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

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

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.