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IV. 35 IV. THREE PROBLEMS 111 Great gnomon (Sama-śaṅku) and its shadow 35. When the Sun is in the northern hemisphere, (i.e. in the six signs, Aries etc.), multiply the sine of the longitude of the Sun by the sine of the maximum declination, (i.e. by 48′ 48″), and divide by the sine of latitude. The minutes so obtained are called the minutes of the ‘Great gnomon’ or Śaṅku, (i.e. sine of altitude), (and in this case, the sine of Prime vertical altitude). From this the shadow of the Sun on the prime vertical must be calculated. (i) Sin prime vertical altitude = sin Sun’s long × 48′ 48″ ÷ sin latitude. This is the Great gnomon, and the radius is the Great hypotenuse. The square root of the square of the hypotenuse lessened by the square of the gnomon is the shadow. Therefore the Great shadow = √radius² – sin ² prime vertical alt. Therefore, by the similarity between the Great shadow and the shadow triangles, we have the proportion, Great gnomon: Great shadow :: Twelve unit gnomon: shadow. From this, the required, (ii) Shadow = 12 × √120² – sin² prime vertical alt. ÷ sin prime vertical altitude. Example 13. The longitude of the Sun is rāśi 1-0. The latitude is 30°. Find the Great gnomon of the Sun at prime vertical, and thereby the gnomonic shadow at that time. (i) The Great gnomon = sin prime vertical altitude = Sin Sun’s longitude × 48′ 48″ ÷ sin latitude = 60′ × 48′ 48″ ÷ 60′ = 48′ 48″. (ii) Shadow = 12 × √120² – 48′ 48″² ÷ 48′ 48″ = 12 × 109′ 38″ ÷ 48′ 48″ = 12 × 109 19/30 ÷ (61/150) = 1644 – 30 ÷ 61 = 26 units and 58 parts, aṅgulas and vyaṅgulas The equation (i) can be written as, Sin prime vertical alt. = sin Sun’s long. × sin max. dec. ÷ sin lat. = sin Sun’s long. × sin max. dec × radius ÷ (sin lat × radius) [Fig. IV. 13] = (sin Sun’s long. × sin max. dec ÷ radius) × (radius × sin lat.) Here, it can be shown that sin Sun’s long × sin max. dec ÷ radius = sin dec., thus: Sin Sun’s long. × sin max. dec ÷ radius = sin Sun’s long. × 48′ 48″ ÷ 120′ = sin Sun’s long. × 61/150 = sin dc. (by IV. 16). Or, from Fig. 13, thus: In the triangle right-angled at R, rS is the Sun’s long. and SR is the declination of the Sun. SrR is the maximum declination. By fundamental formula II, sin rS × sin SrR ÷ radius = sin SR. ∴ sin Sun’s long × sin max. dec ÷ radius = sin dec. 35. Quoted by Utpala on BS 2, p.42 35b. A. काष्ठान्तरगुणा d. A. मण्डलछाया; U. मण्डले छाया