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

252 as may be seen by taking δ = 20°, φ > 61° - 6'. Thus Bhāskara gave the minimum latitude which could enjoy four Koṇa-Śaṅkus. Fig. 41 Fig. 42

253 From fig. 41, it is clear that there are two Koṇa- Śaṅkus at S₁ and S₂ during the forenoon and similarly two at S₁' and S₂' in the afternoon where S₁' and S₂' are the symmetrical points of S₁ and S₂. From fig. 42, it is clear that if Agrā < H sin 45° when δ is south, there will be one Koṇa-Śaṅku in the forenoon at S₁ and one in the afternoon at the symmetrical S₁'. Verses 31 and 32. H cos z at noon known as Dinārdha-Śaṅku. By 'northern hemisphere' it is meant that the Sun is in the northern hemisphere ie. his Sāyana longitude ie. modern longitude lies between 0° and 180°, and 'the southern hemisphere' means that the Sun's longitude lies between 180° and 360°. The direction of δ may be got from the above convention. The latitude and colatitude are always deemed as south and north respectively. The latitude and colatitude being 'added to subtracted from or being decreased by' as the case may be, the declination, we have the zenith-distance and the altitude of the celestial body at Noon. The zenith-distance and the altitude are mutually complements. Comm. In Hindu Astronomy the words “उत्तरगोले” “दक्षिणगोले” are very often used to connote that the Sun is on the north or the south of the celestial equator respec- tively, so that the declination could be automatically known to be north or south respectively. Regarding the latitude, the peculiarity in Hindu Astronomy is that what we call north latitude in modern astronomy is construed as south in as much as the celestial equator gets depressed south in northern latitudes. The colatitude SQ in fig. 41 on the other hand extends north from the south, so that, it is construed as north. The word 'Saṃskāra' is used in Hindu Astronomy in the meaning given above in the translation. Fo

254 example in the equation A = S + B, we say that the Bhuja is had by a Samskāra between A and S. The meaning of Samskāra given by Bhāskara is “समदिशोर्योगः भिन्नदिशोरन्तरम् संस्कारः” Latitude being regarded as southern, if the Sun's declination is 12° north and the latitude 20°, then as they are of opposite direction, effecting the Samskāra as directed 20 - 12 = 8 = zenith-distance (South) = Nata as it is called similarly 70 + 12 = 82 = Altitude = Unnata; here we have added because, both lamba and declination are north. Similarly when δ = 24° north, and φ = 20° as before (south) 24 - 20 = 4° = zenith-distance (north) = Nata 70 + 24 = 94 = unnata (north). But, we take 180 - 94 = 86°. In the above working in the first case we found φ - δ, whereas in the second we found δ - φ. This difference in treatment is not taken objection to, since, the word Antara is used to take the positive value of the difference alone and so in the first instance the nata is pronounced as south, whereas in the second it is pro- nounced north. In modern astronomy, however, we have the formula z + δ = φ, considering z as positive if south, δ and φ positive if north. Here 8° + 12° = 20° (first case cited above) and (- 4°) + 24° = 20° (2nd case, z being negative, for, it is north. In the Hindu symbolism we have to pronounce separately when z is south or north, whereas in modern symbolism the sign alone informs its direction. Similarly in the equation A = S + B, we have to pro- nounce ‘north bhuja’ or ‘south bhuja’ as the case may be, whereas having a convention that δ is + ve when north, and also the Hindu azimuth (measured from the East point) the sign of bhuja indicates its direction. In other words we differentiate the two cases A - S and S - A giving them signs and deducing the direction of the bhuja

255 from the sign itself without an appeal to a picture or without ascertaining whether the northern Agrā prevails over the Southern Saṅku-tala or the Southern Saṅku-tala prevails over the northern Agrā. Thus the Dinārdha Saṅku in symbolism = H cos (φ ± d) (20). Verse 33. Here at noon, Dṛg-jyā is the H sine of nata and the Saṅku is H sine of unnata. Second half of 33 and first half of Verse 34. The product of R and the unmandala-Saṅku divided by Charajyā is called Yaṣṭi. The Yaṣṭi increased by Un-mandala-Saṅku gives H cos z according as the Sun is north or south of the equator. Comm. Unmandala-Saṅku is H cos z when the Sun is on the unmandala. From the sixth latitudinal triangle, wherein Unmandala-Saṅku is Bhuja and Krāntijyā Karṇa, so by comparing with the second latitudinal triangle (or rather operating with the second triangle to signify the Hindu method). (Krāntijyā × Bhuja) / Karṇa = Unmandala-Saṅku = (H sin δ × H sin φ) / R (already derived under (19)). We saw before Charajyā = R tan φ tan δ. Hence as directed in the verse (R × H sin φ H sin δ) / (R × R tan φ tan δ) = Yaṣṭi = (H cos φ H cos δ) / R (21). ∴ H cos z (at Noon) = (H cos φ H cos δ) / R ± (H sin φ H sin δ) / R according as the Sun is on the north or south of the equator.

256 Hence H cos z (at Noon) = Natajyā = (H cos φ H cos δ ± H sin φ H sin δ) / R or in modern symbolism cos (φ ∓ δ) already derived under (20). We shall now show how the formulation is done by the simple rule of three (Ref. fig. 39). If a parallel through o₄d is drawn to cut Aa at a₁, then Aa₁ is called the Yaṣṭi, which is vertical. The triangles Aoa₁ and Do₄d are similar so that Aa₁ / Ao = Dd / Do₄ ∴ Aa₁ = (DD × Ao) / Do₄ . But Ao / Do₄ = R / Charajyā for, all the lines of the diurnal circle and the equator stand in the ratio (H cos δ) / R = Ao / R = Do₄ / Charajyā = Kujyā / Charajyā ∴ Ao / Do₄ = R / Charajyā ∴ Aa₁ = (Unmandala Śaṅku × R) / Charajyā = Yaṣṭi as formulated. Now Dinardha Śaṅku = Aa = Aa₁ + a₁ a = Aa₁ + Dd = Yaṣṭi + Unmandala Śaṅku. It is evident from fig. 21 why in the northern sphere the sum is to be taken whereas in the southern, the difference is to be taken. Latter half of verse 34. Definition of Hṛti and Antyā. The sum or difference of Dyujyā and Kujya will be similarly Hṛti, whereas the sum or difference of Charajyā and radius will be Antyā. Comm. We defined formerly Hṛti and Antyā under our commentary on the latitudinal triangles. From fig. 39 Hṛti = o₁A = o₁o + oA = o₄D + oA = Kujya + H cos δ (Dyujyā) (22).

257 In the parallel great circle, the Equator we have therefore Antyā = Charajyā + R (already derived). Verse 35. Antyā = (Hṛti × R) / (H cos δ) = (Hṛti × Charajyā) / Kujyā and ∴ Hṛti = (Antyā × H cos δ) / R = (Antyā × Kujyā) / Charajyā by what is called Guṇa-ccheda-Viparyaya ie. alternando. Comm. Evident. Verse 36. To obtain Dinārdha-Śaṅku from Antyā and Hṛti (Antyā × un-maṇḍala Śaṅku) / Charajyā = (Hṛti × 12) / K = Dinārdha- Śaṅku. Comm. The second formula is derived from the similarity of △ Ao₄a (fig.) with the first latitudinal triangle. From the similarity of Aao₄, Ddo₄ fig. 39, Ao₄/Do₄ = Aa / Dd ∴ Aa = (Hṛti × unmaṇḍala-Śaṅku) / Kujyā . But Aa = Dinārdha-Śaṅku ∴ Dinārdha Śaṅku = (Hṛti × U.S.) / Kujyā (U. S. = Un- maṇḍala-Śaṅku.) But Hṛti / Kujyā = Antyā / Charajyā ∴ D.S. (Dinārdha-Śaṅku) = Charajyā / Antyā × U.S. = (Hṛti × 12) / K VIII. Verse 37. Meridian Zenith distance. Agrā ± (Hṛti × bhuja of a lat. triangle) / (Karṇa of a lat. triangle) = H sin z where z is the meridian zenith distance. 33

258 Comm. This formula is a special case of the formula A = S + B since the H sine of the meridian zenith distance is the Śaṅku-Bhuja at noon. From fig. 39 (Hṛti × bhuja of a latitudinal triangle) / (Karṇa of a latitudinal triangle) = O₁a = Dinārdha Śaṅkutala. The operation of sign has been already explained. Verse 38. An alternative method. The meridian zenith distance of the Sun can be had also by the formula (Hṛti ± Taddhṛti) B.L.T. / K.L.T. where B.L.T. and K.L.T. are the bhuja and karṇa of any latitudinal triangle. Comm. (Ref. figures 43 and 21). Let E₁ be the centre of the armillary sphere so that QE₁R is the diameter of the celestial equator which is on the median plane. Let S₁ F₁ S be the diameter of the diurnal circle of the Sun, which is also on the meridian plane so that S₁ F₁ is the Taddhṛti, S₁ S is the Hṛti and S the position of the Sun on the meridian. H sin z = SM = SF₁ sin F̂₁ = SF₁ sin ϕ = (Hṛti − Taddhṛti) × s/K where s/K can be replaced by B.L.T. / K.L.T. (s=equinoctical shadow and k the Viṣuvat-Karṇa). In the Southern sphere, H sin z = S′N = S′F₁′ sin ϕ = (S′S₂ + S₂F₁′) sin ϕ = (Hṛti + Taddhṛti) × s/k. Verse 39. Still another way of obtaining the m. z. d. (meridian-zenith-distance). R − H versin (altitude) = H sin z

259 Fig. 43 This formula gives not only the meridian zenith distance but H sine of the zenith-distances of the Sun when he is on the Koṇa-Vṛtta or prime-vertical or unmaṇḍala. Comm. From fig. 43, H sin z = SM = LE₁ = E₁s − sL = R − H versine (Ss) = R − H versine (altitude) as given. Since R − H versine (altitude) = R − {R − H cos (90 − z)} = R − (R − R sin z) = H sin z, so this formula applies wherever the Sun be. This is almost begging the question as H sine of z is being sought through H versine of (90 − z).

260 First half of the verse 40. To obtain the shadow S and K the Chāyakarṇa of any shadow (H sin z × 12) / (H cos z) = S and (R × 12) / (H cos z) = K. Comm. (Ref. fig. 44). (12. H sin z) / (H cos z) = 12 tan z = S. Also 12 / K = cos z = (H cos z) / R so that (12 R) / (H cos z) = K. The Hindu method of looking at this through the similarity of ΔS OM☉ and Ogn the gnomonic triangle. is as follows. ☉M is called Mahā-Śaṅku ie. H cos z ; ☉L is Dṛkjya or H sin z = OM. 12 / (H cos z) = S / (H sin z) so that S = 12 H sin z/H cos z. Also, On / O☉ = 12 / (H cos z) ie. K / R = 12 / (H cos z) ∴ K = (12 R) / (H cos z) . It will be noted that fig. 44 pertains to any vertical plane. Second half of Verse 40. The Dinārdha-Karṇa is equal to (R × k) / Hṛti where k is the Viṣuvat-Karṇa. Comm. The formula is derived through twice apply- ing the rule of three or what is the same, through the similarities of two sets of triangles From fig. 39, O₁A / Aa = Hṛti / Dinārdha-Śaṅku = k / 12 (a) and from fig. 44 On / O☉ = 12 / Dinārdha-Śaṅku = K / R (b) where K is the required Chāyākarṇa. Dividing (a) by (b) Hṛti / 12 = k / 12 × R / K ∴ K = kR / Hṛti

261 Fig. 44 Verse 41. Alternate method of obtaining K 101530/H sin λ = para (say) where λ is the Sāyana longitude of the Sun ; then, (Para × k) / s = K where K is un-mandala-Karṇa First half of Verse 42. To obtain K when the Sun is on the prime-vertical— Para × s/k = Samavṛttakarṇa. Comm. From fig. 19, from the similarity of triangles BD ☉ and CMA, B ☉ / CA = ☉ D / AM ie. ☉ D = (B ☉ × AM) / CA ie. H sin δ = (H sin λ H sin ω) / R (a) Then consider the similarity of the first and the sixth latitudinal triangles ; then Unmandala Sanku / Krāntijyā = s / k (b) where s is the equinoctial shadow and k the Viṣuvat- Karṇa. Again taking that ☉ the Sun lies on the unman-

262 dala in figure 44, (H cos z) / 12 = R / K = (Unmandala Śaṅku) / 12 (c) Eliminating Krāntijyā and Unmandala Śaṅku from (a), (b) and (c) (Unmandala Śaṅku) / (H sin λ H nis ω/R) = s / k ∴ (12R / K) / (H sin λ H sin ω / R) = s / k ie. 12R² / (KH sin λ H sin ω) = s / k ∴ K = (12R² × k) / (s H sin λ H sin ω). Here 12R² / (H sin ω) = (12 × 3438²) / 1397 = 101531 ; but Bhāskara has taken 101530 taking a more correct value of R. Then 101530 / (H sin λ) is symbolized as para so that para × k / s = K = Unmandala Karṇa. Regard- ing the Samavṛttakarṇa, in the place of (b) above we have (Sama-Śaṅku) / Krāntijyā = k / s (b') by the similarity between the first and the fifth latitudinal triangles. Equation (c) holds good with respect to any H cos z and the corres- ponding K since 12 R = K × Śaṅku and 12 R is a cons- tant. Noting therefore R / K' = (Sama-Śaṅku) / 12 (c') eliminating Krāntijyā and Sama-Śaṅku among (a), (b'), (c'), we shall have K = (12 R² s) / (k H sin λ H sin ω) = para × s / k as stated. Second half of Verse 42. To obtain the Dinārdha- karṇa from the Unmandalakarṇa. (Un-mandalakarṇa × Charajyā) / Antyā = Dinārdhakarṇa. Comm. We have equation (c) above stating 12 R = K × Śaṅku. (c) But

263 Iṣṭa Śaṅku / Iṣṭa Hṛti = cos φ = constant = Dinārdha Śaṅku / Hṛti = Sama-Śaṅku / Taddhṛti = Unmaṇḍala Śaṅku / Kujyā (d) (23) Again by virtue of the proportionality of Iṣṭa Hṛti / Iṣṭāntyā = Hṛti / Antyā = Kujyā / Charajyā (e) (24) We have Ishta-Śaṅku / Ishtāntyā = Dinārdha Śaṅku / Antyā = Unmaṇḍala Śaṅku / Charajyā ∴ Iṣṭa Karṇa × Iṣṭāntyā = Dinārdha Karṇa × Antyā = Unmaṇḍala Karṇa × Charajyā (f) (25) ∴ Dinārdha Karṇa = (Unmaṇḍala Karṇa × Charajyā) / Antyā as stated in the verse. Verse 43. (Unmaṇḍala Karṇa × Kṣitijyā) / Hṛti = (Sama Vṛtta Karṇa × Taddhṛti) / Hṛti = Dinārdha Karṇa. Comm. From (c) and (d) above Dinārdha Karṇa × Hṛti = Sama Karṇa × Taddhṛti = Unmaṇḍala Karṇa × Kujyā (g). Khitijyā is the same as Kujyā. From this the statement follows : Verse 44. The ancient Achāryas found the gnomonic shadows when the Sun is on the meridian, prime-vertical and the Kona-Vṛtta (ie, Vertical when the northern or southern Hindu azimuths are 45°) by different methods. I consider him to be the very Sun illuminating the lotus- faces of aitronomers, if anybody could give a method to find the shadow in any required direction, which holds good in all cases universally.

264 Comm. Evident. Verse 45. Definition of Dikjyā H sin (azimuth). The angle between any vertical and the Prime-Vertical measured on the horizon is what is called Digamsa and its H́ sine is known as Dik-jyā either in the Eastern hemi- sphere or the Western. Comm. In modern astronomy azimuth is measured along the horizon from the north point towards the east point round the horizon. In Hindu Astronomy however, the azimuth is measured from the East point on either side and from the West point also on either side specifying whether it is north or south. Verse 46 and first half of 47. To obtain the gnomonic shadow in any arbitrary direction. Assume Rs / (H sin a) as the equinoctial shadow and obtain the H sine of the corresponding latitude L. Then the product of that H sin L and H sin δ divided by H sin ϕ will give H sin D where D is a hypothetical decli- nation. With the new L and this D, as the hypothetical latitude and declination, obtain the meridian zenith distance by the formula Z + D = ϕ, and through this m.z d. obtain the shadow, which will be the shadow in the required direction namely 12 tan (ϕ ± D). Comm. Let gL be the gnomonic shadow on the equinoctial day in a given direction given by a° Digamsa (the Hindu azimuth) and let gN be the shadow in the same direction on any day. (fig. 45) We know that the extremity of the gnomonic shadow on the equinoctial day traces a straight line parallel to the East-West line Eω at a distance of the equinoctial shadow s because the Equatorial plane passing through the foot of the gnomon

265 Fig. 45 and that passing through the top of the gnomon being parallel planes cut the horizontal plane in parallel straight lines. (This will be also proved analytically subsequently). Hence LM = s. Now from the figure LM / gL = AB / gA = H sin a / R ∴ gL = H sin a / Rs I This gL is spoken of as Iṣṭa-Drikmandala palabhā because it is the shadow on the equinoctial day in any vertical. LN is the increment in the shadow on account of declination and we have to compute this and correlate gL and LN. For this refer to figs. 46 and 47. In fig. 46, QRT is the equator, so that when the Sun is on the equator on the equinoctial day in the direction given by ZS, ZT is the zenith-distance. Let ZS be the zenith- distance of the Sun in the same direction on any day From the analogy of finding H sin δ from H sin λ, from this figure 34

266 H sin SR = (H sin ST × H sin T̂) / R II and H sin φ = (H sin ZT × H sin T̂) / R III so that (H sin SR) / (H sin φ) = (H sin ST) / (H sin ZT) ∴ H sin ST = (H sin SR / H sin φ) × H sin ZT. Noting that SR = δ and putting ST = D H sin D = (H sin δ / H sin ϕ) × H sin ZT. [Diagrams: Fig. 46 and Fig. 47 showing spherical triangles with vertices P, Q, N, R, S, T, D, δ] Fig. 46 Fig. 47 The same formulae are derivable from fig. 47 also; only in fig. 46 while there is a decrement in the shadow of the day as compared with the shadow on the equinoctial day, in fig. 47, there is an increment. This is seen from the decrease and increase of ST in the zenith-distance ZT of the equinoctial day in the given direction. Now corre- lating fig. 45 with figures 46 and 47, the shadow gL pertains to the zenith-distance ZT on the equinoctial day whereas the shadows gN pertains to the zenith-distance on the day concerned in the same direction. We have, S / √(12² + S²) = (H sin z) / R so that RS / √(12² + S²) = H sin z where S is the shadow at any instant when the zenith-distance is z. The process indicated by saying ‘Obtain H sin φ

267 construing Rs / (H sin a) as the equinoctial shadow', means computing H sin ZT and there from ZT from the shadow gL of fig. 45. Then the process indicated by saying “Obtain H sin D = (H sin ZT × H sin δ) / (H sin φ) and therefrom D'' means computing ST. Then clearly ZS = ZT ± ST, ie. the required zenith- distance is got by what is technically called Samskāra between ZT and ST as is stipulated between φ and δ to obtain Z from the formula Z ± δ = φ (The word Samskāra was defined as meaning addition when the directions are the same and difference when they are opposite). Then the gnomonic shadow is got from this zenith-distance using the formula S = (12 H sin z) / (H cos z) . Thus the procedure adopted by Bhāskara was con- ceived by him first having Fig. 45 before him and then using figures 46 and 47. In this particular process, H sin a is given and H sin δ also, which means that it is sought to find the shadow on a given day in a given direction. Incidentally we shall find the locus of the extremity of the gnomonic shadow during the course of a day. Let in fig. 48 g represent the gnomon's foot, and S the shadow whose extremity is p. Required to find the locus of p. Take the gnomon to be of unit length so that the length of the shadow S = 12 tan z becomes tan z here. Take Eω and sn the east-west line and the north-south as the axes. Then we have x² + y² = tan² z (1). But we have from the triangle PZS sin δ = sin φ cos z + cos φ sin z sin a ie. sin δ / (cos φ cos z) = tan φ + tan z sin a = tan φ + y (2) ie. sec z / A = y + tan φ when A = cos φ / sin δ .

268 Fig. 48 But sec z = √(1 + tan² z) = √(x² + y² + 1) ∴ √(x² + y² + 1) / A = y + tan Φ which reduces to x² + y² (1 - A²) - 2A² y tan Φ + 1 - A² tan² Φ = 0 (3) From this it is evident that the locus is an ellipse or parabola or hyperbola according as A ⋚ 1; also it will be seen that the eccentricity is A. The locus is wrongly stated to be always a hyperbola in some text books. For it to be an ellipse A < 1 ie. cos Φ < sin δ ie. δ > 90 - Φ ie. Φ + δ > 90. In such latitudes and under such decli- nations, it will be an ellipse ie. at a place just north of the place where the perpetual day just begins the locus will be an ellipse. Hence in the arctic region it will be always an ellipse; and in the place just at which the perpetual day begins it will be a parabola and in the lower latitudes it will be a hyperbola, ie. it will be a parabola where the latitude Φ is given by 90 - δ. When Φ = 90°, A = cos Φ / sin δ = 0 provided δ ≠ 0. If, however, in addition δ = 0, A becomes indeterminate, but we may note then, that the Sun will be circling round the horizon on that equinoctial day at the north pole. We

269 may further note that at the north pole, the altitude of the Sun is always δ so that the length of the shadow cast is always equal to cot δ and this will be infinite when δ = 0. If now φ = 90°, and δ ≠ 0, though A = cos φ / sin δ be comes zero, A tan φ will not be zero because A tan φ = cos φ / sin δ × tan φ = sin φ / sin δ = 1 / sin δ (∵ φ = 90°). On the other hand A² tan φ = A × A tan φ = / sin δ = 0 because A = 0. Thus the term containing y in eqn. (3) vanishes. ∴ The equation reduces to x² + y² = A² tan² φ - 1 = cosec² δ - 1 = cot² δ (ie. +ve) ie. at the north pole the locus will be a circle with radius cot δ. When A = ∞ the locus √(x² + y² + 1) / A = y + tan φ becomes y = - tan φ which means that the hyperbola degenerates into the straight line which is parallel to the east-west line and is in the north at a distance of tan φ ie. s, the equinoctial shadow since the length of the gnomon is taken to be unity. In particular when A tan φ = 1 ie. φ = δ, the constant in (3) is zero, so that the locus passes through the foot of the gnomon as is also evident from the fact that the Sun passes through the zenith. Taking a northern latitude say 17°, the loci of the extremity of the shadow are shown in fig. 48A (page 270) on important days when δ=ω, when δ=0, when δ = - ω, when δ = φ. The maximum mid-day shadow is tan (φ + ω), when δ = - ω taking the gnomon's length to be unity; this shadow is cast north of the gnomon along the south-north line through the gnomon, on Dec. 23rd of the year. The minimum length of the mid-day shadow occurs when δ = φ, the shadow being zero and being at the foot of the gnomon, the Sun being then just overhead. The maximum shadow cast south of the gnomon at mid-day is tan (ω - φ).

270 Shadow on Dec. 23 No Shadow in this direction on Dec 23 Immediately before 21st March B C A On 21st March Immediately after 21st March No Shadow in this direction on 22nd June W B C A E JUNE when δ=φ On 22nd JUNE Fig. 48A Showing the locus of the extremity of the shadow on different days at a latitude of 17° Note. OA, OB are the shadows computed by Bhāskara when the Sun is on the prime-vertical. In the method of finding the shadow under verse 46, we perceive Bhāskara's genius in (1) looking upon the shadow as being made up of two segments namely that due to φ and that due to the declination (2) in conceiving what he calls Iṣṭa-drik-mandala palabhā, and Iṣṭa-drik- mandala Krānti and (3) in deriving the equation H sin D = (H sin ZT × H sin δ) / (H sin φ) Latter half of verse 47 and verse 48. Something to be noted. In computing the shadow in a given direction, there may be two shadows at times in the northern hemi- sphere. When H sin a < Agrā and there will be none in the southern. To compute the second shadow we have to take 180 − L also as the latitude where L is the latitude computed, and proceed in the same way as we have done before.

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