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

271 Fig. 49 Comm. This too exhibits Bhāskara's genius. (Ref. fig. 49). Let MQR₂ be the equator whose pole is p. Let T₁ S₁ Z S₂ T₂ be the circle of azimuth a (Hindu azimuth). Let SS₁ S₂ be the diurnal circle of the Sun cutting the above circle of azimuth at S₁ and S₂, so that ZS₁ and ZS₂ are the two solutions giving the two zenith-distances which give two shadows in the given direction. H sin MS = Agrā ; evidently MẐS > MẐS₁ ie. H sin a < Agrā as stipulated. ZT₁ and ZT₂ give the zenith-distances in the given direction when the Sun is on the equator. S₁ T₁ and S₂ T₂ are the decrements in the zenith-distances on account of declination δ (= S₂ R₂ or S₁ R₁). If MẐS₁ were greater than MẐS ie. if H sin a > Agrā, we would have lost the position S₁ ie. we would have had only one shadow

272 in the afternoon in the given direction and no shadow in the morning. Analytically, the event of having two shadows arises on account of the following circumstance. When we are asked to find Iṣṭākṣajyā from Rs / (H sin a) the shadow in the given direction on the equinoctial day (Ref. gL fig. 45) the L for this given value of the shadow is given by H sin L = RS / √(12² + S²) where S = Rs / (H sin a). We know sin θ = a has two solutions, θ₁ and 180 − θ₁. Hence L will have two values L₁ and 180 − L₁. So, Bhāskara has asked us to compute D₁ and D₂ from L₁ and L₂ and thus have the two solutions. Fig. 50

273 Note (1) When Bhāskara said 'If H sin a < Agrā' he had in mind evidently the azimuth circle MZN which cuts the diurnal path A Q₁ R₁ at S₁ and S₂. At S₁ the azimuth EZM < EZA so that he stipulated that H sin a should be less than the Agrā. But, let the diurnal path of the Sun be Q₂ R₂ where Q₂ falls in between z and p. In such a case we know that the azimuth does not take all values but has a maximum where the vertical touches the diurnal path at T. From PTZ where T is a right angle, taking PZT = 90 - a, a being the Hindu azimuth we have by Napier’s rule sin PT = sin ZP sin (90 - a) or sin (90 - δ) = sin (90 - φ) sin 90 - a or cos δ = cos φ cos a. If a has a lesser value than is given by this equation, the diurnal path does not cut the azimuth circle ie. if cos a > cos δ / cos φ, there will be no shadow in the given direction even though the situation satisfies Bhāskara’s condition namely H sin a should be less than Agrā. Bhāskara has overlooked this case. This may be seen analytically also as follows. We have from the spherical triangle PZS, sin δ = sin φ cos z + cos φ sin z sin a = A cos z + B sin z (say). We know, the maximum value of A cos z + B sin z is √(A² + B²) which is here √(sin² φ + cos² φ sin² a) = √(sin² φ + cos² φ - cos² φ cos² a) = √(1 - cos² φ cos² a). Thus there will be no solution for z if the quantity on the left hand side namely sin δ > the above max. value ie. if sin δ > √(1 - cos² φ cos² a) ie. if sin² δ > 1 - cos² φ cos² a ie. if cos² δ > cos² φ cos² a ie. if cos δ > cos φ cos a ie. if cos a > cos δ/cos p as derived above. Hence even if H sin a > Agrā, there need not be a shadow at all in the given direction. In other words when the Hindu azimuth given is very small and when the decli- nation is too great north or south, there may not be a 35

274 shadow in the given direction. Bhāskara gives an example where he gets two shadows on a day taking the moments when the Sun is on the prime-vertical. In fact having this case of the East-West shadows alone, he conceived that two shadows could be had in a given direction under particular conditions. He chooses a place of s = 5″ ie. a place of latitude 22° – 37′ (Bhāskara often gives this latitude which night indicate that he was probably residing in that latitude which passes through approximately Itarsi). He takes a day when the Sun’s declination is given by H sin δ = 780 ie. δ = 13° – 7′. Then the Sama-Śanku is given by R sin δ ───────── (comparing the second and the fifth latitudinal sin φ triangles Sama-Śanku Krantijyā H sin δ ──────────── = ─────────── = ───────── R H sin φ H sin φ RH sin δ R sin δ ∴ Sama-Śanku = ────────── = ─────────) H sin φ sin φ 3438 × 780 = ────────── = 2028 approximately. I 1322 – 18 R sin δ Also Agrā = ───────── = 845. cos φ Knowing the Sama-Śanku, the East-West shadow may be taken to be determined. Now Bhāskara proceeds to show that at the time of having the second shadow also, in the same East-West direction, the Śanku will be the same Sama-Śanku itself. For this, proceeding according to the Rs method indicated in the verse, “Taking ─────── to be the H sin a equinoctial shadow etc.” we have Rs 3438 × 5 ─────── = ──────────── = what is called Kha-hara Rāsi. H sin a 0

275 Taking this to be the equinoctial shadow H sin L = Rs / √(12² + s²) = R itself (Dealing this way with Kha-hara Rās'is is prohibited in modern mathematics but Bhāskara adds at the end of the commentary that dealing with them cautiously does not effect computations which is of course true, for when the equinoctial shadow is infinity φ = 90° so that H sin φ = R as got). Hence L = 90° and 180° - 90° = 90° = L'. Then H sin D = (R × 780) / (1322 - 18) = same as Sama-S'anku obtained in I = H cos z so that D = 90 - z. Now from the equation z + δ = φ, z = φ - δ = L' - D = 90° - (90 - z) where z is the zenith-distance when the Sun is on the prime-vertical. ∴ The zenith-distance is again the same z. In other words, the second zenith-distance is also that when the Sun is on the prime-vertical. Bhāskara has given this example just to obtain the second shadow as well and he has chosen the event of the Sun being on the prime- vertical to show that the procedure indicated by him may be verified to hold good. Verses 49, 50. Alternate method to find the shadow. Let R² s² + H sin² a × 12² = prathama where s = equinoctial shadow and a the Hindu azimuth. Let Anya = RsA where A is the Agrā. Divide the prathama and Anya by (H sin² a - A²) and still call them prathama and Anya. Then K = √(Adya + Anya²) ± Anya where K is the Chayā- Karṇa. Comm. Let K be the required Chayā-Karṇa. Then Karnāgrā = KA / R = s + b where b is the bhuja.

276 ∴ b = KA / R - s. But (H sin a) / b = R / S where S is the shadow so that S = bR / (H sin a) = (KA / R - s) R / (H sin a) = √(K² - 12²) ∴ (KA - sR) = H sin a √(K² - 12²) ie. K² A² + s² R² - 2 AsRK = H sin² a (K² - 12²) ∴ K² (A² - H sin² a) - 2 AsRK = - s² R² - 12² H sin² a ∴ K² (H sin² a - A²) + 2 AsRK = 12² H sin² a + s² R² I This quadratic is of the form ax² + 2bx = c ie. x² + 2 (b/a) x = c/a II Here 'c'/a is called Adya and b/a Anya. The solution of the above equation is given by x = - b/a ± √(b²/a² + c/a) ie. - Anya ± √(Anya² + Adya). III When b = KA / R + s, putting - s in the place of s in I, K = Anya ± √(Anya² + Adya) IV Out of the four solutions given by III and IV we have taking the positive solutions K = √(Anya² + Adya) ± Anya as stated. Verse 51. If H sin a < A, then in the northern hemisphere ie. where δ is north, ± √(Anya² - Adya) + Anya = K. Comm. We have initially put Anya = H sin² a - A². If H sin a < A, them to avoid a negative value for the

277 Anya, we could put Anya = A² - H sin² a. As a matter of fact in verse 50, we are asked to take H sin² ~ A², as Bhāskara wanted that the second case also be included. Thus putting Anya = A² - H sin² a, equation I becomes K² (A² - H sin² a) - 2 AsRK = - (Adya) so that K = Anya ± √(Anya² - Adya) as given. Verse 52. The Bhuja is to be obtained through Karṇāgra from the equation a = b + s (already proved) and Rb / S = H sin a {ie. S = bR / H sin a as already proved}. This H sin a will be the same in the case of obtaining two values of K ie. two shadows one in the morning and the other in the afternoon, (the only difference being that they will be on alternate sides of the East-West line). Verses 53 and 54. Obtaining the shadow when time is given. In the two previous examples the magnitude of the shadow was obtained in a given direction ; now we shall obtain the same when time is given. The word unnata stands for the time that has elapsed after Sun-rise or that which is the balance of the day time. The unnata sub- tracted from half-the-day gives what is called Nata. The H sine of the unnata minus Chara or increased by the Chara according as the Sun's declination is north or south, is called Sūtra ; this multiplied by the H cos δ and divided by the radius, is called Kalā. Comm. (Ref fig. 51) The time measured by the arc MN, that is the time in between the moment when the Sun S is on the horizon and the moment when he is at L is called the unnata ie. the time measured after rising and the time measured by the arc NQ ie. the time in between

278 Fig. 51 the moments when the Sun is at L and when he reaches the meridian. is called Nata. In modern astronomy this Nata is known as the hour angle h and unnata = H - h where H is the rising hour angle. The time measured by the arc ME is called Chara. Thus unnata-chara = EN and H sin EN H sin (90 - h) = H cos h is called Sutra which is BN shown in fig. 52 representing the Equator. The corresponding line bn in fig. 53 which represents the diurnal circle, is known as Kalā. Thus Sūtra = H cos h (26) and Kalā = (H cos h × H cos δ) / R (27). When the Sun is in the Southern hemisphere, unnata is measured by the arc SL or MN ie. (H - h) where H is

279 Fig. 52 Fig. 53

280 the rising hour angle and ℎ = L P̂ Q and Chara by the arc EM and Sūtra = H sin (MN + EM) = H sin (unnata + chara) = H sin EN = H sin (90 - ℎ) = H cos ℎ. Verse 55. Sūtra multiplied by Kujyā and divided by Charajyā will be also Kalā and Kalā multiplied by the Koti and divided by the Karṇa of any latitudinal triangle will be Iṣṭa yaṣṭi. Comm. In as much as Kalā is the corresponding line in the diurnal circle to the Sūtra in the plane of the celestial equator Sūtra / Kalā = R / (H cos δ) = Charajyā / Kujyā ∴ (Sūtra × Kujyā) / Charajyā = Kalā In verse 33, we saw that Dinārdha Sanku — Unmandala Sanku = Yaṣṭi. This Iṣṭa-yaṣṭi will be therefore the perpendicular dropped from 𝑛, the Sun's position in the diurnal circle on the plane parallel to the horizon and passing through the head of the Unmandala Sanku ie. passing through B and parallel to the horizon in fig. 21. Since the angle between the diurnal plane and the vertical plane of the yaṣṭi is equal to φ the latitude, the Iṣṭayaṣṭi forms a latitudinal triangle with the Kalā, it being the Koti or side opposite to the angle 90 - φ and the Kalā being the hypotenuse, ∴ Yaṣṭi / Kalā = cos φ = Koti of a latitudinal triangle / Karṇa of a latitudinal triangle ∴ Iṣṭayaṣṭi = Kalā × (Koti of a latitudinal triangle) / Karṇa When the Sun is on the meridian, Iṣṭayaṣṭi becomes yaṣṭi of verse 33. The formula for Iṣṭayaṣṭi is therefore (Kalā × H cos φ) / R = (H cos δ · H cos ℎ) / R × (H cos φ) / R from formula (27) = (H cos φ · H cos δ · H cos ℎ) / R² 27'

281 First half of verse 56. The Unmandala S'anku multiplied by the Sūtra and divided by Charajyā is also Iṣṭayaṣṭi. Comm. The Unmandala S'anku and Iṣṭayaṣṭi are the lines in vertical planes corresponding to Charajyā and Sūtra in the Equatorial plane. Hence the proportion. It will be noted that the Unmandala S'anku and Iṣṭayaṣṭi are not in the same vertical plane but parallel vertical planes. None the less the proportionality holds good. Latter half of verse 56 and first half of verse 57. The Sūtra increased or decreased by the Charajyā according as the Sun is in the northern or southern hemi- sphere is what is known as Iṣṭāntyā ; similarly the Kalā increased or decreased by Kujyā is what is known as Iṣṭa- Hṛti. Comm. In fig. 52, Iṣṭāntyā = AN = AB + BN = ME + BN = Charajyā + Sūtra. Similarly in fig. 53, Iṣṭa Hṛti = an = ab + bn = sg + bn = Kujyā + Kalā. ∴ Iṣṭāntyā = H cos h + R tan δ tan φ = R (sin φ sin δ + cos φ cos δ cos h) / (cos φ cos δ) in modern terms (28) Latter half of verse 57. Similarly Iṣṭayaṣṭi increased or decreased by the Unmandala S'anku is Iṣṭa S'anku or H cos z. Comm. Let in fig. 54 which represents the plane of the prime-vertical AA', EW, BB', FF', qq' represent the lines of intersection of this plane with planes parallel to the horizon and passing through A, B, F, q of fig. 21. Then Oa = Unmanda-S'anku, Oβ = Sama S'anku, Or = Dinārdha S'anku. If xx' be the line of intersection of this plane of the prime-vertical with a plane passing through an arbitrary position of the Sun in the diurnal circle and 36

282 Fig. 54 parallel to the horizon. then Ox = Iṣṭa-S'anku = Oα + ax = Un-mandala S'anku + Iṣṭa yaṣṭi. ar = yaṣṭi. In the Southern hemisphere Oα, the Unmandala S'anku will be below the horizon so that Iṣṭayaṣṭi decreased by the Unmandala S'anku will be Iṣṭa-S'anku. Thus we have the method of obtaining the Iṣhta- S'anku from the Unnata Kāla as detailed above. We shall see what this process means in modern terms. Unnatakāla-Charakāla = ⊙P̂A − AP̂E (Fig. 21) = ⊙P̂E where ⊙ is the foot of the declination circle of the Sun e in any arbitrary position in his diurnal path. But ⊙P̂E = QP̂E − QP̂⊙ = 90−h where h is the hour angle of the Sun. Thus Sutra = H sin (90−h) = H cos h I

283 ∴ Kalā = (H cos h × H cos δ) / R II ∴ Iṣṭa-yaṣṭi = [(H cos h × H cos δ) / R] × [(H cos φ) / R] III Now Unmaṇḍala-Śaṅku is derivable from the sixth latitudinal triangle in which Krāntijyā is the Karṇa and Unmaṇḍala-Śaṅku is the Bhuja. Comparing it with the second latitudinal triangle Krāntijyā / R = U. S. / (H sin φ) where U. S. is Unmaṇḍala-Śaṅku. ∴ U. S. = (Krāntijyā × H sin φ) / R = (H sin φ H sin δ) / R IV ∴ Iṣṭa-Śaṅku as per the above formulation is given by Iṣṭa-Śaṅku (I. S.) = (H sin δ H sin δ) / R + (H cos φ H cos δ H cos h) / R² = H cos z V [चित्र: गोलीय त्रिभुज PZS (Spherical triangle PZS) — शीर्ष: Z (zenith), P (celestial pole), S (celestial body / Sun); भुजाएँ: ZP = 90 - φ, ZS = z, PS = 90 - δ; कोण: ∠PZS = 90 - A, ∠ZPS = h, ∠PSZ = η] (Ref. fig. 55) Fig. 55 Formulae for △PZS Z = zenith ; P = celestial pole. S = celestial body, say, the Sun. z = zenith-distance of the celestial body S.

284 PZ = colatitude ; PS = north-polar-distance or co-decli- nation. η is called the parallactic angle ; h = hour-angle of S. (90 — a) = The complement of the Hindu azimuth a being measured from the East point. cos (90 — δ) = cos (90 — φ) cos z + sin (90 — φ) sin z cos (90 — a) sin δ = sin φ + cos z + cos φ sin z sin a (1) cos z = cos (90 — φ) cos (90 — δ) + sin (90 — φ) × sin (90 — δ) cos h = sin φ sin δ + cos φ cos δ cos h (2) cos (90 — φ) cos (90 — a) = sin (90 — φ) cot z — sin (90 — a) cot h ie. sin φ sin a = cos φ cot z — cos a cot h (3) cos (90 — φ) cos h = sin (90 — φ) cot (90 — δ) — sin h × cot (90 — a) ie. sin φ cos h = cos φ tan δ — sin h tan a (4) sin z sin (90 — δ) ─────── = ───────────── ie. sin z cos a = sin h cos δ (5) sin h sin (90 — a) This means in modern terms cos z = sin φ sin δ + cos φ cos δ cos h V which we derive from the triangle PZS. Verse 58. To get H cos z from h the hour-angle or nata Kāla. The H. vers (Nata) is called Sara (CQ of fig. 52) (29) Antyā — Śara = Iṣṭāntyā ie. FQ — CQ = FC = AN (fig. 51) H cos δ Śara × Kujyā Śara × ───────── = ───────────── = phala (CQ) (fig. 53) R Charajyā (30)

285 Hṛti — phala = Iṣṭa-Hṛti ie. fq — cq = fc = an (fig. 53) Comm. (Ref. fig. 52). Nata = arc QN ∴ H. vers (Nata) = QC. (Called Śara) = H. vers (h) Antyā — Śara = FQ — CQ = FC = AN = Iṣṭāntya II Śara × (H cos δ / R) = phala = cq (fig. 53) = (Hvers h × H cos δ) / R Hṛti — phala = fq — cq = fc = an = Iṣṭa-hṛti IV Verse 59. (Phala × Koti of a latitudinal triangle) / Karṇa = Ūrdhwa V = βr (of fig. 54) Comm. (Ref. fig. 54) βr is the corresponding line in the plane of the meridian corresponding to phala in the plane of the diurnal circle. As the angle between these two planes is the latitude itself, by the principle of ortho- gonal projection namely. Magnitude of a projected segment = cosine of the dihedral angle × the magnitude of the segment projected, since Ūrdhwa is the orthogonally projected segment of phala, so, Ūrdhwa = phala × cos φ = (phala × H cos φ) / R VI (31) = (phala × Koti of a latitudinal triangle) / (Karna of the latitude triangle) as stated Thus Ūrdhwa = βr (of fig. 54). Verse 60. Ūrdhwa is also given by Ūrdhwa = [U.S. (Unmandala Śanku) × Śara] / Charajyā Dinārdha-Śanku (D.S.)—Ūrdhwa = Iṣṭa-Śanku (I.S.) = H cos z.

286 Comm. Since U.S. is the projected segment of Charajyā on a vertical plane and since the diahedral angle between the planes is φ the latitude U.S. / Charajyā = Cos φ and so Śara × cos φ = (Śara × H cos φ) / R = Ūrdhwa. From fig. 54, Dinārdha-Śanku — Ūrdhwa = o'q — βr (fig. 54) = o'β' = yX = Iṣṭa-Śanku. In modern times, this means, Śara = H. vers (h) = (R — cos h) = 1 — cos h in modern terms phala = (Śara × H cos δ) / R = (1 — cos h) × cos δ ,, ,, phala × (H cos φ) / R = Ūrdhwa = (1 — cos h) cos δ × cos φ ,, ,, Dinārdha Śanku — Ūrdhwa = Iṣṭa Śanku H cos z = cos z (in modern terms) = H cos (φ — δ) (taking northern declination and following Hindu convention with respect to signs) = Dinārdha·Śanku = cos (φ — δ) in modern terms. ∴ cos (φ — δ) — (1 — cos h) cos φ cos δ = cos z ie. cos φ cos δ + sin φ sin δ — cos φ cos δ + cos φ cos δ cos h = cos z ie. cos z = sin φ sin δ + cos φ cos δ cos h as before. Verse 61. Computation of H cos z (Mahā Śanku) through Antyā and Hṛti. Let the Dinārdha Śanku (D.S.) be computed through Iṣṭāntyā and Iṣṭa Hṛti and therefrom Iṣṭa Śanku. From the Śanku, Drik·jyā ie. H sin z and the shadow (KH sin z) / R could be computed —H sin z should not be computed from Hṛti.

287 Comm. We computed Dinārdha Śaṅku by the formula. D.S. = (Antyā × U.S.) / Charajyā = (Hṛti × Koṭi of a L.T.) / (Karṇa of L.T.) under verse 36 ; similarly Iṣṭa Śaṅku (I.S.) will be given by Iṣṭa Śaṅku = (Iṣṭāntya × U.S.) / Charajyā = (Iṣṭa Hṛti × Koṭi of a L.T.) / (Karṇa of the L.T.) But Iṣṭāntyā = [R (sin φ sin δ + cos φ cos δ cos h)] / [cos φ cos δ] (as under verse 56) ∴ Iṣṭa Śaṅku = [R (sin φ sin δ + cos φ cos δ cos h)] / [cos φ cos δ] × (R sin δ sin φ) / (R tan φ tan δ) from formulae (13) and (19) = R (sin φ sin δ + cos φ cos δ cos h) = R cos z = H cos z Or again Iṣṭa Hṛti = (R cos z) / (cos φ) from formula (11) ∴ Iṣṭa Śaṅku = (R cos z / cos φ) × (H cos φ / R) = R cos z = H cos z Having got H cos z, using the formula H sin² z = R² — H cos² z, H sin z ie. Dṛk-jyā can be computed. Also K = 12R / (H cos z) and S = (KH sin z) / R give the Chāyā Karṇa and Chāyā. Bhāskara cautions us that H sin z cannot be computed from Hṛti as mentioned in verse 37. because there in that verse, the H sin z computed is that at noon alone. Verse 62. Alternate method of obtaining K. The Chāyākarṇa when the Sun is on the unmaṇḍala multiplied by Kujyā or that when the Sun on the prime- vertical multiplied by Taddhṛti, or again that when the Sun is on the meridian multiplied by Hṛti, divided by Iṣṭa Hṛti, will be equal to the Iṣṭa-Karṇa K.

288 Comm. Equation (23) under verse 41 is (Iṣṭa Śaṅku / Iṣṭa Hṛti) = (D.S. / Hṛti) = (S.S. / Taddhṛti) = (U.S. / Kujyā) = cos φ I and 12 / K = (H cos z) / R (under verse 40) ie. K = 12R / Iṣṭa Śaṅku which means Dinārdha Karṇa = 12 R / D.S. ; Sama Karṇa = 12R / S.S. and unmaṇḍala Karṇa = 12 R / U.S. Substituting for the numerators in I 12R / (K × Iṣṭa Hṛti) = 12R / (DK × Hṛti) = 12R / (S.K × Taddhṛti) = 12R / (U.K. × Kujyā) II ∴ Iṣṭa Karṇa × Iṣṭa Hṛti = S.K. × Taddhṛti = U.K. Kujyā ∴ Iṣṭa Karṇa = (Uumaṇḍala Karṇa × Kujyā) / Iṣṭa Hṛti = (Sama Karṇa × Taddhṛti) / Iṣṭa Hṛti = (Dinārdha Karṇa × Hṛti) / Iṣṭa Hṛti Verse 63. Just a caution. If in any context where the word ūna-yuta has been used, the quantity to be subtracted exceeds the quantity from which it is to be subtracted, it goes without saying that subtraction should be reversely effected and in the place of addition subsequently prescribed subtraction should be done and Vice-versa. Comm. Bhāskara gives three examples to illustrate his point. In verse 54, while defining Sūtra (H sin 90 – h) we are asked to subtract chara from unnata when δ > 0

289 Fig. 56 and add Chara to Unnata when δ < 0 and take the H sine of the result. Let us first consider the case when δ > 0. (Refer fig. 56) when the Sun is at ☉, Iṣṭa Śaṅku is H sin ☉L; and Unmandala Śaṅku is H sin BM. When the Sun is at ☉₁, Iṣṭa Śaṅku = H sin ☉₁N. In the first case Iṣṭayaṣṭi = (H sin ☉L — H sin BM) which will be the orthogonal projection of ☉B on the meridian plane. Unmandala Śaṅku and the Iṣṭa Śaṅku in the position ☉₁ are similarly the orthogonal projections of BM and ☉₁N on the same plane. In the position ☉ Iṣṭa Śaṅku = Unmandala Śaṅku + Iṣṭayaṣṭi, whereas in the position ☉₁, Iṣṭa Śaṅku = Unmandala Śaṅku — Iṣṭayaṣṭi which is now downwards. Thus in the place of addition we have subtraction of Iṣṭayaṣṭi. This reversion has arisen out of 37

290 the fact that in the position ☉, Sūtra is the H sine of (KA — KE) whereas in the position ☉₁ Sūtra is the H sine of EC ie. H sine of (KE — KC) ie. in the former position Sūtra = H sine (Unnata-Chara) and in the lattter Sūtra = H sin (Chara-Unnata). Thus a reversion in subtraction here, effects a reversion of addition of the Iṣṭayaṣṭi. Similar is the case in the other cases cited by Bhāskara. Analytically this happens so because cos h, when h > 90, becomes negative and adding cos h tantamounts to subtracting sin θ where h = 90 + θ. Verse 64. Another point to be observed. Hvers (90 + θ) = R — H cos 90 + θ̅ = R + H sin θ. The Unmandala Sanku is not observable when δ is south in as much as it is below the horizon ; none the less it may be computed for the purposes of effecting proportion. Fig. 57 Hvers (CG) = Hvers (CÔG) = GH Hvers (EG) = Hvers (EÔG) = Hvers (90 + θ) = R + OH' = R + EL = R + H sin ☉ as defined by Bhāskara,