Exercise 7A
1 a
 180

 9
20

3 a sin 0.5c  0.479
c
b
c
 180

 12
15

5 180

 75
12

π 180
d

 90
2
π
e
f
g
h
i
7π 180

 140
9
π
7π 180

 210
6
π
5π 180

 225
4
π
3π 180

 270
2
π
3π 
180
 540
π
2 a 0.46 
b 1
180
 26.4

180
 57.3

c 1.135 
d
3
e 2.5 
180
 65.0

180
 99.2

180
 143.2
π
f 3.14 
g 3.49 
180
 179.9
π
180
 200.0
π
b cos 2  0.156
c tan1.05c  1.74
d sin 2c  0.909
e cos3.6c  0.897
4 a 8

2

180
45
b 10 



180 18
c 22.5 



180 8
d 30 



180 6
e 45 
π
π

180 4
f 60 
π
π

180 3
g 75 
π
5π

180 12
h 80 
π
4π

180 9
i 112.5 
π
5π

180 8
j 120 
π
2π

180 3
k 135 
π
3π

180 4
l
200 
π 10π

180
9
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1
m 240 
π
4π

180 3
n 270 
π
3π

180 2
o 315 
π
7π

180 4
p 330 
π 11π

180
6
5 a 50 
b 75 

 0.873 rad
180

 1.31rad
180
c 100 

 1.75 rad
180
d 160 

 2.79 rad
180
e 230 

 4.01rad
180
f 320 

 5.59 rad
180
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2
Exercise 7B
1 a Using l  r :
i
2
l  6  0.45  2.7
ii l  4.5  0.45  2.025
iii l  20 
b Using r 
i
r
3
  7.5π (23.6 to 3 s.f.)
8
l

:
10
50

0.6
3
ii r 
1.26
 1.8
0.7
iii r 
1.5π
12 18
 1.5 

 3.6
5
5
5
12 π
c Using  
l
:
r
i  
10
4

7.5 3
ii  
4.5
 0.8
5.625
iii  
The total angle at the centre is 6x so
6 x  2π
π
x
3
Using l  r to find the minor arc AB :
π 10π
l  10  
cm
3
3
3
12
2 3

2
3
3
Triangle OAB is equilateral, so AOB 
π
3
Using l  r :
l  6
π
 2π
3
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1
4 r  10 cm and   5 rad
Using l  r :
l  10  5  50 
7
25  2  5 2
5 a Using l = rθ :
length of shorter arc  3  0.8  2.4 cm
length of longer arc  (3  2)  0.8  4 cm
Perimeter  2.4 cm  2 cm  4 cm  2 cm
 10.4 cm
b Length of shorter arc  3 cm
Length of longer arc  5 cm
So perimeter  (3  5  2  2) cm
As the perimeter = 14 cm,
8  4  14
8  10
10

 1.25 rad
8
Using l  r :
the arc length of the sector  15 cm
So the perimeter  (15  30) cm
As the perimeter  42 cm,
15  30  42
15  12
12

 0.8
15
8 a COA  π 
2
π
π
3
3
b The perimeter of the brooch
 AB  arc BC  chord AC
AB  4 cm
6
2
π
3
2
4
So length of arc BC  2  π  π cm
3
3
π
As COA 
(60), triangle COA is
3
equilateral.
So length of chord AC  2 cm
4
So perimeter  4 cm  π cm  2 cm
3
4
  6  π  cm
3 

l  r with r  2 cm and  
Using l  r , the arc length  1.2r cm.
The area of the square  36 cm 2 , so each
side  6 cm and the perimeter is, therefore,
24 cm.
The perimeter of the sector
 arc length  2r cm
 (1.2r  2r ) cm  3.2r cm
Perimeter of square  perimeter of sector, so
24  3.2r
24
r
 7.5
3.2
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2
11
9
Length of arc = rθ
Perimeter  2r  r
2r  r  2r
2r  r
  2 rad
2π
π

24 12
3π
r 
2
3π
π
r

 18 m
2 12
d  36 m
12 a  
Using the circle theorem, the angle subtended
at the centre of a circle  2  angle subtended
at the circumference:
AOB  2ACB  0.8 rad
Using l  r :
length of minor arc AB  8.5  0.8
 6.8 cm
OC  R  r
sin 
( R  r ) sin 
R sin   r sin 
R sin 
13 a SR  7  0.5  3.5 m
b Using the cosine rule:
QR 2  7 2  122  2  7  12  cos 0.5
10 a OC  R  r
b
b C = πd = 36π
36π  60  60
Speed 
30  1000
 13.6 km/h
r

Rr
r
r
 r  r sin 
 r (1  sin  )
c R sin θ = r(1 + sin θ)
3
3
R  r 1  
4
4

3
r R
7
3
sin      0.848...
4
2 R  2 R  21
QR  6.75 m
SQ  PQ  PS  12  7  5 m
Perimeter  6.75  5  3.5
 15.3 m (3s.f.)
14 a XOZ 
2π  1.1
 2.59 rad
2
b Using the cosine rule:
XZ 2  52  152  2  5  15  cos 2.59
XZ  19.44 mm
Arc length YZ  5  1.1  5.5 mm
Perimeter  19.44  2  5.5  44 mm
2 R  1.696 R  21
3.696 R  21
R  5.681cm
3
r   R  2.43 cm
7
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3
Exercise 7C
1 a Area of shaded sector
1
  82  0.6  19.2 cm2
2
2 a
b Area of shaded sector
1
 27
  92  
 6.75 cm2
2
6
4
c Angle subtended at C by major arc
 9
 2  
5
5
Area of shaded sector
The triangle is equilateral, so the angle

at C in the triangle is .
3
Angle subtended at C by shaded sector
 2
  
3
3
Area of shaded sector
1
2 16
  42 

 cm 2
2
3
3
1
9 162
  1.22 

 1.296 cm 2
2
5
125
d Area of shaded segment
1
=  102 (1.5  sin1.5)  25.1cm2
2
e Area of shaded segment
1


=  62   sin 
2
3
3
 6  9 3  cm2  3.26 cm2
b
f Area of shaded segment
1

 
   62    62   sin  
4 
4
2
36 36
2


8
2
2
63
 
 9 2  cm 2  111.7 cm 2
 2

 36 
The triangle is isosceles, so the angle at C
in the shaded sector is 0.4 rad.
Area of shaded sector
1
  52  0.4  5 cm 2
2
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1
3 a
c
Area of shaded sector
1
  x 2  1.2  0.6 x 2
2
So 0.6 x 2  12
Area of shaded sector
1
  4.52  x
2
1
So 2   4.52 x
2
40
x
 1.98 (3 s.f.)
4.52
x 2  20
x  4.47 (3 s.f.)
b
4
Area of shaded sector
1
  1 2 23
 x 2   2 
 x 
2
12  2
12

23 2
So 15 
x
24
24  15
x2 
23
x  3.96 (3 s.f.)

Using l  r :
4  6
2

3
So area of sector 
1
2
 62   12 cm 2
2
3
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2
5
7
a
102  102  18.652
2  10  10
 0.739 (3 s.f.)
cos  
b cos   0.739...    2.4025...
1
Area   102  2.4025...
2
 120 cm2 (3 s.f.)
6 Using area of sector 
1 2
r:
2
1
 12 2 
2
100 25
 

rad
72
18
100 
The perimeter of the sector
 12  12  12  12(2   )
61 122
2
 12 

 40 cm
18
3
3
a The perimeter of minor sector AOB
 r  r  0.5r  2.5r
So 30  2.5r
30
r
 12
2.5
b Area of minor sector AOB  1 r 2
2
1
2
2
  12  0.5  36 cm
2
c Area of segment
1
 r 2 (  sin  )
2
1
  122 (0.5  sin 0.5)
2
 72(0.5  sin 0.5)
 1.48 cm 2 (3 s.f.)
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3
8
9
π
12l
x
12
π
1
Area of sector  r 2
2
2
1
12l 
π
  
 
2  π  12
1 12l 2
 
2
π
2
6l

π
a l  r  l  x 
b
6l 2
 24  3600π
π
l 2  25π 2
l  5π
The arc length of AB is 5π cm.
c x
Using the formula,
1
ab sin C :
2
1
area of triangle COB  r 2 sin 
(1)
2
AOC  π   , so area of shaded segment
1
 r 2 (π   )  sin(π   ) (2)
2
As (1) and (2) are equal:
1 2
1
r sin   r 2  π    sin  π   
2
2
sin   π    sin(π   )
But sin(π   )  sin  ,
so sin   π    sin 
Hence   2 sin   π
area of a triangle 
12l 12

 5π  60
π
π
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4
10
11
1 2
r
2
with r  8 cm and   1.6 rad
Area of sector OBC 
So area of sector OBC
1
  82  1.6  51.2 cm 2
2
Using area of triangle formula:
area of triangle OAD
1
  5  5  sin1.6  12.49...cm 2
2
So area of shaded region
 51.2  12.49...  38.7 cm 2 (3 s.f.)
In right-angled triangle OBA :
π
AB
π
tan 
 AB  3.6  tan
3
3.6
3
So area of triangle OBA
1
π
  3.6  3.6  tan
2
3
and area of quadrilateral OBAC
π
 3.62  tan  22.447...cm 2
3
Area of sector
1
2
  3.62    13.57...cm 2
2
3
So area of shaded region
 22.447...  13.57...  8.88 cm 2 (3 s.f.)
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5
12
13 Area of sector  A cm2  1  282  
2
Perimeter of sector  P cm
 r  2r  (28  56) cm
2
1
r  8  cm2
2
1 2
Area of sector OAD  r  cm 2
2
So area of shaded region ABCD
As A  4 P :
392  4(28  56)
98  28  56
70  56
56

 0.8
70
a Area of sector OBC 
1
1
  (r  8) 2   r 2  cm 2  48 cm 2
2
2

So P  28  56  28  0.8  56  78.4
14
 (r 2  16r  64)  r 2   96
 (16r  64)  96
 (r  4)  6
r  4  6
(1)
r  6  4
r
6

4
b Substituting r  10 in equation (1):
6  10 2  4
Rearranging:
5 2  2  3  0
(5  3)(  1)  0
3
3
So   and r  10   6
5
5
Perimeter of shaded region
  r  8  (r  8)  8 cm

18
42
8
 8  28 cm
5
5
a Using the cosine rule:
102  122  142
cos BAC 
 0.2
2  10  12
BAC  cos 1 0.2
 1.369...  1.37 rad (3 s.f.)
b Area of triangle ABC
1
  12  10  sin A  58.787... m 2
2
Area of sector (lawn)
1
 62  A  24.649... m 2
2
So area of flowerbed

 58.787...  24.649...  34.1m 2 (3 s.f.)
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6
b Length of arc AB = 12 × 1.2 = 14.4 cm
OD  12  cos1.2  4.348...cm
BD  12  sin1.2  11.184...cm
AD  12  4.348...  7.651...cm
Perimeter of DAB
 AB  AD  BD
 14.4  7.651...  11.184...  33.236...
 33.24 cm (2 d.p.)
15
a RP2 = 2.52 + 102 − 2 × 10 × 2.5 × cos 0.3
 58.48...
17
RP  7.65 cm
QP  10  0.3  3cm
So perimeter of S
 3  7.5  7.65  18.1cm (3 s.f.)
b Area of S
1
1
  102  0.3   2.5  10  sin 0.3
2
2
2
 11.3cm (3 s.f.)
BE  5  sin 0.6  2.823...
2.823...
12
hence BCE  0.237...
and BCD  0.474...
so sin BCE 
16 a
Shaded area to left of BD
1
  122  (0.474...  sin 0.474...)
2
 1.271...
Shaded area to right of BD
1
  52  (1.2  sin1.2)
2
 3.349...
AC  12  tan1.2  30.865... cm
Area of triangle AOC
1
  12  30.865...  185.194... cm 2
2
So area of R
1
 185.194...   122  1.2  98.794...
2
2
 98.79 cm (2 d.p.)
So total shaded area
 1.271...  3.349...  4.62 cm2 (3 s.f.)
Challenge
l
r
1
1
l
1
So area  r 2  r 2    rl
2
2 r 2
Arc length l  r   
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7
Chapter review 7
1
2
a Using Pythagoras’ theorem to find OM :
a In the right-angled triangle ABC :
BA
5
1
cos ∠BAC =
=
=
AC 10 2
π
so ∠BAC =
3
b Area of triangle ABC
1
= × AB × AC × sin ∠BAC
2
1
π
= × 5 × 10 × sin =
21.650...cm 2
2
3
Area of sector DAB
1
π
=
× 52 × =
13.089...cm 2
2
3
Area of shaded region
= area of  ABC − area of sector DAB
= 21.650. . . − 13.089. . . = 8.56 cm 2 (3 s.f.)
OM 2 =17 2 − 152 =64 ⇒ OM =8cm
1
Area of OCD = × CD × OM
2
1
= × 30 × 8 = 120 cm 2
2
b Area of shaded region R
= area of semicircle CDA1
− area of segment CDA2
Area of semicircle CDA1
1
× π × 152= 353.429...cm 2
2
=
Area of segment CDA2
= area of sector OCD
− area of triangle OCD
1
× 17 2 × ∠COD − 120
2
In right-angled triangle COM :
CM
15
sin ∠COM =
=
OC
17
so ∠COM =
1.0808...
hence ∠COD =
2.1616...
So area of segment CDA2
=
1
= × 17 2 × 2.1616... − 120
2
= 192.362...cm 2
So area of shaded region R
= 353.429... − 192.362...
= 161.07 cm 2 (2 d.p.)
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1
3
c 4.65 ⩽ r < 4.75, 5.25 ⩽ p < 5.35
Least possible value for area of sector
1
= × 5.25 × 4.65 =
12.207 cm 2 (3 d.p.)
2
(Note: Least possible value is 12.20625,
so 12.207 should be given, not 12.206)
a Reflex angle AOB = (2π − θ ) rad
Area of shaded sector
1
× 6 2 × ( 2π − θ )
2
= (36π − 18θ ) cm 2
So 80= 36π − 18θ
⇒ 18θ= 36π − 80
36π − 80
=
⇒θ
= 1.839 (3 d.p.)
18
=
d Maximum possible value of θ
max p 5.35
=
= = 1.1505...
min r
4.65
So give 1.150 (3 d.p.)
Minimum possible value of θ
min p 5.25
=
= = 1.1052...
max r
4.75
So give 1.106 (3 d.p.)
5
b Length of minor arc AB
= 6θ= 6 × 1.8387... = 11.03 cm (2 d.p.)
4
a Using l = rθ :
6.4 = 5θ ⇒ θ =
6.4
= 1.28 rad
5
b Using area of sector =
R1 =
a Using l = rθ :
p = rθ ⇒ θ =
p
r
b Area of sector
1
1 2 p 1
= r 2θ =
r × =
pr cm 2
r
2
2
2
1
× 52 × 1.28 = 16
2
1 2
rθ:
2
=
c R2 area of circle − R1
= π × 52 − 16 = 62.5398...
R
16
1
=
So 1
= =
R2
62.5398... 3.908...
1
p
⇒ p=
3.91 (3 s.f.)
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2
7
6
a Area of shape X
= area of rectangle + area of semicircle
= (2d 2 +
a Area of segment R1
= area of sector OPQ
− area of triangle OPQ
1
1
⇒ A1 = × 62 × θ − × 62 × sin θ
2
2
⇒ A1 = 18(θ − sin θ )
1
πd 2 ) cm 2
2
Area of=
shape Y
1
=
(2d ) 2 θ 2d 2θ cm 2
2
=
b A2
Since X = Y :
1
2d 2 + πd 2 =
2d 2θ
2
Divide by 2d 2 :
π
θ
1+ =
4
= π × 62 − 18 (θ − sin θ )
= 36π − 18 (θ − sin θ )
Since A2 = 3 A1:
36π − 18 (θ − sin θ ) = 3 × 18 (θ − sin θ )
b Perimeter of shape X
= (d + 2d + d + πd ) cm with d =
3
= (3π + 12) cm
c Perimeter of shape Y
= (2d + 2d + 2dθ ) cm
with d = 3 and θ = 1 +
area of circle − A1
π
4
θ ) 54 (θ − sin θ )
36π − 18 (θ − sin=
=
π 72 (θ − sin θ ) 36
π
= θ – sin θ
2
π
sin θ= θ −
2
π
=12 + 6 1 + 
4

3π 
= 18 +
 cm
2 

d Difference
3π 
= 18 +
 − (3π + 12)
2 

3π
= 6−
2
= 1.287...cm
= 12.9 mm (3 s.f.)
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3
8
9
a Using the cosine rule in  ABC :
b2 + c2 − a 2
cos A =
2bc
52 + 92 − 102
⇒ cos
=
∠BAC
= 0.06
2×5×9
⇒ ∠BAC =1.50408...
= 1.504 rad (3 d.p.)
b i
Using the sector area formula:
1
area of sector = r 2θ
2
⇒ area of sector APQ
=
1
× 32 × 1.504 = 6.77 cm 2 (3 s.f.)
2
ii Area of shaded region BPQC
= area of ABC − area of sector APQ
1
1
= × 5 × 9 × sin1.504 − × 32 × 1.504
2
2
= 15.681...
= 15.7 cm 2 (3 s.f.)
iii Perimeter of shaded region BPQC
= QC + CB + BP + arc length PQ
=2 + 10 + 6 + (3 × 1.504)
= 22.51...
= 22.5 cm (3 s.f.)
a Area of sector
=
1 2
1 2
=
rθ
r × 1.5 cm 2
2
2
3 2
r = 15
4
60
⇒ r2 =
= 20
3
⇒ r=
20 =
4×5 =
So
4× 5= 2 5
=
AB r=
(1.5) 3 5 cm
b Arc length
Perimeter of sector OAB
= AO + OB + arc length AB
= 2 5 +2 5 +3 5
=7 5
= 15.7 cm (3 s.f.)
c Area of segment R
= area of sector − area of AOB
1 2
r sin1.5
2
= 15 − 10 sin1.5
= 15 −
= 5.025 cm 2 (3 d.p.)
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4
10
11
a Using the right-angled ADC:
a Using the right-angled ABD, with
π
:
∠ABD =
3
3
π
si n =
3
AB
3
⇒ AB=
=
sin π
3
3
2
b Area of badge = area of sector
π
1
= × (2 3 ) 2 θ where θ =
2
3
π
1
=
× 4× 3×
2
3
2
= 2π cm
c Perimeter of badge
= AB + AC + arc length BC
π
= 2 3 +2 3 +2 3×
3
π
= 2 3  2 + 
3

2 3
( π + 6) cm
3
= 1.84 rad (2 d.p.)
3
2
3×
2 3 cm
=
=
3
=
35
sin ∠ACD =
44
35
So ∠ACD =
sin −1  
 44 
35
and ∠ACB =
2 sin −1  
 44 
⇒ ∠ACB =1.8395...
b i
Length of railway track
= length of arc AB
= 44 × 1.8395...
= 80.9 m (3 s.f.)
ii Shortest distance from C to AB is DC.
Using Pythagoras’ theorem:
2
DC
=
442 − 352
DC =
442 − 352 = 26.7m (3 s.f.)
iii Area of region
= area of segment
=area of sector ABC − area of ABC
1
1
× 442 × 1.8395... − × 70 × DC
2
2
2
= 847m (3 s.f.)
=
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5
12
c
Area of the cross-section
= area of rectangle ABCD
− area of shaded segment
Area of rectangle = 4 × 12 × sin
a In right-angled OAX (see diagram):
x
= sin θ
6
⇒x=
6 sin θ
= 2=
= DC )
x 12 sin θ (AB
So AB
Perimeter of the cross-section
= arc length AB + AD + DC + BC
= 6 × 2θ + 4 + 12 sin θ + 4
=(8 + 12θ + 12 sin θ ) cm
So 2(7 + π) = 8 + 12θ + 12 sin θ
⇒ 14 + 2π = 8 + 12θ + 12 sin θ
⇒ 12θ + 12 sin θ − 6= 2π
π
6
= 24 cm 2
Area of shaded segment
= area of sector − area of triangle
1
π 1
π
= × 62 × − × 62 × sin
2
3 2
3
2
= 3.261...cm
So area of cross-section
= 20.7 cm 2 (3 s.f.)
13
Divide by 6:
π
2θ + 2 sin θ − 1 =
3
b When θ =
π
,
6
π 
1
+  2 ×  − 1
3 
2
π
=
3
2θ + 2 sin θ − 1 =
a O1A = O2A = 12, as they are radii of their
respective circles.
O1O2 = 12, as O2 is on the circumference
of C1 and hence is a radius (and vice
versa).
Therefore AO1O2 is equilateral
π
So ∠AO1O2 =
3
2π
and ∠AO1 B = 2 × ∠AO1O2 =
3
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6
13 b Consider arc AO2 B of circle C1.
Using arc length = rθ :
2π
=
arc length AO2 B =
12 ×
8π cm
3
Perimeter of R
= arc length AO2 B + arc length AO1 B
= 2 × 8π= 16π cm
c Consider the segment AO2 B in circle C1.
Area of segment AO2 B
area of sector O1 AB − area of O1 AB
1
2π 1
2π
× 122 ×
− × 122 × sin
2
3
2
3
2
= 88.442...cm
Area of region R
= area of segment AO2 B
=
+ area of segment AO1 B
= 2 × 88.442...
= 177 cm 2 (3 s.f.)
14 a The student has used an angle measured in
degrees – it needs to be measured in
radians to use that formula.
b
50
× π rad
180
1 2
1
5
π
r θ = × 32 ×
2
2
18
5
=
π cm 2
4
50
=
°
When is small:
LHS ≈ 4θ

(2θ ) 2 
and RHS ≈ 37 − 2 1 −

2 

so 4θ = 37 − 2 + 4θ 2
4θ 2 − 4θ + 35 =
0
b 2 − 4ac < 0
So there are no solutions.
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7
Challenge
a Let the centre of the larger circle be O and
the midpoint of AB be M.
The right-angled triangle OAM has sides
OA = 10 cm and AM = 7 cm
To find the size of the angle AOM,
7
sin AOM =
10
AOM = 0.7753… radians
Since AOB = 2AOM
AOB = 1.5507… = 1.551 radians (3 d.p.)
Similarly, let the centre of the smaller circle
be P.
The right-angled triangle PAM has sides
PA = 8 cm and PM = 7 cm
1
b Area of sector APB = r 2 ∠APB
2
1
= × 82 × 2.1308...
2
= 68.187… (cm2)
1
Area of triangle APB = ab sin ∠APB
2
1
= × 8 × 8 × sin 2.1308...
2
= 27.110… (cm2)
1
Area of triangle AOB = ab sin ∠AOB
2
1
= ×10 ×10 × sin1.5507...
2
= 49.989 (cm 2 )
Finally, the shaded area R is found by adding
the area of triangle APB and the area of
triangle AOB and subtracting the area of
sector APB:
Area R = 27.110… + 49.989… − 68.187…
= 8.91 cm2 (3 s.f.)
To find the size of the angle APM,
7
sin APM =
8
APM = 1.0654… radians
Since APB = 2APM
APB = 2.1308… = 2.131 radians (3 d.p.)
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8