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

IV. 54 IV. THREE PROBLEMS 127 (iv) (Sūryāgrā ∓ Agrā × shadow hyp. ÷ 120 = ‘Perpendicular’ (of ∓, the upper sign is for north declination, and the lower for south. If the ‘Perpendicular’ got is positive then it is north, if negative, south.) (v) √(shadow² − Perpendicular²) = Base Here, steps (ii), (iii) and (iv) can be simplified and put in the form: ‘Perpendicular’ = (12 × sine latitude ∓ shadow hypotenuse × sin declination) ÷ sin colat. (of ∓, the upper sign is for north decli- nation and the lower for south. As already said, the Perpendicular obtained is north if positive and south if negative. If, when the declination is north. Shadow hypotenuse × sin declination > 12 × sin latitude, then deduct the less from the greater and take it as negative, i.e. take the resulting Perpendicular as south.) C |
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| \ Shadow | \ 5-0 Perp. 2-37 |
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| \ Shadow angle A | 90° \ B +---------) 4-16 Base Fig. IV. 15 Example 22. The latitude of a place is 30°, whence sin lat = 60', and sin colat = 103' 55". The Sun at the time of taking the shadow = rāśi 1-15, whence sin Sun’s long = 84' 51", sin declination = 48' 48" × 84' 51", ÷ 120 = 34' 30", (north, as the Sun is in the first 6 signs). For this place and time if the shadow is 5 digits, find the direction of the shadow. (i) shadow hypotenuse = √(5² + 144) = 13. (ii) Sūryāgrā = 12 × 120 × 60 ÷ (13 × 103' 55") = 63' 57".2 (iii) Agrā = 48' 48" × 84' 51" ÷ 103' 55" = 34' 30" × 120 ÷ 103' 55" = 39' 51".4 (iv) ‘Perpendicular’ = (63' 57" − 39' 51") × 13 ÷ 120 = aṅg. 2-36.6 (The ‘Perpendicular’ is north, as the result is positive) (v) The ‘Base’ = √(5² − (2 − 36.6)²) = aṅg. 4-16. Or, using the simplified form, the Shadow-hypotenuse, 13 aṅg, being known, ‘Perpendicular’ = (12 × 60' − 13 × 34' 30") ÷ 103' 55" = 271' 30" ÷ 103' 55" = aṅg. 2-36.6. Then the ‘Base’ is calcu- lated as done above. Using the ‘Base’ and the ‘Perpendicular’, the direction of the shadow is found thus graphically. (see Fig. 15).