पञ्चसिद्धान्तिका (आचार्य वराहमिहिर - सूर्य, रोमक, पौलिश, वासिष्ठ एवं पितामह सिद्धान्त)

94 PAÑCASIDDHĀNTIKĀ IV. 21 Therefore the latitude of the place = NNP = ZO. (What we have said is for places in the northern hemisphere, i.e. north latitudes. In the southern hemisphere, i.e. at places of south latitudes, SP is lifted up from S, and the celestial equator is depressed northward by the same amount.) The complement of ZO, OS, is called the co-latitude (Lamba). Thus in triangles formed by great-circle- arcs of the stellar sphere and the sky-sphere, the latitude is involved directly or indirectly. The for- mulae for the solution of these triangles have been already given. Now, the two formulae for latitude can be proved by using the meridian, thus: see Fig.7 Ob:observer N: north point S: south point NS₂ZS₁OS₀S₃: The meridian Z: zenith O: point of intersection of meri- dian and celestial equator. S = S₂, S₁, S₀, S₃,: four positions of the mid-day sun. OS: Sun's declination Fig. IV. 7 As already described, OZ = latitude. On the equinoctial days at mid-day the Sun, S₀ is at O. ∴ ZS₀ (the south zenith distance of the Sun) = ZO = latitude (first formula). On other days, the Sun may be (i) south of O, (S₃), or (ii) north of O but south of Z, (S₁), (iii) north of O and north of Z, (S₂). i. Here, the latitude = OZ = S₃Z − S₃O = the south zenith distance of the Sun − the declina- tion. (second part of second formula). ii. Here the latitude = OZ = ZS₁ + S₁O = the south zenith distance of the sun + the declination (first part of the second formula). iii. Here the latitude = ZO = S₂O − S₂Z = the declination − the north zenith distance of the Sun. (This case is not given by the author). The zenith distance of the midday Sun used in the formulae is to be found thus: see Fig.8. EG = gnomon of 12 units ET = The midday shadow, TG = The shadow hypotenuse, ZGS = the zenith distance = angle TGE. Sin zenith distance (ZD) = sin ZGS = sin TGE = TE × 120′ ÷ TG = Shadow × 120′ ÷ Shadow hypotenuse = Shadow × 120′ ÷ √(shadow² + gnomon²) = Shadow × 120′ ÷ √(shadow² + 144)., (where shadow is in the units taken). From sin ZD, arc ZD is found. Fig. IV. 8