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IV. 56 IV. THREE PROBLEMS 135 (iv) As ‘the sine’ is north, the difference is to be taken. As ‘the sine’ is greater, Agrā = 66′ 30″ − 36′ = 30′ 30″, (and the Sun is in the southern hemisphere). (v) Sine latitude of Sun = 30′ 30″ × 96 × 5 ÷ 244 = 60′. (vi) The degrees of Bhuja = rāśi 1-0. As the Sun is in the southern hemisphere, the Sun is rāśi 6-0

• rāśi 1-0, i.e. rāśi 7, or rāśi 12-0 − rāśi 1-0, i.e. rāśi 11. As it is Uttarāyaṇa, the Sun’s longitude must be rā. 11. Applying the simplified method, since the lower sign is to taken as the distance is north, sin long = (12 × 72 ∼ 96 × 16-37.5) × 150 ÷ (61 × 30). Here since distance × sin colat is greater, Sin long = (96 × 16-37.5 − 12 × 72) × 150 ÷ (61 × 30) = 60′, and the Sun must be in the south- ern hemisphere. The rest of the work is the same. The proof of the above rules is as follows: In the previous work, the ‘Perpendicular’, i.e. the dis- tance of the tip of the shadow from the east-west line, was calculated, given the Sun and the shadow, and from that the ‘Base’ and the direction were calculated. Here, given the distance and the ‘Perpendicular’, the Sun is computed. Therefore this is the converse of the previous work, and can be derived from that. Steps (i) and (ii) are the same as steps (i) and (ii) of the previous work, and have been derived there. We shall therefore derive (iii), (iv) and (v) from (iii), (iv) and (v) there. In the previous work in (iv), ‘Perpendicular’ = (Sūryāgrā ∓ Agrā) × shadow hypotenuse ÷ 120. ∴ ‘The sine = (Sūryāgrā ∓ Agrā) = Perpendicular × 120 ÷ Shadow hypotenuse, as in (iii) here. Since, ‘the sine’ = (Sūryāgrā ∓ Agrā), when the Sun is in the northern and southern hemispheres, respectively, Agrā = Sūryāgrā ∼ ‘the sine’. It has been mentioned that Sūryāgrā is always north, ‘the sine’ is either south or north according to the line to the tip of the shadow from the east-west line, and Agrā is south if the Sun is in the northern hemisphere and vice versa. Therefore, when Agrā is north, (i.e. when the Sun in the southern hemisphere,) ‘sine’ is north, and greater than Sūryāgrā. Therefore, in using (‘the sine’ − Sūryāgrā), we get that the Sun is in the southern hemisphere. If Agrā is south, and therefore to be got negative by the addition of Sūryāgrā, (i.e. when the Sun is in the northern hemisphere), and ‘the sine’ is north, Sūryāgrā is greater than ‘the sine’. Here we have to use (Sūryāgrā − ‘the sine’), and we get that when the Sun is in the northern hemisphere. If Agrā is south again, (i.e. the Sun is in the northern hemisphere, again), and ‘the sine’ is also south, then we have the case, Agrā = Sūryāgrā + ‘the sine’, in which case also the Sun is in the northern hemisphere. From the Agrā, the sine of Sun’s longitude is got thus: In step (iii) of the previous work, Agrā = Maximum declination × sine longitude of the Sun ÷ sin colatitude. ∴ sin long. of the Sun = Agrā × sine colatitude ÷ max. dec. = Agrā sin colat ÷ 48′ 48″, as we get here in step (v). The explanation of getting the Sun’s longitude from its sine has already been given in connection with getting the sines for degrees (IV.1-15). Another point to be noted in this connection is this: In what the author gives, there is nothing to say about the addition of ‘the sine’ and Sūryāgrā when they are of different directions, and therefore