सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

306 positive. In other words in the equation a = b + s, a is positive when δ is north, b is +ve when the azimuth of the Sun is north of the East point, and s is always positive. Since in north latitudes, Śaṅkutala will be always south of the Udayāstasūtra and considered positive, the E.S.L. will be on the north of the East-west line so that s is considered positive. We should have had to consider s negative in southern latitudes, as per the above convention but as the Hindu astronomers did not have to concern themselves with south latitudes, the question of sign for s did not arise except taking it as always posi- tive. Hence the equation a = b + s will hold universally with the same conventions of sign which we stipulated with respect to the equation A = B + S. The foregoing analysis is on the modern lines. [चित्र: वृत्त के अंतर्गत W-E व्यास, E·S·L समानांतर रेखा, तथा बिन्दु O, C, P, R, M, L, A, N, B, K युक्त ज्यामितीय आरेख] Fig. 60

307 Now let us see how the convention of signs is stipulated in Hindu astronomy with respect to the equation a = b + s. We have said that ‘s’ is always north of the East-west line and considered positive. Regarding ‘a’, it is said by Bhāskara व्यस्तगोला which means that when δ is north and the Sun is said to be in northern hemisphere, a is said to belong to the southern hemisphere. Also when the Sun is on the north of the prime-vertical and Śankubhuja is considered north, the Chāyābhuja being south of the East-west line is considered south. With these conventions of directions (we say of directions, and not signs because the Hindu astronomers do not speak of signs but only of directions) it is stipulated in Hindu astronomy that quantities of like directions are to be added, otherwise their difference is to be taken ‘तुल्यदिशोर्योगः, भिन्नदिशोः अन्तरम्’. This convention stipul- ating addition or difference is technically called ‘संस्कार, Samskāra’. That is why it is said simply “पलच्छायया सौम्यया संस्कृता” ie. ‘Samskāra (on the aforesaid lines) is to be effected between a and s to get bhuja b’. Here it may be reiterated that the word ‘अन्तरम्’ ie. ‘difference’ is used in its restricted sense namely that the positive difference alone is to be taken. Thus the ‘antaram’ of 8 and 5 is 3 as well as that of 5 and 8 is also 3. Then it might be asked how to decide the direction of the bhuja, if we were to take s as equal to a ~ b and not a-b or b-a. The answer is that between a and s whose directions are known as per the aforesaid convention, equating their difference ie. a ~ s to b, we have to take b as having that direction which is indicated by that quantity either a or s which has a larger numerical value. Thus while on modern lines we take a = b + s to hold universally with the convention of signs which we have agreed to on modern lines namely that a is +ve if δ is north and b is positive if the Hindu azimuth is north of the East point and s is always +ve, we take on the Hindu lines a ± s = b with the conventions stipulated with respect

308 to directions (not of signs) namely that a is south if δ is north, s is always north and b is to be taken to belong to that direction to which the numerically bigger quantity of a and s belongs in the case of difference. Also it is to be taken to belong to that direction of a and s when both of them have the same direction. With this convention in mind, Bhāskara clarifies the convention by citing examples. (1) S=5, δ is north, Agrā=916'-48'' K=30 so that a = KA / R = (916⅘ × 30) / 3438 = 8 units (south, because it should be taken to be व्यस्तगोला ie. belonging to the direction opposite to that of δ). Question. "What is b and of what direction?" Answer. b = a +~ s = 8 (south) ~ 5 (north). (We are taking the difference because Samskāra is to be construed as addition of quantities having the same direction and difference of quantities of opposite direction ∴ b = 3 and is on the south because the numerically bigger quantity of 8 and 5 belongs to south. Q. 2. δ is north S=5, A=916'-48'' ; K=15, 'what is b and in what direction?' Answer. a = KA / R = 4; b = a +~ s = 4 +~ 5 = 4~5 (here a is south δ being north and s is north so that difference is stipulated as above) = 1 (north because 5 belongs to north. We add here two more examples to illustrate addition by saying that δ is south in the above examples so that a is north in both the examples. Hence in (1) b=8+5=13 (north) and in (2) b=4+5=9 (north). Refering to Fig. 60, we see there three cases depicted namely the extremi-

309 ties of the shadows being A, B and C. In the first case AL = Karnāgra = a (Karnāgra is the distance of the extremity of the shadow from the E.S.L. namely K.L.R. whereas bhuja is the distance of the same from the East- west line namely PMN) AM = bhuja and ML = s so that b=a+s addition being justified since both a and s are of the same direction namely north. In the second case BK = Karnāgra, BN = bhuja and NK = s so that b=s-a, the difference being justified because a is south and s north. Here we have taken the difference as s-a and not a-s because s is numerically greater and being oriented north, the bhuja is north. In the third case, PR=s, CP=b and CR=a so that b=a-s, the difference being justified be- cause s is north and a south. Also we have taken the difference as a-s and not s-a because a is numerically greater and as such lends its direction namely 'south' to the bhuja. Thus in the three examples cited, addition is prescribed between a and s only when the extremity of the shadow is to the north of E.S.L. In the case of A=S+B or B=A-S also, addition is prescribed only when δ is south, which means that the corresponding a ie. AL is north. In fact the prescription of addition or differ- ence accord in the cases of both the equations either A=B+S or a=b+s. Verses 74 and 75. Hereafter questions are being set and answered on diurnal problems. Seeing the shadow of the gnomon, the azimuth and longitude of the Sun or seeing two shadows with their respective directions, whoever knows the equinoctial shadow of the place, I consider him as the Garuda or Eagle who could overcome the false pride of puffed up snakes of astronomers. Given that when K=30 units, the bhuja is 3 units south, and when K=15, the bhuja is 1 unit north, com- pute the latitude, or again given H sin δ = 846 and given K and b of a shadow, compute the equinoctial shadows.

310 Comm. The questions are clear the second verse illustrating the first. Verse 76. Answer of the first question. (b₁ K₂ ⁺~ b₂ K₁) / (K₂ ~ K₁) = s according as the bhujas are of the same or opposite directions. Comm. Suppose b₁ and b₂ are of the same direction so that b₁ = a₁ - s and b₂ = a₂ - s, taking the modern convention of signs. But a₁ = (K₁A) / R and a₂ = (K₂A) / R ∴ b₁ = (K₁A) / R - s and b₂ = (K₂A) / R - s ∴ (b₁ + s) / K₁ = A / R = (b₂ + s) / K₂ ∴ K₂ b₁ - K₁ b₂ = s (K₁ - K₂) ∴ s = (K₂ b₁ - K₂ b₂) / (K₁ - K₂) If, however b₁ and b₂ are of opposite directions ie. of opposite signs, writing - b₂ for b₂, we have s = (K₂ b₁ + K₁ b₂) / (K₁ - K₂) . Here we have chosen to follow the modern convention of signs; otherwise we have to con- sider four alternatives, for, bhujas of the same direction might mean both of the type OC (fig. 60) or both of the type of OB or one of the tyye OB and one of the type of OA or both of the type OA. Verses 77 and 78. Answer to the second question. Let Laghu ≡ L = ((KH sin δ) / R)² ; 12² (L - b²) ≡ Ādya ; Para = 12² b. Let Ādya and Para be divided by L ~ 12² ; call them still Ādya and Para; then √(Para² + A) ± Para = s according as b is north or south.

311 Comm. The data are H sin δ, K and b; since K and b are given a is known. Thus from the triangle PZS (fig. 61) we have sin δ = sin ϕ cos z + cos ϕ sin z sin a, all quantities except ϕ are known. Solving this trigonometrical equation which is of the form a cos ϕ

• b sin ϕ = c, we can have ϕ. Fig. 61 We shall now see how it is solved by Bhāskara. Let s be the equinoctial shadow. Then a = b + s = KA / R = (KH sin δ) / (H cos ϕ) I But 12 / K = 12 / √(s² + 12²) = (H cos ϕ) / R so that H cos ϕ = 12 R / √(s² + 12²). Substituting in I b + s = (KH sin δ √(12² + s²)) / 12 R which reduces to 12² R² (b + s)² = K² H sin² δ (12² + s²) ie. s² (12² R² − K² H sin² δ) + 2b 12² R² s = 12² {(H sin² δ) K² − b² R²} ie. s² (12² − (K² H sin² δ) / R²) + 2. 12² b. s) = 12² ((K² H sin² δ) / R² − b²) Here (K² H sin² δ) / R² is symbolized as L 12² ((K² H sin² δ) / R² − b²) put as Ādya and 12² b is put as para.

312 So the equation reduces to s² (12² - L) + 2. Para. s = Ādya ; Divide throughout by 12² - L and put again Para / (12² - L) as Para and Ādya / (12² - L) = Ādya Then the equation reduces to s² + 2 Para s = Ādya ; completing the square (s + Para)² = Para² + Ādya ∵ s + Para = √(Para² + Ādya) ∴ s = √(Para² + Ādya) - Para as one solution. We have taken to start with a = b + s which holds good according to the Hindu convention when b is south ; if, however b is north a = b ~ s so that (b ~ s)² = s² + b² - 2bs. So in the equation we have to write -s for s, so that we have now s² - 2 Para s = Ādya ie. (s - Para)² = Para² + Ādya ∵ s = √(Para² + Ādya) + Para as the second solution. Verse 79. When the Sun's longitude is 135°, the shadow of the gnomon is 12 units and west. What is the latitude? Comm. Here is a method of obtaining the latitude of the place by observing the gnomon's shadow when the Sun is on the prime-vertical. Verse 80. Answer to the question above. (12 R) / K = H cos z ; s = (12 H sin δ) / √(Sama-Sanku² - H sin² δ) Comm. Solution in modern terms. S = 12 ∴ tan z = 1 ∴ z = 45; but when the Sun is on the prime-vertical, we have by Napier's rule Sin δ = sin ϕ cos z = sin ϕ √2. But since λ = 135°

often handles vertical fractions in plain text either as: [L..] triangle (H sin δ) / x = s / 12 where x is the Koṭi OR line-by-line exact layout. Wait! Let's examine how the lines flow: If we look at: triangle H sin δ / x = s / 12 where x is the Koṭi Let's check the rest of the page: - ∴ s = (12 H sin δ) / x. But we are given that the Sama- - Śaṅku ie. - H cos z = R cos z = R / √2 because z = 45° when the - shadow equals the length of the gnomon. - and H sin δ = (H sin λ H sin ω) / R = (H sin 135 H sin ω) / R - = (H sin 45 H sin ω) / R = R/√2 × (H sin ω) / R = (H sin ω) / √2 - ∴ x = √(Sama-Śaṅku² - H sin² δ) = √( R²/2 - (H sin² ω)/2 ) - `= (H cos ω) / √2 ∴ s = (12 H sin δ × √2) / (H cos

314 that moments I would reckon you as one who could be well compared with the goad that could be applied to the head of the wild elephants of puffed up astronomers. Verses 82, 83. Answer to the first question. Assume the H sine of the Unnatakāla to be Iṣṭa Hṛti in the first place. Multiply it by 12 s and divide by k² the square of the Viṣuvatkarṇa. Then you get an approximate value of H sin δ. Then compute with this, H sin δ, Charajyā etc. and thereby obtain a more correct value of the Iṣṭa Hṛti. Multiply this by the H sin δ got before and divide by the first Hṛti assumed. Then we have a nearer approximation of H sin δ. Repeat the process till a stationary value has been reached. That will be the correct H sin δ. Comm. We know that H sine of the Unnatakāla is nearly the Iṣṭāntyakā. It will be noted that Iṣṭāntyakā is the sum of two H sines namely (1) Carajyā (2) H sine of Unnatakāla minus Chara. The second H sine is called Sūtra or H cos h. (Vide page 278). Thus Sūtra + Carajya = Iṣṭāntyaka whereas H sine (Unnatakāla) = H sine of the Cāpas of Carajyā and Sūtra. In other words H sin (Unnatakāla) = H sin (H sin⁻¹ Carajyā + H sin⁻¹ (Sūtra)). Iṣṭa Hṛti is H cos δ Iṣṭāntyaka × ----------- . But we do not know H sin δ so R that Iṣṭa Hṛti could not be got. So we will not be far from truth in assuming the given Unnatakāla to be Iṣṭa Hṛti itself ie. Taddhṛti here, as the Sun is on the prime- vertical. The formula for Taddhṛti is R² H sin δ R². H sin δ K² H sin δ ----------------- = --------------------- = ------------ H cos ϕ H sin ϕ R. 12 s 12 s ----- × R × --- k k

315 ∴ (Taddhṛti × 12 s) / k² = sin δ. Thus assuming H sine of the given Unnatakāla to be Taddhṛti and multiplying it. by 12 s and dividing by k² we have the value of H sin δ. But this is approximate because the given Unnatakāla is not exactly Taddhṛti but only an approximate value. From this H sin δ, compute H cos δ, Charajyā, Kujyā and through the process indicated in verse 54 namely “Subtract the Characāpa from the Unnatakāla. The H sine of the result is called Sūtra. Multiply the Sūtra by H cos δ and divide by R; then we have Kalā. Add Kujyā to Kalā; we get Iṣṭa-Hṛti”, we obtain a more correct value of Taddhṛti. Then here we may cut short the process as follows namely ‘ If by the assumed Taddhṛti we had the previous H sin δ, what shall we have for this more approximate Taddhṛti ’? The result will be a more approximate value of H sin δ. Again form the Taddhṛti with this H sin δ and so repeating the process till we have a stationary value, we have the correct value of H sin δ. Note. This is a beautiful example of the method of successive approximations which is a modern technique but which was so much in vogue and favourite with the Hindu astronomers. (It will be noted how to cut short the method). Verses 84, 85. Answer to the second question. Obtain 12² R²/(R² − H sin² h) s² + 1 and divide R² by this and take the square root which gives H sin δ. Then (R × H sin δ) / (H sin ω) gives H sin λ whose Cāpa gives the longitude of the Sun. Comm. The H sine of the given Natakāla is H sin h and R² − H sin² h = H cos² h. Let H sin δ be x, which is required to be found. Then R² − x² = H cos² δ; H cos h = Sūtra and

316 (Sūtra × H cos δ) / R = Kalā = (H cos h . H cos δ) / R = ∴ (H cos h . √(R² — x²)) / R But Kalā is the Koti of the fifth latitudinal triangle of which H sin δ is Bhuja. Hence (Kalā × H sin ϕ) / (H cos ϕ) = H sin δ = x ie. (H cos h √(R² — x²)) / R × (H sin ϕ) / (H cos ϕ) = x; but (H sin ϕ) / (H cos ϕ) = s / 12 ∴ Squaring both sides [(R² — H sin² h) (R² — x²)] / R² × s² / 12² = x² ∴ 12² R² x² = s² R² (R² — H sin² h) — s² x² (R² — H sin² h) ie. x² {(12² R² + s² (R² — H sin² h)} = s² R² (R² — H sin² h) ∴ x² = [s² R² (R² — H sin² h)] / [12² R² + s² (R² — H sin² h)] = R² / [(12² R²) / (s² (R² — H sin² h)) + 1] ∴ x = R / √[(12² R²) / (s² (R² — H sin² h)) + 1] = H sin δ as given. From H sin δ, the method of obtaining λ is clear from the formula (H sin λ H sin ω) / R = H sin δ. In the given numerical example h = 5 nādīs = 360° / 12 = 30° since 60 nadis of time correspond to 360°. Thus H sin h = R / 2 ; the remaining work follows. Verse 86. Another question. When the Sun is on the prime-vertical the gnomonic shadow is noted to be 16 inches. The Unnatakāla is 8 nādis. If you could give the H sin δ and s, I shall con- sider you nothing short of one who is an adept in solving the totality of the diurnal problems. Verses 87 and 88. Answer to the question.

317 Here also assume H sin (Unnata) to be the Taddhṛti as formerly done. Then as the shadow is 16″, K̇ = √(16² + 12²) = 20″. Then H cos z = (12 R) / K = 12 / 20 × 3438 Unnatakāla = 8 nāḍīs = 48°. ∴ H sin (48°) = assumed Taddhṛti. Then from the fourth latitudinal triangle, (H sin 48) / (H cos z) = k / 12 ∴ Approximate value of k is (12 H sin 48) / (12/20 × 3438) = (20 H sin 48) / 3438 . This is a known quantity from which s could be computed since 12² + s² = k². Again from the fifth latitudinal triangle s / k = (H sin δ) / (S. S.) . Here s, k and S. S. (Samamandala-Sanku) are known ∴ H sin δ could be got approximately. Thus we have found approximately the required quantities s and H sin δ. From this H sin δ and s we have to compute again H cos δ, Carajyā, Kujyā, etc. and applying the procedure of verse 54 [H sin (Unnata-cāra) × H cos δ] / R + Kujyā = Iṣṭa Hṛti; this is nearer value of Iṣṭa Hṛti than the assumed Taddhṛti. From this again obtain as before s and H sin δ; we could not apply the proportion “ If by the assumed Taddhṛti we have the previous H sin δ, what shall we have for this computed Taddhṛti (Iṣṭa Hṛti) ” for the reason given below. So Repeat the entire process till an invariable quantity is got which will be the correct value of H sin δ. Here repeating the entire calculation is correct and not taking the proportion because Taddhṛti = (R sin δ) / (Sin ϕ cos ϕ) where sin δ and sin ϕ are both to be computed. Formerly we could take the proportion in verse 81 because the latitude of Ujjain being known, in the magni-

318 tude of Taddhṛti namely (R H sin δ) / (Sin ϕ cos ϕ) only H sin δ is variable and Taddhṛti is directly proportional to H sin δ. But in the present example both H sin δ and H sin ϕ are both variables so that, that kind of rule of three does not work. Verse 89. Oh! Mathematician! At a place where s = 5'', there 10 nādikas after Sun-rise the shadow S is observed to be 9''. Tell me what the longitude of the Sun would be, if you are an adept in computing as well as understanding the geometry of the sphere. Verses 90, 91. Answer to the question posed. Assume H sine (Unnatakāla) to be Iṣṭāntyakā. Then (K × H cos z × R) / (12 × I. A.) = H cos δ where I. A. = Iṣṭāntyakā. R² − H cos² δ = H sin² δ ; from this approximate H sin δ and the given s compute a more approximate I. A. Repeat the process till an invariable quantity is obtained for H sin δ, which will be its correct value. From this, using the formula H sin δ = (H sin λ H sin ω) / R , λ could be had. Comm. We know the formula for I. A. as (R³ H cos z) / (H cos φ H cos δ) Assuming Unnatakālajyā as I. A. H sin (Unnatakāla) = (R³ H cos z) / (H cos φ H cos δ) ∴ H cos δ = (R³ H cos z) / ((12 R. I.A.) / k) = (k × R × H cos z) / (12 × I. A.) ; from which obtaining H sin δ and proceeding as indicated we have λ. In the above proof we have used our formula. But Hindu Astronomers proceed from first principles. Let us hear Bhāskara. Since S = 9''; K = √(9² + 12²) = 15''

319 ∴ Mahā-Sanku = H cos z = (R × 12) / K = (3438 × 12) / 15 = 2750 - 24. We know that Mahā-Sanku forms a latitudinal triangle with Iṣṭa Hṛti. So (H cos z) / (Iṣṭa Hṛti) = 12 / k ∴ Iṣṭa Hṛti = (H cos z × k) / 12 ∴ Iṣṭāntyā = (Iṣṭa Hṛti × R) / (H cos δ) ie. H cos δ = (Iṣṭa. Hṛti × R) / Iṣṭāntyā = (H cos z × k × R) / (12 × I. A.) substituting the above value of Iṣṭa Hṛti. Here H cos z is got above and I. A. has been assu- med above as H sin (Unnatakāla). Note. (1) Computing H cos δ = (12 R / 15) × (√(s² + 12²) × R) / (12 × H sin (60)) = (R² × 13) / (15 H sin 60) = (13 R × 2) / (15 × √3). Here H cos δ > R which is invalid. (2) This is the only place where Bhāskara gave a numerical example with a slight flaw. In other words, under the given circumstances the shadow must be greater than what is given. However, the procedure indicated is mathematically correct. (3) It is interesting to note that the flaw was noted by a commentator named Lakṣmīdāsa as reported by Munīśwara in his Marīchi Bhāṣya. Munīśwara also noted the flaw but argues away in an untenable way. Another commentator named Gaṇeśa who was the author of the commentary named Śiromaṇipracāśa, does not seem to have noticed the flaw, or even if he did notice, probably he fought shy of pronouncing that there was a flaw. In fact a simple flaw like this in numerical examples, is not in the least derogatory to the prestige of Bhāskara. So, the commentators who happened to notice the flaw need not have pointed the same.

320 (4) Or again in the given place, for the value of H cos δ to be valid the Unnatakāla x must be such that 15 H sin x > 13 R so that H cos δ might be less than R. This means sin x > 13/15 = .8667 so that x > 60°—4' ; so instead of 10 nādīkas, if the time were given to be just even one Vinādika greater, it would have been alright, or again if the latitude were given to be just a little less it would have been alright. Verse 92. Oh ! Mathematician ! please tell me the magnitudes of the equinoctial shadow and the longitude of the Sun if at a place on a particular day, Kujyā is 245 and Taddhṛti 3125. Verse 93. Answer to the questiou above. s = √[ 144 Kujyā / (Taddhṛti—Kujyā) ] and H sid δ = (12 Kujyā) / s and H sin λ = (R H sin δ) / (H sin ω) . Comm. From the fifth latitudinal triangle compared with third, Kujyā / Krāntijyā = Krāntijyā / (Taddhṛti — Kujyā) = Agrā / S. S. = s / 12 (1) (2) (3) (4) Multiplying (1) by (2) Kujyā / (Taddhṛti — Kujyā) = s² / 12² ∴ s = √[ 144 Kujyā / (Taddhṛti — Kujyā) ] Also Equating (1) and (4) Krāntijyā = (12 / s) Kujyā. Verse 94. Given that H sin δ + S. S. + Taddhṛti—Kujyā = 6720, and Kujyā + Agrā + H sin δ = 1960. Then I shall consider him who finds s and the longitude of the Sun as the very Sun illuminating the lotuses of astronomers.

321 Verse 95. Answer to the question above. Divide 12 × Second sum by the first sum, that will be s. Again (12 × Second sum) / (12 + s + k) = H sin δ. From H sin δ, λ could be had as before. Comm. Comparing the third and fifth latitudinal triangles Kujyā / Krāntijyā = Krāntijyā / (Taddhṛti — Kujyā) = Agrā / S. S. = s / 12 (Kujyā + Krāntijyā + Agrā) / (Krāntijyā + Taddhṛti + S. S. — Kujyā) = 1960 / 6720 = 7 / 24 ∴ s = 7/24 × 12 = 7/2 = 3½″. Again comparing the third and the first latitudinal triangles s / Kujyā = 12 / Krāntijyā = k / Agrā = (s + 12 + k) / (Kujyā + Agrā + Krāntijyā) (1) (2) (3) (4) = (s + 12 + k) / 1960 (5) Equating (2) and (5) Krāntijyā = (12 × 1960) / (12 + s + k) = (12 × 1960) / (7/2 + 24/2 + 25/2) = (12 × 1960) / 28 = 840 since when s = 7/2 k = 25/2 which is the hypotenuse of the triangle formed by the equinoctial shadow with the gnomon. Equating (1) and (5) Kujyā = 245 ; equating (3) and (5) Agrā = 875. Now from the fourth latitudinal triangle compared with the first Agrā / s = S. S. / 12 = Taddhṛti / k (1) (2) (3) 41

322 From (1) and (2) S. S. = 12 / (7/2) × Agrā = 24/7 × 875 = 3000 From (1) and (3) Taddhṛti = k / s × Agrā = 25/2 × 2/7 × 875 = 3125. Note. This is a beautiful example exhibiting Bhās- kara's dexterity in algebra. Verse 96. Given that the sum of H sin δ, S. S. and Taddhṛti – Kujyā = 1440, and the sum of Agrā, S. S. and Taddhṛti = 800, I shall deem him whoever finds s and the longitude of the Sun, as the very Sun illuminating the lotuses of astronomers. Verse 97. Answer to the problem above. The second sum divided by the first and multiplied by 12 gives k from which s could be got. Then the first sum divided by s + 12 + k̄ gives H sin δ from which the longitude of the Sun could be got. Comm. Comparing the third and the fifth latitudinal triangles, we have Agrā / Krāntijyā = S. S. / (Taddhṛti – Kujyā) = Taddhṛti / S. S. (1) (2) (3) = k / 12 = (Agtā + S. S. + Taddhṛti) / (Krāntijyā + Taddhṛti – Kujyā + S. S.) = 1800 / 1440 = 5 / 4 I (4) (5) (6) Equating (4) and (6) k = (12 × 5) / 4 = 15 ∴ k² = 225 = 12² + s² ∴ s = 9. Again comparing the fourth latitudinal triangle, with the fundamental,

328 Agrā / s = S. S. / 12 = Taddhṛti / k = (Agrā + S. S. + Taddhṛti) / (s + 12 + k) (1) (2) (3) = 1800 / (9 + 12 + 15) = 1800 / 36 = 50 II (4) Equating (1) and (4) Agrā = 9 × 50 = 450 Equating (2) and (4) S. S. = 12 × 50 = 600 Thirdly Taddhṛti = 15 × 50 = 750 Again Equating (1) and (6) of I Agrā / Krāntijyā = 5 / 4 = 450 / Krāntijyā ∴ H sin δ = (450 × 4) / 5 = 360 from which λ could be computed. Verse 98. The chara at a place where s = 9, is equal to 3 nādis. If you could compute the longitude of the Sun, then certainly you are a leader among astronomers, Oh ! Scholar ! Verse 99. Answer to the problem above. 12 Carajyā / √((12 × Carajyā / R)² + s²) = H sin δ where from λ the longitude of the Sun could be computed. Comm. Let H sin δ = x; then from the third lati- tudinal triangle Kujyā / Krāntijyā = s / 12 = 9 / 12 = 3 / 4 ∴ Kujyā = 3 x / 4 since Krāntijyā means H sin δ ∴ Carajyā = 3 x / 4 × R / (H cos δ) = 3 R x / (4 √(R² - x²)) = H sin (3 × 6) = H sin 18° ∴ Squaring 9 R² x² = 16 (R² - x²) H sin² 18 = 16 Carajyā² (R² - x²)

324 ∴ x² (9 R² + 16 Carajyā²) = 16 R² Carajyā² ∴ x² = (16 R² Carajyā²) / (9 R² + 16 Carajyā²) ∴ x = (4 R Carajyā) / √(9 R² + 16 Carajyā²) = (12 Carajyā) / √(81 + (12² Carajyā² / R²)) = (12 Carajyā) / √(9² + (12 Carajyā / R)²) Here Carajyā being known, H sin δ could be computed. Verse 100. If you studied what is known as Madh- yamāharaṇa, then compute λ the longitude of the Sun given that H sin δ + H cos δ + H sin λ = 5000. Verse 101. Answer to the problem above. Let the given sum multiplied by 4 and divided by 15 be Ādya ; then H sin δ = Ādya − √(910678 − (2 square of the given sum / 337)) . Comm. Let H sin δ = x ; then H cos δ = √(R² − x²) and since H sin δ = (H sin ω H sin λ) / R ∴ H sin λ = (x R) / (H sin ω) = (x R) / 1397 ∴ The given sum = x + √(R² − x²) + (x R / 1397) = 5000 ∴ √(R² − x²) = 5000 − x (1 + R / 1397) = 5000 − (4835 / 1397) x ∴ R² − x² = 5000² + x² (4835 / 1397)² − (2 × 5000 × 4835 / 1397) x ∴ x² {1 + 4835² / 1397²} − (2 × 5000 × 4835 / 1397) = R² − 5000²

325 ie. x² (1397² + 4835²) - 2 × 4835 × 1397 × 5000 x = 1397² (R² - 5000²) ie. x² (25328834) - 2 × 4835 × 1397 × 5000 x = 1397² (R² - 5000²) ∴ x² - (2 × 4835 × 1397 × 5000 x) / 25328834 = (1397² (R² - 5000²)) / 25328834 ∴ x² - 2 × 5000 x × 6754495 / 25328834 = ,, Converting 675 / 2533 into a continued fraction we have 1/(3 +) 1/(1 +) 1/3 = 4 / 15 so that the equation could be written as x² - 2 × 5000 (x 4) / 15 = (1397² (R² - 5000²)) / 25328834 Here (5000 × 4) / 15 is symbolized as Ādya so that we have x² - 2 Ādya x = (1397² (3438² - 5000²)) / 2532883 ∴ (x - Ādya)² = Ādya² + (1397² (3438² - 5000²)) / 2532883 = 5000² × 16 / 225 - (5000² × 1397²) / 2532883 + (1397² × 3438²) / 2532883 = 5000² (16 / 225 - 1397² / 2532883) + (1397² × 3438²) / 2532883 Here 16 / 225 - 1397² / 2532883 is approximated to -2 / 337 and (1397² × 3438²) / 2532883 is approximated to 910678 so that we have x = Ādya ± √(910678 - (2 s²) / 337) where s is the given sum. Since the positive sign of the radical is invalid because H sin δ ≯ R, so the negative sign is taken.