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

271 Fig. 49 Comm. This too exhibits Bhāskara's genius. (Ref. fig. 49). Let MQR₂ be the equator whose pole is p. Let T₁ S₁ Z S₂ T₂ be the circle of azimuth a (Hindu azimuth). Let SS₁ S₂ be the diurnal circle of the Sun cutting the above circle of azimuth at S₁ and S₂, so that ZS₁ and ZS₂ are the two solutions giving the two zenith-distances which give two shadows in the given direction. H sin MS = Agrā ; evidently MẐS > MẐS₁ ie. H sin a < Agrā as stipulated. ZT₁ and ZT₂ give the zenith-distances in the given direction when the Sun is on the equator. S₁ T₁ and S₂ T₂ are the decrements in the zenith-distances on account of declination δ (= S₂ R₂ or S₁ R₁). If MẐS₁ were greater than MẐS ie. if H sin a > Agrā, we would have lost the position S₁ ie. we would have had only one shadow