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Published online by Cambridge University Press: 12 April 2016
In Fig.1, let U be the rising point of the ecliptic (udayalagna), T be the nonagesimal (tribhona-lagna) and M be the meridianecliptic point (madhya-lagna). Since T is at a distance of one quadrant from U along the ecliptic, the complement of the zenith distance of, that is the arc TK, will be the required angle between the ecliptic UTM and the horizon NUEKS. The equivalent problem in Indian astronomy is therefore to find what is called ‘dṛkkṣepa-jyā’ or the sine of the zenith distance of the nonagesimal.