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

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.

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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

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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

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∴ (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

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(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.

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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.

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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.

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Verse 102. In a place where s = 5″, the sum of
H sin δ, S. S., Taddhṛti, Kujyā and Agrā is 6500 ; find
them individually oh, mathematician, if thou art adept in
understanding the sphere and dealing with the latitudinal
triangles.
Verse 103. Answer to the problem above.
Assuming H sin δ to be equal to 12 s and computing
the various quantities cited ; take their sum. Then by
rule of three “ If for this sum got, the individual magni-
tudes are such and such what will they be for the given
sum ” each can be had.
Comm. The cited magnitudes are respectively H sin δ,
(R H sin δ) / (H sin ϕ), (R² H sin δ) / (H sin ϕ H sin ϕ), (H sin δ H sin ϕ) / (H sin φ) and
(R H sin δ) / (H cos ϕ) which are all proportional to H sin δ, ϕ being
given through ‘ s ’. With this idea of proportionality at
the back of his mind, Bhāskara sets this ingenious ques-
tion, and gives an easy way of solving it by assuming
H sin δ to be 5 × 12 = 60, so that the others can be got
rationally.
With this H sin δ, S. S. (3438 × 60) / (3438 × 5/13) = 156, Taddhṛti =
(3438² × 60) / (3438 × 5/13 × 3438 × 12/13) = 169 ; Kujyā = (60 × 5) / (13 × 12/13) = 25
Agrā = (3438 × 60) / (3438 × 12/13) = 65
The sum of these is 475. So, by the rule of three men-
tioned above, H sin δ = 1200, S. S. = 3120, Taddhṛti =
3380, Kujyā = 500 and Agrā = 1300.
Or alternatively given s = 5, k = 13 so that H sin ϕ =
(3438 × 5) / 13, H cos ϕ = (3438 × 12) / 13. Hence the values of

327
the various magnitudes are H sin δ, (H sin δ × 13) / 5 ,
(H sin δ × 13²) / 60 , (H sin δ × 5) / 12 and (H sin δ × 13) / 12
The sum of these is H sin δ (1 + 13/5 + 169/60 + 5/12 + 13/12) =
H sin δ ((60 + 156 + 169 + 25 + 65) / 60) = 475/60 H sin δ = 95/12
H sin δ = 9500 ∴ H sin δ = 1200 from which by substi-
tution the remaining magnitudes could be obtained.
Verse 104. If the sum of Agrā, H sin δ and Kujyā be
2000 find them individually, oh ! mathematician if thou
be an adept in the geometry of the sphere and compu-
tation.
Comm. Here the quantities are respectively
(R H sin δ) / (H cos φ), H sin δ and (H sin δ H sin φ) / (H cos φ) so that their
sum is H sin δ (1 + R / (H cos φ) + (H sin φ) / (H cos φ))
= (H sin δ (H sin φ + H cos φ + R)) / (H cos φ)
Here also we are to presume s = 5 so that the above
sum is (H sin δ (s + 12 + k)) / 12 (by proportion of the first and
second latitudinal triangles) = (H sin δ (5 + 12 + 13)) / 12 =
5/2 H sin δ = 2000 ∴ H sin δ = 800. Substituting this
value in the above formula, Agrā
= (RH sin δ) / (H cos φ) = (3438 × 800 × 13) / (3438 × 12) = 10400 / 12 = 866-40 ;
Kujyā = (H sin δ H sin φ) / (H cos φ) = (800 × 5) / 12 = 4000 / 12 = 333-20

328
OG = Gnomon ; OP = Shadow of the gnomon ;
PM = the Bhuja drawn from the extremity of the shadow
P perpendicular on the East-west line ; OM = Koti of the
shadow extending along the East-west line. AB is the
Nalaka placed along the Chayakarṇa PG. The eye is
placed at A and the planet ☉ is visible through the tube
of the Nalaka AB.
Fig. 62
Verses 105, 106 and 107. The method of observing
through the instrument called Nalaka, the planetary
position.
On a horizontal plane mark a point and through it
draw the East-west line and also the North-south ; if the
planet is in the East mark off the computed Koti of the
shadow towards on the East-west line; if the planet is in
the Western hemisphere, mark this Koṭi towards the East.
From the extremity of the Koti mark the computed Bhuja
perpendicular to the East-west line and draw the computed
shadow from the point so as to form a right-angled

329
triangle with the Bhuja and Koti. Extend a thread from
the point of intersection of the bhuja and shadow to meet
the gnomon's top so as to form the Chāyākarṇa or the
hypotenuse of the right-angled triangle of which the other
sides are the gnomon and the shadow. Along this thread
place the Nalaka such that the lower extremity of the
Nalaka coincides with the eye. Seeing through the Nalaka,
the planet is to be seen. I shall tell how the planet could
be seen in water as well.
Comm. The Nalaka is a simple tube formed generally
of bamboo. The purpose of this is to verify the correct-
ness of the computation of the shadow and its bhuja. If
the computation is wrong the planet will not be seen in
that direction. It might be asked how the shadow and
bhuja are pertinent with respect to a planet, whose shadow
cannot be observed as that of the Sun. True, but the
computation of the shadow and bhuja are done as will be
done with respect to the Sun, knowing the declination etc.
as in the case of the Sun. Computation does not depend
on the observation of the actual shadow. Computing the
magnitudes of the Bhuja and Koti, the direction of the
Chāyākarṇa points to the planet in the sky.
Verse 108. Observing the planet through the Nalaka
in water.
Fig. 63
42

330
Place the Śaṅku at the point of intersection of the
Bhuja and shadow and holding the Nalaka along the join
of the top of Śaṅku and the point, the planet could be seen
in a basin of water placed at the point.
Comm. Let P be the planet casting the shadow AC
of the gnomon AB. C the extremity of the shadow is the
point of intersection of the shadow and the Bhuja.
Though we have shown the gnomon in the position AB,
it need not have been placed there in as much as we have
the computed magnitudes of the shadow, the Bhuja and
the Koti. Now we are directed to place the Śaṅku act-
ually at C the point of intersection of the shadow and the
Bhuja. Thus CD is the Śaṅku. Since CD=AB and
and both are vertical evidently Δs DBA and DCB
are congruent. Hence DĈB = DÂB. But DĈB = zenith-
distance of the planet and as such is equal to BÂD (also
the zenith-distance of the planet) ∴ DÂB = BÂP.
Hence if a tray of water is placed at A, the planet will be
visible as seen through DE, the Nalaka since the angle
DAB is the angle of incidence and BÂP the angle of
reflection are equal.
Verse 109. The planet is to be shown to the king,
who has an eye of appreciation for the same, either direct-
ly (as shown in fig. 62) in the sky or through water as
shown in the fig. 63, having finished the preliminaries
indicated.
Comm. Clear.
End of the Tripraśnādhyāya.