पञ्चसिद्धान्तिका (आचार्य वराहमिहिर - सूर्य, रोमक, पौलिश, वासिष्ठ एवं पितामह सिद्धान्त)

III.37 III. PAULIŚA-SIDDHĀNTA 75 It must be noted here that when even the person with correct knowledge gets so much, the writer will get more. It must also be noted that only the writer of bad astronomy goes to hell, not the reader, whose sin is small. In this verse there is a jumbling of words and phrases and induction into the text extraneous words intended as commentary. The words, sphuṭagaṇitavid etc. seem to be the first half of the verse because in the first foot there are twelve and in the second eighteen syllables. Therefore what comes before that is the second half. In that, there are may syllables more than the required twenty- seven. Selecting the required words alone, we have reconstructed the third and fourth foot. For the observations of K S. Shukla on 32-37 vis-a-vis NP, see his paper ‘The PS of VM (1)’, JIHS 9 (1974) 62-76. [इति पञ्चसिद्धान्तिकायां वराहमिहिरविरचितायां पौलिशसिद्धान्तो नाम तृतीयोऽध्यायः ||]¹

1. A.B. पौलिशसिद्धान्तः; C.D. इति पौलिशसिद्धान्तः || Thus ends Chapter Three entitled ‘Pauliśa-Siddhānta: Planetary Computations etc.’ in the Pañcasiddhāntikā composed by Varāhamihira

Chapter Four THREE PROBLEMS — TIME, PLACE AND DIRECTION ४. चतुर्थोऽध्यायः त्रिप्रश्नाधिकारः Introductory Problems on Time, Place and Direction, involving spherical trigonometry, are dealt with in this chapter. The first fifteen verses are devoted to the construction of a table of sines. As this kind of matter does not involve constants specific to any siddhānta and is commonly found in all siddhāntas, we cannot say which Siddhānta this belongs to, Pauliśa or Saura, the only two siddhāntas meant to be expounded in detail by the author. Probably it is the author’s own, meant for both, or taken from both. Two things point to this conclusion: In the part of the work dealing with the Saura, viz., chs. IX, X, XI, XIII, XIV, XV, XVI, and XVII, no space is given to the sine tables, though required, and to the problems dealt with here, and therefore if these are not meant for Saura, it would be imperfect though almost full. On the other hand, certain redundant and crude rules point to this chapter’s connection with the Pauliśa, as also its position in the chapter distribution in the PS text. [ज्यानयनम्] षष्टिशतत्रयपरिधेर्वर्गदशांशात् पदं स विष्कम्भः | तदिहां(शच)तुष्कं संप्रकल्प्य रा(श्य)ष्टभागज्या || १ || Table of R Sines

1. Take the circumference as measured in 360 units, square it, take the tenth part of the square, and find its square root. The result is the diameter of the circle in the units taken. We assume the diameter to be 4°, (i.e., 240′) and hereunder give the tabular sines of angles for 3° 45′ interval. The rule is: diameter = √circumference²/10. It comes to this: d = c/√10. The formula, d = c/π is well known, and the author has taken √10 as an approximation for π which is incommensurable and usually represented by the approximate values, 22/7, 355/133, 3.1416 etc. The Sūrya Siddhānta too gives √10 as the value of π in its instruction to find the circumference of the earth from its diameter (I.59.): “The earth’s diameter is 1600 yojanas. Square this, multiply by 10, and find the square root. This is the earth’s circumference.” By thus taking √10 for π, an error of about 0.0067% results, and for a circumference of 21,600′, we get the radius 3415′, instead of the well-known 3438′. But it must be mentioned here that this error does not affect the computation of the sines 1a-b. A.B. परिधे वर्ग c. A.B. तदिहांशाश्चतुष्कं (B. ॰ष्क) b. A. विष्कुम्भः d. A. संप्रकल्प्य; B. प्रकल्प्य. A.B. राश्याष्ट०

IV.2 IV. THREE PROBLEMS 77 mentioned in the succeeding verses, because it can be shown that the author derives the sines from a correct formula, (not dependent on this wrong ratio of the diameter to the circumference), based on 120' as the radius of the circle. If he had depended on the wrong value, the first tabular sine, i.e. sin 3° 45' would be 7' 54", (being the 96th part of the circumference, where the sine is indistin- guishable from the arc), and not the correct 7' 51" as given by the author. Taking the diameter as 4°, and thereby the maximum sine (i.e. the radius) as 120', is arbitrary. In general, the Siddhāntas give the maximum sine, 3438', as arrived at from taking the circumfer- ence as 360° or 21600'. The Vākyakaraṇa makes it 43°. In actual work, the sines enter only as a ratio to the maximum sine, and therefore no harm, will result by taking these different maximum sines. TS and NP have not understood the meaning of the second half of the verse, and mis-interpret aṁśacatuṣkam as quadrant. व्यासार्ध[स्य] कृतिर्ध्रुवसंज्ञिका कृतांशस्ततः स मेषस्य | ध्रुवकरणी मेषोना द्वयोस्तु राश्योः पदं ज्याः स्युः ॥ २ ॥ 2. The square of the radius, (i.e. 14,400), is called dhruva (karaṇī), (literally, ‘Fixed Irrational’). The fourth part of it, (i.e. 3600), is the karaṇī (Irrational) related to the first sign, (or 30°). Dhruvakaraṇī minus the karaṇī of Meṣa, (i.e. 14,400 − 3600 = 10,800), is the karaṇī of two signs, (or 60°). The square root of a karaṇī is the tabular sine. Being square of tabular sines given in minutes, the karaṇīs are squares of minutes, which is their peculiarity as given by the author, though this is not mentioned explicitly. The other well-known characteristic of a karaṇī, viz. irrationality, is found in all karaṇīs except 14,400 and 3600, though the author calls these also karaṇīs in a general way. In modern terminology the word sine used in connection with the angle is defined thus: [Figure: Right-angled triangle ABC with right angle at C and angle marked at B] Fig. IV. 1-a In the right angled triangle, (fig. 1-a), sine ∠ B = AC/AB, or sine ∠ A = BC/AB, i.e. as the ratio of the opposite side to the hypotenuse. In tabulating the sines, the hypotenuse is taken as unity, and the ratio expressed as a decimal fraction. 2a. A.B. कृते ध्रुव०; C.D. कृतिध्रुव० c. B1.3. ये योना; B3. येषोना b. A.B. ०ज्ञिता. A.B. कृताशाःस्ततः. A.B. सशेषस्य d. A. दयोस्तु; B. दयो सु

78 PAÑCASIDDHĀNTIKĀ IV. 2 The ancients however expressed the sines in minutes-length or, more accurately, in minutes and seconds-lengths, the maximum sine called Trijyā (meaning ‘the sine of three signs’, i.e. 90°), occur- ring separately in the work to make up the ratio. This is the way in which they conceived the sine (meaning ‘bow-string’ from its Sanskrit equivalent śiñjinī, synonymous with jyā). In Fig. 1-b. A₃ E F₃ D is the circumference of the circle, centre B. A part of the circumference like ADF, A₁D F₁, etc. is called dhanus (literally, ‘bow’) or arc. Fig. IV. 1-b The straight lines ACF, A₁C₁F₁, etc. forming the ‘bow-strings’ of the respective ‘bows’ are the jyās or full sines. But in actual practice, the halves of the full sines AC, A₁C₁, etc. above are used with the name of ‘sines’, with respect to the half-bows or arcs, AD, A₁D₁, etc. Because the arcs AD etc. are as the angles ABD etc., the sines AC etc. are spoken of with respect to the angles ABD (= ABC) etc. also. Thus, AC is the sine of ∠ ABD or arc AD, A₁C₁ is the sine of ∠ A₁BD or arc A₁D₁ and so on. It is this connection of the sine with the arc that has given it the nature of a length, which is expressed in minutes and seconds on account of the connection of the arc with the angle at the centre. It may be mentioned here that CD, C₁D, C₂D etc., appearing like the arrows on the respective bow-strings, are called śara (meaning ‘arrow’). If A₃BD is a right angle, i.e. three signs, then, obviously, A₃B is the sign of this angle, i.e. it is the sine of three signs, and therefore called trijyā. Its length is clearly half the diameter A₃BF₃, i.e. the radius, equal to 120′. Now, let the angle ABD be equal to one sign, i.e. 30°. ABD = DBF = 30°. ∴ ∠ ABF = 60°. AB = BF, being radii. ∴ ∠ BAF = ∠ BFA = 60°. Thus ABF is an equilateral triangle, and AF = AB = 120′. ∴ AC = AF/2 = 60′. Thus sine 30° = 60′. Its karaṇī is its square, viz. (60′)² = 3600, the karaṇī of Meṣa as mentioned by the text. Then, let ∠ A₂BD be equal two signs, or 60°. A₂BD = DBF₂ = 60°. ∴ C₂ is a right angle. So the karaṇī of 2 signs = A₂C₂² = A₂B² − BC₂² = A₂B² − AC², (∵ △ A₂B C₂ ≡ △ BAC), = 120² − 60² = 10,800, A₂B being the radius. This also agrees with what the text says. (The square root of 10800 minutes, i.e. 103′ 55″, is the sine of 2 signs, which agrees with the value given in the table.)

IV. THREE PROBLEMS 79 Incidentally, we shall derive the karaṇī and sine of one and a half signs, i.e. 45°, mentioned in verse 4. Let A₁BD be equal to 45°. ∠ A₁ = 45°, and ∠ C is a right angle. ∴ A₁C = C₁B. But, A₁B² = A₁C₁² + C₁B² = 2 A₁C₁². ∴ A₁C₁² = 120²/2 = 14400/2 = 7200 = the karaṇī of one and a half signs as mentioned in the text. Its root, 84'51", is sine 45°, agreeing with what is given in the tables. शेषेष्विष्टेषु धनु-र्द्विगु(णं) पदात् प्रोज्झ्य शेषगुणहीना [त्] । [व्यासस्याऽर्धार्द्धर्गं] द्विगुणकरण्यां समायोज्यम् ॥ ३ ॥ तत्पादोऽभिमता [स्याद्] ध्रुवा तदूनाऽवशेषपिण्डस्य । ध्रुवकरणीदलमध्यर्धसंज्ञमन्योऽत्र विधिरुक्तः ॥ ४ ॥ इष्टांशद्विगुणोनत्रिभज्ययोना त्रयस्य चापज्या । षष्टिगुणा सा करणी तया ध्रुवोनाऽवशेषस्य ॥ ५ ॥ 3-5. The other tabular sines, (i.e. sine 3° 45′, sin 7° 30′ etc. other than the four mentioned of the total 24) are formed successively in the following manner: Let the angle or arc for which the sine is required be θ. I. Sin²θ = ¼[sin²2θ + {120 − sin(90° − 2θ)}²] II. Sin²θ = 60 × {120′ − sin (90° − 2θ)}, where the sines are in minutes etc. Of the 24 sines, the karaṇī of the nth sine = 14400 − the karaṇī of the (24 − n)th sine. 7200 is the karaṇī of one and a half signs, i.e. 45°. Thus, as karaṇīs 8, 12, and 16 are known, those of their halves etc. and (24 − halves) etc. can be found successively. Thus all the sines from 1 to 24 can be found. Of the two formulae, the first is suited to geometrical representation, and the second to computation. Example 1. Given the 8th karaṇī (i.e. of 30°) 3600 and its sine 60′, the 16th karaṇī (i.e. of 60°) 10800, and its sine 103′ 55″.33, find the 4th and 20th karaṇīs and sines, using each of the two formulae. The desired sine is the 4th, i.e. of 4 × 3° 45′ = 15°. 2θ = 30°, 90° − 2θ = 60°. 3a. A. धनुर्द्वि. A.B.C.D. ॰गुणपदा॰ 4a. A. तपदो; B. पदो; C.D. त[स्य] पदो. b. A. पदायोज्य; B. पदायोज्य्; C.D. पदायॊग॰ A.B.C.D. ॰भिमतज्या A.C.D. गुणहीना; B. गुणाहीना b. A. तदुना. B. ॰विशेषे c. A.B. तृव्यासपादार्द्धाद्धर्गं; (B. om तु; B2. ०र्गं); d. A. र्द्धसंज्ञा; B1.2. र्द्धसंज्ञां; B3. र्द्धं संज्ञा C.D. [त्रिज्या तदर्धवर्गौ] द्वि C.D. संज्ञकोऽन्योऽत्र. A. विधिनुक्तः d. A. कारथो; B. कारयो 5a. A.C. इच्छांशद्विगुणेन (C. ॰णोन) A. समायोज्यं; B. समाप्रोज्यन्त; b. B3. त्रयंस A.B. वायज्या C. द्विगुणज्यार्धस्य संयोज्यः; D. द्विगुण [र्ध] करणी c. A. स कारणी; B. स करणी समायोज्यः d. B. षपा. A.B. ॰नामशेषस्य

80 PAÑCASIDDHĀNTIKĀ IV. 5 I. The 4th karaṇī = ¼ [sin² 30° + (120 − sin 60°)²] = ¼ [60² + (120 − 103′ 55″.33)²] = ¼ [3600 + (16′ 4″.67)²] = ¼ (3600 + 258 97/144) = ¼ (3858 97/144) = 964 385/576 ∴ the fourth sine, i.e. sine 15° = √964 385/576 = 31′ 4″. II. The 4th karaṇī = 60 (120 − sin 60°) = 60′ (120′ − 103′ 55″.33) = 60′ × 16′ 4″.67 = 964 385/576 From this, sin 15° = √964 385/576 as before = 31′ 4″. We shall prove the first formula geometrically, and derive the second from the first. Fig. IV. 2 In Fig. 2, FBD = θ, and DF is its arc of which the sine wanted is DE. DE² = the wanted karaṇī (i.e. sine² θ). DFA = 2DF. ∴ DE = ½DA. ∴ sin² θ = DE² = ¼DA² = ¼(AC² + CD²). Now, ∵ AC = sine 2 arc FD = sin 2θ, AC² = sin² 2θ; and ∵ CD² = (BD − BC)² = (BD − AG)² = [120′ − sin (90° − 2θ)]², sin²θ = ¼ (AC² + CD²) = ¼ [sin² 2θ + {120′ − sin (90° − 2θ)}²] From this, sinθ = √sin²θ. From I, we can derive II thus: ¼ [sin² 2θ + {120′ − sin (90° − 2θ)}²] = ¼ {sin² 2θ + 120² + sin² (90° − 2θ) − 2 × 120 × sin (90° − 2θ)} = ¼ {120² + 120² − 2 × 120 × sin (90° − 2θ)} (∵ sin² 2θ + sin² (90° − 2θ) = sin² 2θ + Cos² 2θ = radius²) 2 × 120′ = ──────── {120′ − sin (90° − 2θ)} 4 = 60′ {120′ − sin (90° − 2θ)} Now, of TS, Thibaut alone proves the formula, while Sud. Dvivedi uses it. The form of the first formula given by them differs from that given by us, and is as follows:- sin²θ = (½ sin²2θ) + [½ {120′ − sin (90° − 2θ)}]² Though their formula is correct, it entails more work, and to give the formula in this form they have made many unwarranted changes in the already correct readings. We have made only one correc- tion, and that grammatical, for the sake of syntax, viz. dhanurdviguṇapadāt into dhanurdviguṇam padāt, which entails the occurrence of an extra syllable, which can be explained, as before, by rules of prosody. Or, let the reading be śeṣe tviṣṭe dhanurdvi-guṇam padāt projjhya etc. Now follow six verses giving the sines, computed by the author himself.

IV. 9 IV. THREE PROBLEMS 81 शेषज्याः 'स्वरतिथयो' 'गुण-शिव-धृति'भिश्च 'विंशतिः' सहिता । 'पञ्चनरकं' 'शतार्धं त्रिसमेतं' 'षष्टि'रिति लिप्ता ॥ ६ ॥ सैकाऽजे पञ्चाशत् 'पञ्चाष्टकं' 'पञ्चवर्गवेदा' श्च । 'त्रिंशच्चतुर्भिरधिका' 'षट्पञ्चाशच्छराः' शून्यम्' ॥ ७ ॥ 6-7. The other sines are the following: In the first sign, the minutes parts are successively 7, 15, 20 + 3, 20 + 11, 20 + 18, 5 × 9, 50 + 3, and 60. The seconds, respectively, are: 50 + 1, 5 × 8, 25, 4, 30 + 4, 56, 5 and 0. Thus we have for the first sign –

82 PAÑCASIDDHĀNTIKĀ IV. 11 Adding the minutes and seconds, and 60' for the end of the first sign, the sines are:-

IV. 14 IV. THREE PROBLEMS 83 there is one second more. By this mistake the 24th, i.e. the radius, has become 120' 1", and even this obvious mistake they have failed to note. The promised sine-intervals are here given: मुनयोऽब्जे व्येकान्ते 'रसत्रयं' 'पञ्च(कौ)' 'कृता' (श्र) गवि । 'शिखि-पक्षचन्द्र-शून्याः' द्वित्रिमिथुने कला ज्या [नाम्] ॥१२ ॥ मेषे विकलार्धशतं सैकं 'व्येकेन्द्रि(ये)-श्वरं त्रिंशत् । (द्वा)विंशतित्रिवर्गः ------------------------------ ॥ १३ ॥ ---------------------------------------------------- । ------ 'खगुणकृतार्णवयमनवकसमुद्रशिखिवर्गैः ॥ १४ ॥ 12. Of the intervals the minute parts are, in the first sign, 7, 7, 7, 7, 7, 7, 7, 6; in the second sign: 6, 6, 6, 5, 5, 4, 4, 4; and in the third sign, 3, 3, 2, 2, 1, 1, 0, 0. 13-14. The seconds in the first sign are, 50 + 1, 50 − 1, 50 − 5, 50 − 11, 30, 22, 9, (and here is a break resulting in the loss of the 4th foot of the 13th verse and the first two feet with 4 mātrās of the 3rd foot of the 14th verse. From the values of the sines we can compute that the seconds in the 8th interval must be 55, which must have been given in the missing part. By examining the remaining part we can construct the meaning of this verse thus). The seconds of the intervals in the second sign are, respectively 10 × (4, 2, 0, 4, 2, 5, 3, 0)

• (0, 3, 4, 4, 2, 9, 4, 9), added each to each. TS have not been able to see that in what is left of the 14th verse, the digits in the unit places of the eight numbers giving the seconds of the intervals of the second sign are given. This is because they have made mistakes in the 14th and 16th sines resulting in an error of + 1" in the 14th interval, − 1" in the 15th and + 1" again in the 16th. This has prevented them from finding the correct values by comparison, so that they make many changes in the readings here, saying that the text is very corrupt, here. It may be seen that not a single word is wrong here. 12a. A. मुनयोज्ये; B. गुनयोज्ये to 14c. D suggests for 13d [पञ्चाशच्च विषयसंयुक्तम्] b. A. त्रयं को; B. त्रयपञ्चको;. C. त्रयं (त्रिः) शशः C. indicates the gap by dots as done by A. कृताच्चे गवि; B. कृता-गेवि; C. कृताब्धी गवि; us. D suggests for 14: D. कृताग्निगवि ख[समुद्र] गुण [द्विकृताः] .......... c. A. शिखिपकृत चन्द्रः; B. शिखिपक्ष कृतार्णव [द्वि] यमनवे [न्द्रि] यसमुद्रशिखिवर्गे[शाः] || d. C. द्वौद्विमिथुने. A-B. ज्या |; C. ज्यार्द्धे; D. ज्या [सु] 13a. A. विकलार्द्धशत° d. A. समुद्रा b. B1. सैकां; B2. सैव्यं; B1.2. वर्गे; B2. वर्गे A1.B1. °न्द्रिय स्वरं C. खगुणकृतार्णवयमनवसकसमुद्रा शिखिवर्गैः (?) || c. A1.B2. द्विविंशति 13d. A.B. Unindicated gap of 13d. upto खगुण of 14c.

84 PAÑCASIDDHĀNTIKĀ IV. 15 ‘मनुविषयतिथिरसा:’ स्युस्त्रिगुणाः पञ्चाष्टकं ‘स्वरोपेतम्’ | सप्तदशनवपञ्चकं षोडश चेति क्रमान्मिथुने || १५ || 15. The seconds of the intervals in the third sign, are, 3 × 14, 3 × 5, 3 × 15, 3 × 6, 5 × 8 + 7, 17, 9 × 5 and 16. The intervals can be got by deducting the previous sines from the succeeding ones, and com- pared with the author’s concern for correctness, they are correct. It is to be noted that the intervals of TS also come correct in the third sign, because there is a uniform error of 1″ in all sines. The intervals are as tabulated:

IV. THREE PROBLEMS 85 Fig. IV. 3 AK = sine AG (neglecting the first sign) because, EF, HJ, JK, GF, are all equal, as already men- tioned. Therefore, for an arc or angle in the second quadrant, (3 to 6 signs), deduct it from six signs and get the sine of the remainder. For that in the third quadrant, (6 to 9 signs), deduct six signs and find the sine of the remainder. For that in the fourth quadrant, deduct it from twelve signs and find the sine of the remainder. Example 2 (a). Find the sine of the arc or angle equal to 4 signs. It is in the second quadrant. Therefore, sine 4 signs = sine 6 signs − 4 signs = sine 2 signs = 103′ 55″. Example 2 (b). Find the sine of signs 7-11-15. This is in the third quadrant. Therefore sine of sign 7-11-15 = sine (7-11-15 − 6-0-0) = sine 1-11- 15 = 79′ 7″. Example 2 (c). Find the sine of signs 9-15-0. This is in the 4th quadrant. Therefore sine of signs 9-15-0 = sine (12 signs − sign 9-15-0) = sine 2-15-0 = 115′ 55″. Declination From here to the end of the chapter, problems based on the solution of spherical triangles are dealt with, being problems involving position, time and direction. As a preliminary, the declina- tion of the Sun and the Moon are required, which are given first. Stellar sphere (Bhagola) The ancient astronomers speak of three spheres, the Terrestrial sphere or Earth sphere on which we live, the Stellar sphere or the Sphere of the stars, and the Sky sphere or the Sphere of the sky. We shall describe the terrestrial sphere in connection with the Sāura chap. XIII-XV. Of the other two, we shall now describe the stellar sphere, a knowledge of which is immediately required. It is the apparent sphere on which the stars appear to be fixed, and form a frame of reference for

86 PAÑCASIDDHĀNTIKĀ IV. 15 the motion of bodies like the Sun, Moon, planets etc. Actually the stars are at widely different dis- tances from us and are moving in various directions at speeds of several miles per hour. But with all this they appear to be, and can be represented, as being fixed on a sphere, because of their enormous distances from us, so much so that we are practically viewing the same sphere as people several generations ago did. But the ancients believed that the stars were luminous bodies fixed on the under-surface of a sphere of radius only sixty times that of the orbit of the Sun (actually the earth) round the earth, with the centre of the sphere at the earth's centre. This sphere seems to be rotating about once a day, by the actual rotation of the earth, about once a day. Fig. IV. 4 On this sphere (Fig. 4) there is an important great-circle called the ecliptic (Krānti-vṛtta) (EC), marked by the twenty-seven asterisms, Aśvinī etc., on which the Sun (S) moves, (actually appears to move on account of the motion of the earth), completing a revolution once a year. The Moon and the planets move in orbits inclined to the ecliptic at small angles. Their longitudes are reckoned along the ecliptic from a fixed point called the first point of Meṣa or Aśvinī. Latitudes are measured on secondaries to the great circle, meeting at a point called the pole of the ecliptic (Kadamba). Another great circle called the Celestial Equator (AOB) cuts the ecliptic at r, called the First point of Aries or Vernal Equinox point, which, instead of being fixed, has a slow westward motion on the ecliptic. At the period of the authors of the first Siddhāntas, this point coincided with the first point of Meṣa, this being the reason why that particular point was taken by them for reckoning from. The angle between the two great circles (SrR) is called the Obliquity of the ecliptic. It was about 24° at the time of Varāhamihira, and now it is about 23° 27'. The declination (SR,) (Krānti) of a body like the Sun, is measured along the secondary passing through the body, (NPSRSP,) called the Declina- tion circle, all the secondaries meeting at the celestial poles, the poles of the stellar sphere, (SPNP), on the axis joining which the sphere apparently rotates. The declination is found by solving the spherical right angled triangle SrR. Spherical triangles were in general solved by the Hindu astronomers using the properties of plane right angled triangles formed by the sections of the sphere along the great circle arcs forming the spherical triangle (The

IV. 16 IV. THREE PROBLEMS 87 Greeks solved them by an extension of Manelau's Theorem to figures on the sphere.) We shall con- tent ourselves with giving the formulae for solution and refer to them whenever necessary by way of proof. Let ABC (fig. 5) be a spherical triangle, right angled at C, and R the radius of the sphere. I. Cos AB = Cos AC × Cos BC ÷ R II. sin BC = sin AB × sin A ÷ R III. Cos A = R × Cos AB. sin AC/sin AB. Cos AC IV. Cos A = Cos BC × sin B ÷ R V. Cos AB = R. Cos A × Cos B/(sin A × sin B) These, together with the general identities, cosθ = sin (90° – θ), sin² θ + Cos² θ = 1, will suffice for explaining the formulae occurring in the text. [Fig. IV. 5] [रविचन्द्रयोः क्रान्तिः] जीवाऽध्यर्धशतां' (शघ्नै) काषष्टि (र्दि) नेशकाष्ठा (ज्या) । चन्द्रस्य सविक्षेपस्तदपक्र(मो) राशिपादे(भ्यः) ॥ १६ ॥ Declination of the Sun and the Moon 16. The sine of Sun's declination is found by multiplying the sine of its longitude by 61 and dividing by 150. Its arc is the declination. The declination of the Moon found thus is the mean declination. Its true declination is the mean declination plus latitude. The intervals of declinations for intervals of quarter signs are given (in the next two verses, 17-18). This is the formula: sin dec. = sin sāyana long. × 61/150. From this the arc forming the declina- tion is found by using the tables, and then the true declination of the Moon, using this. It must be noted that when the longitude reckoned from r (i.e. sāyana long.) is within 6 signs, the declination is North, and when more than 6 signs it is South. In the case of the Moon, if the declination and the latitude are of the same direction they should be added to get the true declination. If they are of different directions, their difference is the true declination, its direction being that of the greater. The author has not mentioned this because it is obvious. Example 3 (a). Find the maximum declination of the Sun. Obviously, the maximum sine of the Sun's longitude (i.e. when the Sun is 3 signs or 9 signs) will give the maximum declination. It is therefore given by sin dec. = 120' × 61 ÷ 150 = 48' 48". Its arc, 24° is the maximum declination. 16a. A. जीवाध्यार्द्ध०; B. जीवा व्या-र्द्ध; C. जीवाव्यर्द्ध; D. [साङ्कुलिप्ता] . A.B. काष्टान्तः; C. काष्ठान्तः; D. जीवाव्यध्यर्ध. B. सिताशा; A.C.D. शतांशाः D. काष्टान्तः b. A. सैकाःषष्टि; B. सैका षष्ठि; C. सैका षष्टि; d. A.B.C.D. तदपक्रम. A.C. ०पादेन्यः

88 PAÑCASIDDHĀNTIKĀ IV. 17 Example 3 (b). Sāyana Sun is rāśi 4-7-30. Find its declination. Sin 4ʳ 7° 30′ = Sin (6ʳ-0-0 – 4ʳ-7-30) = sin 1ʳ-22-30 = 95′ 12″. Sin Declination = 95′ 12″ × (60 + 1) ÷ 150 = 95′ 12″ × (2/5 + 2/5 × 60) = 38′ 5″ + 38″ = 38′ 43″. Its arc, 18° 50′, is the declination. Since the sāyana Sun is within 6 rāśis, the declination is north. Example 3 (c). The sāyana Moon is 9ʳ0°0′. Its latitude is 4°N. Find its declination. The sāyana longitude being 9ʳ0° 0′, the mean declination is the maximum, south, i.e. 24°S. Its lat. is 4°N, i.e. of opposite direction. ∴ the true declination is 24° – 4° = 20° S. South because South is greater. The author uses the word kāṣṭhā to signify declination, which is uncommon. Sometimes this word itself is used to mean sine declination. The rule for sin declination is explained thus: Our siddhāntas take the maximum declination to be 24°. As the maximun declination occurs when the sāyana longitude is 3 signs, the angle between the ecliptic and the celestial equator (i.e. the obliquity of the ecliptic) also is 24°. In Fig. 5, take AB and AC as parts of the ecliptic and celestial equator. Then, A = 24° and AB is the sāyana longitude. BC is the declination wanted. By formula II under the present verse, sin dec. = long. × sin 24° ÷ 120′. But sin 24° ÷ 120′ = 48′ 48′′ ÷ 120′ = 61/150. Hence, sin dec = sin long × (60 + 1)/150, which is the given rule. Now for the direction: See Fig. 4. At r the Sun moving along the ecliptic crosses the celestial equator, and passes from South to North. As great circles bisect one another, till the longitude is 6 sings it moves north of the celestial equator, for which declinations are reckoned, and then moves south of it. Therefore for longitude 0 to 6 rāśis, the declination is North, and for longitude 6 to 12 rāśīs, it is South. As the Moon and other planets move in their own orbits inclined to the ecliptic, the declinations computed from their longitudes reckoned along the ecliptic are only approximate. To get correct declinations, their distances north or south from the ecliptic points, called their ‘latitudes’, should be combined in the proper manner as instructed. But the result by thus adding or subtracting will be only approximate, because the latitudes are directed towards the pole of the ecliptic (Kadamba), while the declinations are directed towards the Celestial Pole. The maximum error that can occur thus is about 24′. Combined with the error in latitude due to other factors like proportion by degrees of the argument of latitude (advocated by the Pauliśa) instead of the sine etc., the error will be considerable. Now, for the readings. For syntactical purposes, and getting the proper meaning, śatārmśāśsaikā has been corrected as śatāṁśaghnaikā, ṣaṣtidineśa as ṣaṣṭirdineśa, kāṣṭhānta as kāṣṭhā jyā, apakramarāśi as apakramo rāśi and pādenyaḥ as pādebhyaḥ. As for TS, they have not touched this verse and the next two, saying, in so many words, that they cannot interpret them. NP’s interpretation of the three verses has also been affected by the highly corrupt text. लिप्ताशत [क] म (शीत्या) मेषे 'त्रिख' यु (क्त) 'मिन्द्रिय' 'मनू' (न) म् । गवि ('मनु') 'भव' 'मुनि' 'रूपै' - श्र [तु] गुणैः संयुतं च शतम् ॥ १७ ॥

IV. 18 IV. THREE PROBLEMS 89 नवतिस्त्रियुता षष्टिश्चत्वारिंश'च्छिवा'श्च मिथुना(न्ते) । मेषाद् गताऽऽगतमुदग्दक्षिणतोऽदस्तूलादिषु च ॥ १८ ॥ 17-18. The (promised) intervals of declinations in minutes for intervals of quarter-signs, are, in Sāyana Meṣa: 180 + 3, 180 + 0, 180 − 5, 180 − 14, in Sāyana Vṛṣabha, 100 + 4 × 14, 100 + 4 × 11, 100 + 4 × 7, 100 + 4 × 1, and in Sāyana Mithuna, 90, 63, 40 and 11. (Thus the intervals are 183, 180, 175, 166, 156, 144, 128, 104, 90, 63, 40, 11.) These are to be added successively to get the declinations from Meṣa (Aries) to Mithuna (Gemini). Then from Kar- kaṭaka (Cancer) to Kanyā (Virgo), these should be deducted in the reverse order, until at the end of Kanyā, the declination is zero. These declinations from Meṣa to the end of Kanyā are north. Then from Tulā (Libra) to the end of Dhanus (Sagittarius), the south declinations increase in the given order, and from Makara (Capricorn) to Mīna (Pisces) the south declinations decrease in the reverse order, until at the end of Mīna the declination is zero again.) Example 4. Find the declination of the ecliptic point ending (Sāyana) Capricorn. The end of Capricorn is rāśi 10-0-0. This falls between rāśis 6 and 12. ∴ The declination is south. The declination ending Sagittarius (i.e. beginning Capricorn) is 24° S. The declination at the end of Capricorn is 24° − 11′ − 40′ − 63′ − 90′ = 24° − 3° 24′ = 20° 36′S. The author has perhaps computed the declination by applying the formula of verses 16 and got the intervals by deducting the previous from the next. Or, these verses are taken in toto from the original Pauliśa and given here, for there are small differences from the computed values. Or, the differences are scribal errors. Both are given hereunder for comparison:

90 PAÑCASIDDHĀNTIKĀ IV. 19

19. (Plant a gnomon at the centre of) a circle having a diameter equal to four times the gnomon. Mark the two points where the shadow of the gnomon enters the circle and emerges from it. The line joining the points is the east- west line. The line drawn perpendicular to this by means of equal intersecting circles, is the north-south line. Though the east-west line is first asked to be drawn and the north-south next, as perpendicular bisector to the east-west, it will be better if the north-south line is first drawn by means of equal intersecting circles with the two points as centres. Then using the points of intersection of the north-south line and the original circle as centres, by the same means, the perpendicular bisector forming the east-west line can be drawn, which will pass through the centre as required. On the other hand, if the east-west line is first drawn by joining the first two points, another line parallel to it and passing through the centre is to be drawn as the desired east-west line. 19a. A. संकुश्रतुविस्तारे; C. शङ्कइगुलविस्तारे in V. 9c. as noted by the scribes. c. A. अपेरैद्री; C.D. अपेरैन्द्री Thus, B1 adds in the margin, प्रति पत्र एक; d. A. यवाश्च; C.D. यवैश्च B2. adds, प्रतिपत्र एके, and B3. adds प्रति नूं ∙ B. After कार्ये one leaf missing upto दिवा पत्र एक. And all add अग्रो नास्ति ।

IV. 21 IV. THREE PROBLEMS 91 The gnomon should be twelve units in length, not necessarily digits; no harm will result, pro- vided all measurements are given in the same units. The circle also can be of any desired diameter, not exactly four gnomons in length. We are not sure whether the word for ‘four’ occurs at all in the text, it is so corrupt in that part. We can only say it cannot be śaṅkvaṅgula as corrected by TS; the letters are so different. Finding the directions in the manner described is explained thus: The North is directed towards the north pole of the earth. Corresponding to this is the celestial north pole, (from which we can find the north, if we can only observe it correctly, and therefrom the other directions). At mid-day the Sun is on the meridian, and at equal times before and after, its altitudes and directions are equal, provided its declination does not change. As the Sun’s position is thus symmetrical, before and after noon, with the meridian as the line of symmetry, the gnomonic shadow is symmetrical with the north-south line (which corresponds to the meridian) as the line of symmetry. Therefore if two gnomonic shadows, one in the morning and one in the evening, of equal lengths, are marked on a horizontal surface, the bisector of the angle between the two shadows is the line of symmetry, and therefore the north-south line. From this the east-west line, which is its perpendicular bisector, is drawn. The circle, asked to be drawn, serves the purpose of marking the equal shadows. By the same symmetry, the ends of the shadows are at equal distances from the east-west line, and so the line drawn between them is also east-west, being parallel to the east-west line drawn. Therefore the author asks us to draw the east-west line formed by joining the two points first, and proceed. We have said that the Sun’s declination must be the same, i.e. does not change during the interval. But actually it changes. So if we do the work at a time of the year when the change in declination is very little, then the directions found will be nearly accurate. This happens near the solstices, and there- fore the work should be done when the sun is near the solstices. Methods to find the directions accu- rately even when the declinations are changing rapidly, are given by writers like Vaṭeśvara, Parameśvara etc., and also explained by Govindasvāmin in his commentary on the Mahābhāskarīya, III.1. [छायातः अक्षानयनम्] विषुवद्दिन [सममध्य] छायावर्गात् सवेदकृतुरूपात् | मूलेन शतं विंशं विषुवच्छायाहतं छिन्द्यात् || २० || लब्धं विषुवज्जीवा चाप [म] तोऽक्षोऽ [थवैव] मिष्टदिने | मेषाद्यपक्रमयुतस्तुलादिषु विवर्जितः स्वाक्षः || २१ || Latitude from Shadow 20. Measure the mid-day shadow on the day when the Sun is at the equinoxes (the equinoctial shadow), square it, add 144, and find the square root. By this divide the product of the shadow into 120. 21. The result is the sine of the latitude of the place, called Viṣuvajjīvā (or Viṣuvajyā). Its arc is the latitude. Or, do this work on any day and get the arc. If the Sun is in the six signs from sāyana-Meṣa, i.e. if the Sun’s declination is north, add the declination to the arc, the latitude is got. If the Sun is in the six

92 PAÑCASIDDHĀNTIKĀ IV. 21 signs from sāyana-Tulā, i.e. if the declination is south, subtract the declina- tion from the arc, the latitude is got. That is: i. Sine latitude = 120' × equinoctial shadow ÷ √(144 + equinoctial shadow²). The arc from this is the latitude. ii. Sine south zenith distance of the Sun, (SZD) = 120' × mid-day shadow ÷ √(144 + midday-shadow²). The arc of this is the SZD. Using SZD, Latitude = SZD±declination ('plus' should be used if the declination is north, and minus if south.) Example 5 (a). At a certain place the equinoctial shadow is 5 units. Find the latitude of the place. Sin. lat. = 5 × 120' ÷ √(5² + 144) = 600' ÷ 13 = 46' 9''. Arc of 46' 9'' = 22° 37'. The latitude is 22° 37'. Example 5 (b). At a place when the Sun is at the end of sāyana Tulā, the mid-day shadow is found to be 9 units. Find the latitude of the place. From the formula, (the mid-day-Sun's) sin SZD = 9 × 120' ÷ √(9² + 144) = 1080'/15 = 72'. SZD = arc of 72' = 36° 53'. The declination of the Sun at the end of Libra is 11° 44'S (from 16-18). Taking the minus sign, since the declination is south, the latitude = 36° 53' − 11° 44' = 25° 9'. Note: Rule (i) can be used everywhere, while rule (ii) should be used only if the midday sun is south of the zenith. If it is north, having north zenith distance, (NZD), declination = NZD = latitude. But the work being a Karaṇa, the author intends it to be used only in North India, where the midday zenith distance is always south, and hence this has not been mentioned by him. Further the author envisages only north latitudes by his formulae. Sky-sphere (Khagola) Here onwards, explanations require a knowledge of the sky-sphere (khagola) with the stellar sphere imposed on it. Therefore we shall describe the sky sphere. Hindu astronomers describe it as the ‘Casket Boundary of our universe’ (Brahmāṇḍa-kaṭāha-sampuṭa), and marked by the penetration of sunlight. Beyond that there is no sunlight. The measure of a great circle on the sky-sphere is said to be the number of yojanas the Moon, or the Sun or any planet moves in a kalpa (yuga according to the followers of Āryabhaṭa), — though, really, the sphere is only illusory and supposed to have an indefinite radius. As the stellar sphere also is enormous, we can take the surfaces of the two spheres sliding on each other, and forming spherical triangles by arcs on each intersecting those on the other. The problems will entail the solution of these triangles. (The formulae for solution have already been given). This will be understood by examining Fig. 6. 20-21. Quoted by Utpala on BS 2. pp. 59-60. 20a. A.सममधृष्टो; C.दिनमध्याह्नो                          d. A.C.D.छिंद्यम्     b. A. om-रूपात्र-om.                            12b. A. चापतोक्षो. A.C.D. ॰थवा यथेष्टदिने     c. A. शते. A.D.विंशात्                            d. A. खोक्षः

IV. 21 IV. THREE PROBLEMS 93 Fig. IV. 6 NESWZ is the sky-sphere. It is the sky as seen by an observer on the earth who fancies himself stationary, though taking part in the rotation of the earth, and thinks that the stellar sphere is rotating on the axis joining the celestial poles, NP, SP. NESW is the horizon marked by the cardinal points, North, East, South, West. (Note that the east and west points are interchanged so as to appear as we see them when looking up at the sky.) Z is the zenith, corresponding in the sky to the observer's position on the earth, and is the point where the line from the centre of the earth, through the observer, joins the sky-sphere. NNPZOS is the meridian. EZW is called the prime vertical. ENP is part of what is called unmaṇḍalam. ZSH is part of the vertical circle from the zenith to the horizon passing through the sun, moon, etc. (Note that the prime vertical is the vertical circle passing through the east and west points, and that the meridian itself can be considered as a vertical circle passing through the North and South points.) ZS is the zenith distance of S, and HS is the altitude. S₁SS₂ is the diurnal circle, the apparent path of S daily, due to earth's rotation. The stellar sphere (Fig.4) can be recognised here by the celestial equator ErROW, by the ecliptic EcrSC, by the north celestial pole NP, by the position of the Sun S, etc. and by the declination circle. NPSR, of which SR is the arc of declination. Because of the position of the observer on the earth with reference to the terrestrial North pole, the celestial North pole (NP) seems lifted up along the meridian from the north-point (N) so that its altitude is equal to the latitude of the place, and by this the celestial equator is depressed southward by the same amount from the prime vertical.

94 PAÑCASIDDHĀNTIKĀ IV. 21 Therefore the latitude of the place = NNP = ZO. (What we have said is for places in the northern hemisphere, i.e. north latitudes. In the southern hemisphere, i.e. at places of south latitudes, SP is lifted up from S, and the celestial equator is depressed northward by the same amount.) The complement of ZO, OS, is called the co-latitude (Lamba). Thus in triangles formed by great-circle- arcs of the stellar sphere and the sky-sphere, the latitude is involved directly or indirectly. The for- mulae for the solution of these triangles have been already given. Now, the two formulae for latitude can be proved by using the meridian, thus: see Fig.7 Ob:observer N: north point S: south point NS₂ZS₁OS₀S₃: The meridian Z: zenith O: point of intersection of meri- dian and celestial equator. S = S₂, S₁, S₀, S₃,: four positions of the mid-day sun. OS: Sun's declination Fig. IV. 7 As already described, OZ = latitude. On the equinoctial days at mid-day the Sun, S₀ is at O. ∴ ZS₀ (the south zenith distance of the Sun) = ZO = latitude (first formula). On other days, the Sun may be (i) south of O, (S₃), or (ii) north of O but south of Z, (S₁), (iii) north of O and north of Z, (S₂). i. Here, the latitude = OZ = S₃Z − S₃O = the south zenith distance of the Sun − the declina- tion. (second part of second formula). ii. Here the latitude = OZ = ZS₁ + S₁O = the south zenith distance of the sun + the declination (first part of the second formula). iii. Here the latitude = ZO = S₂O − S₂Z = the declination − the north zenith distance of the Sun. (This case is not given by the author). The zenith distance of the midday Sun used in the formulae is to be found thus: see Fig.8. EG = gnomon of 12 units ET = The midday shadow, TG = The shadow hypotenuse, ZGS = the zenith distance = angle TGE. Sin zenith distance (ZD) = sin ZGS = sin TGE = TE × 120′ ÷ TG = Shadow × 120′ ÷ Shadow hypotenuse = Shadow × 120′ ÷ √(shadow² + gnomon²) = Shadow × 120′ ÷ √(shadow² + 144)., (where shadow is in the units taken). From sin ZD, arc ZD is found. Fig. IV. 8