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

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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,
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