig. 32, S sin a = b where b is called the Chāyābhuja ie. Chāyābhuja = K sin z sin a (3). Thus K sin δ / cos ϕ = 12 tan ϕ + b. But again from Fig. 33, when ☉ the Sun at vernal equinox is on the meridian and as such has a meridian zenith-distance equal to ϕ, 12 tan ϕ = s where s is called the Vishuvat-chāyā or equinoctial shadow. Thus we have K sin δ / cos ϕ = s + b (4). Again if the Sun be on the horizon, from Fig. 34, E ☉ is called the Agrā, A, so that from the triangle P ☉ N, 30