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

229
sidereal, we subtract the rising time of SA from the given
Sāvana time and the result will be the same, for,
— SA = — (S′A — S′S) = S′S — S′A. which means
subtracting SA from the time tantamounts to increasing
by S′S and subtracting S′A. This increase by S′S is add-
ing what have been defined as gati-kalas so that automati-
cally the Sāvana units got converted into sidereal units,
by taking the position S and subtracting SA instead of
taking the position S′ and subtracting S′A. The result is
the same. So, it is said ‘तात्कालिकार्कस्य’ in the beginning
of the verse 2.
In the second case mentioned in verse 4, if the time
given, falls short of the Bhogyāsus, then simply (x × 30) / T
where x is the time given in asus and T the rising time of
the Rāsi in which both the Sun and the lagna are then
situated gives the arc in degrees which if added to the
longitude of the Sun gives that of the lagna. Here also
the position ‘S’ counts.
Verses 5 to 6½. To find conversely the time that has
elapsed after Sun-rise given the lagna.
The Bhogyasus of the Sun and the Bhuktāsus of the
Lagna together with the rising times of intermediate
Rasis gives the time required.
If the Sun and the Lagna both be in the same Rāsi,
then the arc in between them, multiplied by T and divide
by 30, gives the time required.
If, however, the longitude of the Lagna falls short of
that of the Sun, ie. if the Sun be below the horizon, (in
this case SL > 180°) then finding the time of rising of SL
and subtracting from a day, we have the time of the Lagna
before Sun-rise.

230
However, here. there is one complication if we
consider the तात्कालिकार्क ie. s the Sun at the given time.
This position of s cannot be had unless we know the time,
which is itself required. So we get in the first place the
time pertaining to S'L, which is the time measured in
sidereal units the position S' being that of the Sun at Sun-
rise. If we don't take recourse to convert this time in
sidereal units to Sāvana units using the proportion
between them, then the alternative is to obtain the position
S using the time obtained and then calculate the time
again in Sāvana units.
If the time given to find the Lagna, be sidereal, it
goes without saying that we find it from S'. Also S'
being given and if the time after sun-rise is required in
sidereal units for a given Lagna, the method of succes-
sive approximation is unnecessary.
Verse 7. To find the Lagna before Sun-rise called
Vilōmalagna.
Suppose it be required to find the lagna before Sun-
rise, given the time before Sun-rise. Obtain the then posi-
tion of the Sun and find his Bhuktāsus; subtract them from
the given time; from the remainder, subtract the rising
times of as many Rāsis as could be, rāsis behind the Sun's
position. If R be the remainder in the time after these
subtractions, then R × 30/T where T is the rising time of
is the next preceding Rāśi, together with n × 30, where n
the number of integral Rāśis subtracted and the arc
of the next Rāśi by which the Sun has advanced in his
Rāśi at the time of the Lagna (known from the position
of S found) the sum total of these three items being
subtracted from the position of S gives the longitude of L.
Comm. Easy.
Verse 8. To obtain the East-West line.

231
The East-West line is roughly the join of the
extremities of the morning shadow as well as that in
the after-noon of a gnomon placed at the centre of a
circle drawn on a plane with any arbitrary radius,
when those shadows equal the radius of the circle. But
this line is to be deflected keeping its western point ie.
the extremity of the morning shadow fixed through a
distance K (sin δ₁ ~ sin δ₂) / cos ϕ at the eastern end perpendi-
cular to it, where the above distance is measured in units,
which measure K.
Comm. The east and west points are where the
celestial equator cuts the horizon. The east point is
thus the point where the Sun-rises when he is exactly at
the vernal equinox. The question is how to draw the
east-west line on a plane. For this we are asked to
draw a circle with any radius on that plane. The plane
is described here as अम्भःसुसमीकृतक्षिति that kind of sur-
face as is determined by the surface of water there. Such
a kind of surface forms approximately a horizontal plane
not of course exactly because such a surface is really
spherical, the earth being a sphere. But because the
radius of the earth is sufficiently large, we can take such
a surface to be a horizontal plane for all practical purposes.
Having drawn a circle place the gnomon vertical at the
centre. In the morning note the extremity of the shadow
when it equals the radius. In the afternoon also mark
the point when the shadow equals the radius. Join those
two points. It represents roughly the east-west line,
roughly because the Sun's declination changes in between
the two moments however small the change might be.
Ignoring the change in the declination, this line will
be east-west because of the following reason. The length
of the shadow is 12 tan z where 12 units are the measure
of the gnomon. But since the shadow on both the occasions
is equal to the radius of the circle z, the zenith-distance

232
will be the same on both the occasions. Then the spherical
triangles PZS₁ and PZS₂ where P is the celestial pole, Z the
Zenith and S₁ and S₂ are the positions of the Sun on the
two occasion, are congruent their three sides being respe-
ctively equal, provided we take PS₁ = PS₂ ie. 90 — δ₁ =
90 — δ₂ ie. δ₁ = δ₂ on both the occasions. When the
two triangles are thus congruent, PẐS₁ = PẐS₂ ie. S₁ and
S₂ are equidistant from the plane of the Prime-Vertical.
Hence the extremities of the shadows will be equidistant
from the East-West line ; or this may be seen in another
way ; S₁ S₂ will be perpendicular to the meridian plane
and as such parallel to the plane of the Prime-Vertical.
The correction mentioned in the verse is known as
the Agrāntara correction which was originally given by
Chaturvedā chārya and then accepted by Sripati. Why
it is called Agrāntara is because it is a change in what
are called Karnavrittāgras of the two occasions where we
shall see in due course that the formula for Karnavrittāgra
is (K sin δ) / (cos φ) where K is hypotense of the gnomonic triangle
formed by the gnomon and its shadow S at any place and
time. This correction is a very minute correction and
as a matter of fact could be ignored. But the fact that the
correction was cognized and correctly formulated testifies
to the knowledge of the sphere which the above
acharyas had. Assuming the formula of the Karnāgra
here (it will be proved by us later in this chapter) if
δ₁ and δ₂ be the declinations on the two occasions
respectively the Karnāgras will be (K sin δ₁) / (cos φ) and (K sin δ₂) / (cos φ) ,
K being equal on the two occasions because the shadows
are equal. Hence the correction being the difference of
the Agrās, it is [K (sin δ₁ ~ sin δ₂)] / (cos φ) as stated by Bhās-
kara,

233
We shall now prove it in modern terms from the
spherical triangle PZS. We have the formula sin δ =
sin ϕ cos z + cos ϕ sin z sin a where PZS = 90 — a, a
being the Hindu azimuth measured from the East point.
Multiply the above equation by K and divide throughout
by cos ϕ where K is called the Chāyākarṇa equal to
√(12² + S²), S being the gnomonic shadow at the moment.
Hence
K sin δ / cos ϕ = K cos z tan ϕ + K sin z sin a (1)
Fig. 31 Fig. 32
But from Fig. 31, K cos z = 12, K sin z = S, (2)
and from Fig. 32, S sin a = b where b is called the
Chāyābhuja ie. Chāyābhuja = K sin z sin a (3). Thus
K sin δ / cos ϕ = 12 tan ϕ + b.
But again from Fig. 33, when ☉ the Sun at vernal
equinox is on the meridian and as such has a meridian
zenith-distance equal to ϕ, 12 tan ϕ = s where s is called
the Vishuvat-chāyā or equinoctial shadow. Thus we have
K sin δ / cos ϕ = s + b (4).
Again if the Sun be on the horizon, from Fig. 34,
E ☉ is called the Agrā, A, so that from the triangle P ☉ N,
30

234
Fig. 33 Fig. 34
cos (90 — δ) = cos ϕ cos (90 — A) ie. sin δ / cos ϕ = sin A or
in the Hindu form H sin A = R (sin δ / cos ϕ) = Agrajyā (5).
This Agrajyā is in a circle of radius R, and if it be re-
duced to a circle whose radius is K, is will be (K / R) × (R sin δ / cos ϕ)
= (K sin δ / cos ϕ) which is called Karṇāgrā. Hence we have
Karṇāgrā = s + b which we shall write as a = b + s (6).
This is an important formula which is going to be
formulated later in verses 72, 73. In the above formula,
s being constant, by differentiating δa = δb which means
that the variation in the bhuja is on account of the
variation in the Karṇāgrā. If in Fig. 35, Cω', CE″ be
morning and evening shadows when they are equal as per
the verse under comment, Mω' = b the morning bhuja,
NE″ = b', the evening bhuja dE' is the variation in the
bhuja ie. b — b' = δb which is formulated and equal to
δa. But δa = δ(K sin δ / cos ϕ) = K δ (sin δ) / cos ϕ, ϕ being con-
stant and K also being constant because the shadow
S is constant and K = √(S² + 12²) = constant on both
the occasions. ∴ δb = δK = (K / cos ϕ) (sin δ₁ — sin δ₂)
as stated by Bhāskara.

235
Fig. 35
Verse 9. Having determined the East-West line as
mentioned, by drawing with a compass with the same radius
on either sides of Eω (Eω is the East-West line through the
centre of the circle drawn parallel to the rectified East-
West line namely E′ ω′ where E″ ω′ is the approximate
East-West line obtained by the join of the extremities
of the shadows) two arcs which intersecting each other
form a fish-like enclosure as shown in the figure and
as such called Matsya meaning a fish, and joining the
ends of the fish namely f, g and producing fg, we have
the south and the north points s and n. Or again,
the north-south line could be had from Fig. 36, where
EC is a rod held in the direction of the north-pole
as seen by the eye at e and EA and CB the lines indicated
by the plumb-line called अवलम्बकसूत्र A and B being on
the plane and AB joined passing through N, the north-
point. Or again the directions could be determined as
follows from a single shadow S namely Cω′ in Fig. 35,

236
calculating the bhuja ω'M and Koti
CM; Bhuja and Koti being known,
and the shadow Cω' being drawn,
holding two rods whose lengths are
equal to the bhuja and Koti perpendi-
cular to each other, one extremity of
the bhuja-rod being held at ω' and
one extremity of the Koti-rod being
held at C, and the rods making a
right angle at M, then the bhuja-rod
determines the north-south direction
and the Koti-rod the East-West dire-
ction.
Fig. 36
Verse 10. The Bhuja is defined as the distance of the
extremity of the shadow from the east-west line where
the S'anku or gnomon is placed at the intersection of
Eω and NS. Koti = √(S² − b²) ∴ Koti will be in the
East-West direction. So Chayā-koti = K sin z cos a (7).
Comm. Easy.
Verse 11. The Chayākarṇa, K is equal to √(S² + 12²),
so that √(K² − 12²) = S or √[(K + 12) (K − 12)] = S.
Comm. Easy.
Verse 12. The S'anku is also called Nara or Nā.
The zenith-distance of the Sun at Noon when the Sun
is in ♈ is the latitude of the place, called pala or
Aksha; the altitude then is called lamba or colatitude.
Comm. The word S'anku we have previously used
for the gnomon. It is also used for the H cosine of
the zenith-distance and to differentiate it from the
previous S'anku called Dwādasāṅgul'a-S'anku or
twelve-unit-length S'anku, it is termed Mahā S'anku
and occasionally Iṣṭa-S'anku. Mahā S'anku or Iṣṭa
S'anku = H cos z (8). Thus in figure (37) ☉M =
H cos z. In the fig. where gn = gnomon, go = S

237
the shadow, ☉ = the position of the Sun whose zenith-
distance is ☉ z. If z' be taken as the zenith ☉ z' meas-
ures the zenith-distance whereas if z be taken as
the zenith ☉ z is the zenith-distance. The apparent
inconsistency that both ☉ z' and ☉ z are taken
as the zenith-distance is not there if we consider the
^ ^
zenith-distance as the angle ☉ nz' = ☉ oz. O ☉ = R,
☉ z = z, so that ☉ L = H sin z, and LO = ☉ M =
H cos z = Śanku or Nara or Nā. it is called Nara or
Nā which means man, because
a man may consider himself
as a gnomon, which is called
Śanku. So the word is also
applied to the parallel H cos z
parallel to the gnomon and
called Mahāśanku. H sin z is
called Drigjyā (9) zenith-dis-
t

238
Fig. 38
plane of the meridian - where E₁ is the centre of
the sphere.
(4) The fourth latitudinal triangle is E₁ S₁ F₁ the
projection of ESF (Fig. 21) on the meridian plane
where E₁ F₁ is the Sama-Sanku or the Sanku of the
celestial body when it is on the prime vertical. E₁ S₁
is the Agrajyā as mentioned, S₁ F₁ is what is called
Taddṛti.
(5) The fifth latitudinal triangle is E₁ B₁ F₁ where
E₁ B₁ is Krāntijyā or H sin δ, E₁ F₁ is the Sama-
Sanku defined above and B₁ F₁ is what is called the
higher segment of the Taddṛti which is equal to
Taddṛti minus Kujyā or Kshitijyā.
(6) The sixth latitudinal triangle is E₁ D₁ B₁
where E₁ D₁ is called the first segment of Agrajyā, D₁ B₁ is
what is called un-mandala Sanku or H cos z of the
celestial body when it is on the unmandala or the
Equatorial horizon and E₁ B₁ Krāntijyā.

239
(7) The seventh latitudinal triangle is D₁ S₁ B₁
where D₁ S₁ is the second segment of the Agrajyā, S₁ B₁ is
the Kujyā, and D₁ B₁ is the unmandala Sanku.
(8) The eighth latitudinal triangle is B₁ L₁ F₁
where B₁ L₁ is equal to the first segment of Agrajyā,
L₁ F₁ is the higher segment of Agrajyā, L₁ F₁ is the
higher segment of the Sama-Vritta-S'anku, and F₁B₁ is
the upper segment of Taddṛti mentioned before.
Comm. It was already mentioned that a latitudinal
triangle is such a right-angled tri-angle constituted by
the chords of the celestial sphere where the angles in the
triangle are ϕ, 90 − ϕ, 90°. The side opposite to ϕ is
called the Bhuja, that opposite to (90 − ϕ) is called
Koti and the third Karna. Such triangles are not only
eight as have been mentioned, but many more will be
there as mentioned by Bhāskara. They are all formed
as mentioned by him by the intersections of the diurnal
paths and the celestial equator with the circles of the
sphere namely horizon, prime vertical, meridian,
Equatorial horizon and declination circles. These
circles clearly intersect at ϕ or 90 − ϕ. The eight
triangles mentioned are those whose elements will be
entering computations. There is another important
latitudinal triangle with which we have to deal later
namely that formed by what is called Hriti, S'anku, and
S'ankutala (Fig. 39).
O₃K = Agrā ; O₃C = S'anku-tala ; CN = S'anku-bhuja
= CK ∴ Agrā = Sankutala + Sanku-bhuja
Agrā = R (H sin δ / H cos ϕ) ; S'ankutala = H cos Z tan ϕ
S'anku-bhuja = (H sin z H sin a) / R (Ref. fig. 40')
∴ R (H sin δ / H cos ϕ) = H cos z tan ϕ + (H sin z H sin a) / R I

240
Fig. 39
or in modern terms sin δ = sin ϕ cos z + cos ϕ sin z sin a
as derived from the triangle PZS. Formula I is with
respect to the Mahā Śaṅku ⊙ L of fig. 40; if it be reduced
to the Śaṅku of the gnomon the Śaṅku-bhuja ⊙ N becomes
the Chāyābhuja pr which will be equal to
(H sin z H sin a / R) × (K / R) = K sin z sin a whereas,
Agrā becomes (R H sin δ / H cos ϕ) × (K / R) = (K sin δ / cos ϕ) which is called
Karṇāgrā and Śaṅku-tala becomes
(H cos z tan ϕ × K / R) = K cos z tan ϕ = 12 tan ϕ = s,
which is a constant quantity. It will be noted that
Karṇāgrā differs from point to point since K differs from
time to time during a day. In fact Karṇāgrā is the
perpendicular dropped from the extremity of the shadow

pages! The page itself is in English, with English text. We should transcribe the text as it is (English, transliterated Sanskrit, numbers, symbols).

Let's look at:
^ ^ ^ ^
vs
the angles Â, B̂, Ĉ, D̂, the angles between the vertical
Wait, what if we do:
Wait, let's look at:
If an OCR line detector is used, it detects:
^ ^ ^ ^
Wait, does Tesseract detect ^ ^ ^ ^? Usually Tesseract might detect it as ^ ^ ^ ^ or ^ ^ ^ ^.
Wait, but if it's considered a line:
Let's see what happens if someone DOES consider it a line:
If we omit it, we have 2 fewer lines!
Wait, let's count total lines:
With carets as separate lines:
241
on the line parallel to the East-West line and north of it
...
If we look at the image, those carets are very distinct, printed with substantial space:
^ ^ ^ ^
Wait, let's look at the first caret line:
^ ^ ^ ^
It is indented to match A, B, C, D.
And the second caret line:
` ^ ^

242

Bb = Sama-S'anku, and Dd unmandala S'anku. O₂b is not
actually the Agrajyā but parallel and equal to it, since
Agrajyā is the H sine of SE of fig. 21, which is the per-
pendicular from S on Eω. Similarly do₄ is not the second
segment of Agrajyā but a parallel and equal segment.
Thus Bb is the perpendicular distance between EE' and
FF' ie. Eω and a parallel through F to Eω (fig. 21.)
Dd is the perpendicular distance between DD' and BB';
bo₂ is the perpendicular distance between Eω and SS'; do₄
the perpendicular distance between DD' and SS' (fig. 21) ;
Bo₂ the perpendicular distance between FF' and SS' ; Do₄
the perpendicular distance between BB', SS'. In this
fig. 39, O₁A is called Hṛti, Co₃ = Ishta-Hṛti or any arbitrary
hṛti. Taddhṛti Hṛti and Kujyā are special cases of
Ishtahṛti. Hṛti is the maximum of Ishtahṛti. Calling
Aa, Bb, Cc, Dd S'ankus in general ao₁, bo₂, co₃, do₄ are
called S'ankutalas. Aa is called Dinārdha-S'anku or the
S'anku of the mid-day ; whereas Dd is the unmandala-
S'anku and Bb Sama-S'anku. Cc is called Ishta-S'anku.
Perpendiculars from A, B, C, D on the plane of the prime
vertical are called S'anku-bhujas. Since B is on the
prime-vertical itself, the S'anku-bhuja is zero and at
this point BO₂ is Agrajyā. In the arbitrary case at C,
the perpendicular from C on the plane of the prime-
vertical being S'anku-bhuja, which is equal to the per-
pendicular from C on Eω, and Co₃ being the S'anku-tala,
and since the perpendicular distance between ML and
Eω is the Agrajyā, which is equal to the sum of O₃ C and
the S'anku-bhuja S'anku-tala + S'anku-bhuja = Agrajyā.
(10) which is a different expression of (5). We shall
present this analytically. Putting S'anku = H cos z, and
using the latitudinality of Cco₃ (H cos z) / 12 = Co₃ / s = Co₃ / K
applying similarity with the first fundamental latitudinal
triangle where s = equinoctial shadow, and K the
Viṣuvat-Karṇa. Hence we have

243
Co₃ = Ishta-hṛti = K/12 H cos z = (H cos z)/(H cos φ) = (cos z)/(cos φ) (11)
Co₃ = Śanku-tala = s/12 H cos z = H cos z tan φ
= (H cos z H sin φ)/(H cos φ) = R cos z tan φ (12).
From the triangle PZS, we have the formula
sin δ = sin φ cos z + cos φ sin z sin a written under verse 8
∴ (sin δ)/(cos φ) = cos z tan φ + sin z sin a or in the Hindu form
(R sin δ)/(cos φ) = H cos z tan φ + (H sin z H sin a)/R . I.
We saw under verse 8 that (R sin δ)/(H cos φ) = (RH sin δ)/(H cos φ)
= Agrajyā (formula 5). Also we have from (12) above
H cos z tan φ = Śanku-tala. From fig. 40, if SM be drawn
secondary to the prime-vertical, the right-angled spherical
triangle SMZ gives
[L14

244
Hṛti = O₁A = OO₁ + OA
= Kujyā + H cos δ or
Dhujyā VI since OO₁ =
DO₄ = Kujyā. The line in
the Equatorial plane cor-
responding to Kujyā is
Charajyā which is equal to
R tan φ tan δ (13) so that
we can write Kujyā =
R sin δ tan φ = H sin δ tan φ
(13'). Also from VI
[Fig. 40]
Antyā = R + Charajyā
= R + R tan φ tan δ (14)
which gives us the duration
of half the day where R
gives 6 hrs and Charajyā
the increment in day due
to φ and δ.
[Fig. 40]
Verse 18. To obtain the
magnitudes of the various
chords or the elements of
the latitudinal triangles.
The elements or the sides of all these latitudinal
triangles are mutually derivable from similarity.
Comm. Easy.
Second half of Verse 18. The radius multiplied by
the Kotis or Bhujas and divided by the Karnas gives
H cos φ and H sin φ.
Comm. The seven triangles except the second, of the
eight latitudinal triangles are compared here with the
second. Remembering Bhujas are the sides of the
triangles opposite to φ and Kotis opposite to 90 − φ,

245
Bhuja / Karṇa in any triangle = Bhuja / Karṇa in the second latitudinal
triangle = (H sin φ) / R ∴ (R × Bhuja) / Karṇa = H sin φ Akshajyā
or Palajyā (15) Similarly Koti / Karṇa in any triangle is equal
to the Koti / Karṇa in the second latitudinal triangle = (H cos φ) / R
so that (R × Koti) / Karṇa = H cos φ = lambajyā (16).
Verse. The arcs of H sin φ and H cos φ are respec-
tively the Akshamsas and lambamsas as they are called ie.
latitude and colatitude. H sin φ and H cos φ are also
obtainable thus √(R² - H sin² φ) = H cos φ and H sin φ =
√(R² - H cos² φ) Or again (H sin φ × Koti) / Bhuja = cos φ and
(H cos φ × Bhuja) / Koti = H sin φ where the Bhuja and Koti
may belong to any latitudinal triangle.
Verse 20. Agrajyā can be had by multiplying Krāntijyā
by Karṇa of any lat. triangle and divided by its Koti.
Also Sama-Sanku = Karṇa / Bhuja × Krāntijyā and
(Sama-Sanku × Karṇa) / Koti = Taddhṛti.
Comm. The first of these statements pertains to the
similarity of the third triangles to the others. The second
of the statements pertains to the similarity of the fifth
latitudinal triangle to others whereas the third pertains to
that between the fourth and the others.

246
Thus Sama-Śaṅku or S.S.
= (H sin δ × R) / (H sin φ) or (R sin δ) / (sin φ) (17).
Taddhṛti = R × (H sin δ) / (H sin φ) × R / (H cos φ) = (R sin δ) / (sin φ cos φ) (18).
Verse 21. Taddhṛti = (Karṇa × Agrajyā) / Bhuja
Comm. This pertains to the similarity between the
fourth latitudinal triangle and the others.
Latter half of Verse 21 and first half of Verse 22.
Sama-Śaṅku = (Taddhṛti × Koṭi) / Karṇa = (Agrajyā × Koṭi) / Bhuja
(Sama-Śaṅku × Bhuja) / Koṭi = Agrajyā.
Comm. The first statement is made out of the
similarity between the fourth lat. triangle and the others
whereas the second statement and the third as well are
made out of the similarity between the third and others.
II half of Verse 22. Sama - Śaṅku = Upper
segment of Taddhṛti × Karṇa / Koṭi
Comm. The similarity is between the fifth triangle
and others.
Verse 23. Kujyā = Krāntijyā × Bhuja/Koṭi
Upper segment of Taddhṛti = (Krāntijyā × Koṭi) / Bhuja and
Kujyā + Upper segment of Taddhṛti.
Comm. The first statement is through the similarity
of the third triangle with others, whereas the second is
through the similarity of the fifth with the others. The
third statement is clear from Fig. 21.

247
Verse 24. (Kujyā × Bhuja) / Karṇa = second segment of
Agrajyā (Krāntijyā × Koṭi) / Karṇa = first segment of Agrajyā
and Agrā = Sum of the two segments.
Comm. The first statement is based on the similarity
between the seventh triangle and others and the second on
the similarity between the 6th and the others.
Verse 25. (First segment of Agra × Bhuja) / Koṭi
= Un-mandala-Śaṅku and (Krāntijyā × Bhuja) / Karṇa
= Un-mandala-Śaṅku. (19)
Comm. Both the statements are based upon the
similarity of the sixth triangle and others. Thus un-
mandala śaṅku = U.S. = (H sin δ × H sin φ) / R =
= R sin δ sin φ.
Verse 26. (First segment of Agrā × Koṭi) / Bhūja
= Un-mandala-Śaṅku = Kujyā × Koṭi / Karṇa.
Comm. The first statement is based on the similarity
of the sixth latitudinal triangle and others and second
that between the seventh and others.
Sama-śaṅku − unmandala śaṅku = upper segment of
Sama-śaṅku.
Verse 27. (Agrā × Bhuja) / Karṇa = Kujyā
Taddhṛti − Kujyā = upper segment of Taddhṛti.

248
Comm. The first statement is based on the similarity
between the third triangle and the others and the second
statement is evident.
Second half of verse 27. Other elements could be
derived from what is already known and from what has
been obtained. Also by alternando and invertendo we
could pass from one element to the other and vice-
versa.
Verse 28. Karṇa = √(Bhuja² + Koti²)
Bhuja = √(Karṇa² − Koti²)
Koti = √(Karṇa² − Bhuja²)
Thus the third could be had from the other two in all
the triangles.
Verse. There are sixty-three ways of obtaining H sin ϕ
and H sin ϕ. On account of hundreds of ways of
obtaining Agrajyā etc., there are an infinite number of
ways of obtaining H cos ϕ etc.
Comm. Under verse 23 Bhāskara says that there are
98 ways of obtaining Taddhṛti. Taking the third latitude
triangle, H sin δ could be obtained in seven ways, from
this H sin δ, Kujyā could be obtained in seven ways;
hence, according to the principle of association namely
that when one thing could be done in m ways and another
in n ways, both the operations could be together performed
in mn ways, so Kujyā could be obtained in 7 × 7 = 49
ways; similarly the upper segment of Taddhṛti could be
had in 49 ways; so that adding the two Taddhṛti could be
obtained in 98 ways.
Similarly suppose we have to find H sin δ. H cos ϕ
could be found in seven ways and from H cos ϕ, H sin δ
could be found in seven ways. Thus H sin ϕ could be