Radians Mixed exercise

1 2

a Using Pythagoras’ theorem to find OM :

OM 2 = 17 2 − 152 = 64 ⇒ OM =
8 cm
1
Area of OCD = × CD × OM
2
a In the right-angled triangle ABC : 1
= × 30 × 8 = 120 cm 2
BA 5 1 2
cos ∠BAC = = =
AC 10 2
b Area of shaded region R
π
so ∠BAC = = area of semicircle CDA1
3
− area of segment CDA2
b Area of triangle ABC
1 Area of semicircle CDA1
= × AB × AC × sin ∠BAC
2 1
1 π = × π × 152= 353.429...cm 2
= × 5 × 10 × sin = 21.650...cm 2 2
2 3
Area of segment CDA2
Area of sector DAB
= area of sector OCD
1 π
= × 52 × = 13.089...cm 2 − area of triangle OCD
2 3
1
= × 17 2 × ∠COD − 120
2
Area of shaded region In right-angled triangle COM :
= area of  ABC − area of sector DAB CM 15
sin ∠COM = =
= 21.650... − 13.089... = 8.56 cm 2 (3 s.f.) 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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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)

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
a Reflex angle AOB = (2π − θ ) rad min p 5.25
Area of shaded sector = = = 1.1052...
max r 4.75
1 So give 1.106 (3 d.p.)
= × 6 2 × ( 2π − θ )
2
= (36π − 18θ ) cm 2 5
So 80= 36π − 18θ
⇒ 18θ= 36π − 80
36π − 80
⇒θ
= = 1.839 (3 d.p.)
18

b Length of minor arc AB

= 6θ = 6 × 1.8387... = 11.03 cm (2 d.p.)

a Using l = rθ :
6.4
6.4 = 5θ ⇒ θ = = 1.28 rad
5

1 2
b Using area of sector = rθ:
2
1
R1 = × 52 × 1.28 = 16
2

a Using l = rθ : c R2 area of circle − R1

=
p
p = rθ ⇒ θ = = π × 52 − 16 = 62.5398...
r
R 16 1 1
So 1
= = =
b Area of sector R2 62.5398... 3.908... p
1 1 2 p 1 3.91 (3 s.f.)
⇒ p=
= r 2θ = r × = pr cm 2
2 2 r 2

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 6 7

a Area of shape X
= area of rectangle + area of semicircle a Area of segment R1
1 = area of sector OPQ
= (2d 2 + πd 2 ) cm 2
2 − area of triangle OPQ
1 1
1 ⇒ A1 = × 62 × θ − × 62 × sin θ
Area of=
shape Y = (2d ) 2 θ 2d 2θ cm 2 2 2
2 ⇒ A1 = 18(θ − sin θ )

Since X = Y : =b A2 area of circle − A1

1 = π × 62 − 18 (θ − sin θ )
2d 2 + πd 2 = 2d 2θ
2
= 36π − 18 (θ − sin θ )  
Divide by 2d 2 :
π
1+ = θ Since A2 = 3 A1:
4
36π − 18 (θ − sin θ ) = 3 × 18 (θ − sin θ )  
b Perimeter of shape X
 36π − 18 (θ − sin=
θ ) 54 (θ − sin θ ) 
= (d + 2d + d + πd ) cm with d =
3
= (3π + 12) cm 6π 72 (θ − sin θ )         
 3=
π =θ sin θ  
 –
c Perimeter of shape Y π
sin θ= θ −  
= (2d + 2d + 2dθ ) cm 2
π
with d = 3 and θ = 1 +
4
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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 8 9

a Using the cosine rule in  ABC :

b2 + c2 − a 2
cos A =
2bc
1 2 1 2
52 + 92 − 102 a Area of sector
= rθ
= r × 1.5 cm 2
⇒ cos= ∠BAC = 0.06 2 2
2×5×9 3 2
⇒ ∠BAC =1.50408... So r = 15
4
= 1.504 rad (3 d.p.) 60
⇒ r2 = = 20
3
b i Using the sector area formula:
⇒ r= 20 = 4×5 = 4× 5= 2 5
1
area of sector = r 2θ
2 b Arc length
= (1.5) 3 5 cm
AB r=
⇒ area of sector APQ
1 Perimeter of sector
= × 32 × 1.504 = 6.77 cm 2 (3 s.f.)
2 = AO + OB + arc length AB
ii Area of shaded region BPQC = 2 5 +2 5 +3 5
= area of  ABC − area of sector APQ =7 5
1 1 = 15.7 cm (3 s.f.)
= × 5 × 9 × sin1.504 − × 32 × 1.504
2 2
= 15.681... c Area of segment R
2
= 15.7 cm (3 s.f.) = area of sector − area of  AOB
1
= 15 − r 2 sin1.5
iii Perimeter of shaded region BPQC 2
= QC + CB + BP + arc length PQ = 15 − 10 sin1.5
=2 + 10 + 6 + (3 × 1.504) = 5.025 cm 2 (3 d.p.)
= 22.51...
= 22.5 cm (3 s.f.)

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10 11

a Using the right-angled  ADC :

35
sin ∠ACD =
44
35
a Using the right-angled  ABD, with So ∠ACD = sin −1
44
π
∠ABD = : 35
3 and ∠ACB = 2 sin −1
44
π 3
si n = ⇒ ∠ACB = 1.8395...
3 AB
= 1.84 rad (2 d.p.)
3 3
⇒ AB= π
=
sin 3
3 2 b i Length of railway track
2 = length of arc AB
3×
= 2 3 cm
=
3 = 44 × 1.8395...
= 80.9m (3 s.f.)
b Area of badge = area of sector
1 π ii Shortest distance from C to AB is DC.
= × (2 3 ) 2 θ where θ =
2 3 Using Pythagoras’ theorem:
1 π DC
= 2
442 − 352
= × 4× 3×
2 3
2 DC = 442 − 352 = 26.7m (3 s.f.)
= 2π cm

c Perimeter of badge iii Area of region

= AB + AC + arc length BC = area of segment
π = area of sector ABC − area of  ABC
= 2 3 +2 3 +2 3×
3 1 1
= × 442 × 1.8395... − × 70 × DC
π 2 2
= 2 3  2 +  2
 3 = 847m (3 s.f.)
2 3
= (6 + π) cm
3

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 12 c

Area of the cross-section

= area of rectangle ABCD
− area of shaded segment
a In right-angled OAX (see diagram): π
Area of rectangle = 4 × 12 × sin
x 6
= sin θ = 24 cm 2
6
⇒x= 6 sin θ Area of shaded segment
So AB x 12 sin θ (AB
= 2= = DC ) = area of sector − area of triangle
1 π 1 π
= × 62 × − × 62 × sin
Perimeter of the cross-section 2 3 2 3
2
= arc length AB + AD + DC + BC = 3.261...cm
= 6 × 2θ + 4 + 12 sin θ + 4 cm So area of cross-section
=(8 + 12θ + 12 sin θ ) cm = 20.7 cm 2 (3 s.f.)

So 2(7 + π) = 8 + 12θ + 12 sin θ 13

⇒ 14 + 2π = 8 + 12θ + 12 sin θ
⇒ 12θ + 12 sin θ − 6 = 2π

Divide by 6:
π
2θ + 2 sin θ − 1 =
3

π
b When θ = , a O1A = O2A = 12, as they are radii of their
6
π  1 respective circles.
2θ + 2 sin θ − 1 = +  2 ×  − 1 O1O2 = 12, as O2 is on the circumference
3  2
of C1 and hence is a radius (and vice
π versa).
=
3 Therefore  AO1O2 is equilateral.
π
So ∠AO1O2 =
3
2π
and ∠AO1 B = 2 × ∠AO1O2 =
3

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 13 b Consider arc AO2 B of circle C1. b
Using arc length = rθ :   θ2 
2 1 − 1 −  − 1
2(1 − cos θ ) − 1  2 
arc length AO2 B =12 ×
2π
= 8π cm ≈ 
3 tan θ − 1 θ −1
Perimeter of R θ2
2× −1
= arc length AO2 B + arc length AO1 B = 2
θ −1
= 2 × 8π= 16π cm
θ −1
2
=
c Consider the segment AO2 B in circle C1. θ −1
Area of segment AO2 B (θ − 1)(θ + 1)
=
θ −1
area of sector O1 AB − area of O1 AB
= θ +1
1 2π 1 2π
= × 122 × − × 122 × sin
2 3 2 3
2
= 88.442...cm  (2θ ) 2 
7 + 2 1 − 
Area of region R 7 + 2 cos 2θ  2 
16 a ≈
= area of segment AO2 B tan 2θ + 3 2θ + 3
+ area of segment AO1 B
 4θ 2 
7 + 2 1 − 
=  2 
= 2 × 88.442...
2θ + 3
= 177 cm 2 (3 s.f.) 9 − 4θ 2
=
2θ + 3
14 a The student has used an angle measured in (3 + 2θ )(3 − 2θ )
degrees – it needs to be measured in =
radians to use that formula. 2θ + 3
= 3 − 2θ
50
b 50
= ° × π rad b 3
180
1 2 1 5
r θ = × 32 × π
2 2 18 17 a When θ is small:
5 = LHS 32 cos 5θ + 203 tan10θ
= π cm 2
4
 (5θ ) 2 
≈ 32 1 −  + 203(10θ )
 2 
 θ2 
1 −  −1 = 32 − 16(25θ 2 ) + 2010θ
cos θ − 1  2 
15 a ≈
θ tan 2θ θ × 2θ
θ2 So 32 − 400θ 2 − 2030θ =
182
−
= 2 400θ 2 + 2030θ + 150 =
0
2θ 2 40θ 2 + 203θ + 15 =
0
−θ 2
=
4θ 2 b 40θ 2 + 203θ + 15 = 0
1 (40θ − 3)(θ − 5) =0
= −
4 3
θ= ,5
40

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17 c θ = 5 is not a valid solution, as 5 is not 20 a
‘small’. 403 is ‘small’, so this solution is
valid.

18 cos 4 θ − sin 4 θ
= (cos 2 θ − sin 2 θ )(cos 2 θ + sin 2 θ )
= cos 2 θ − sin 2 θ
1 − sin 2 θ − sin 2 θ
=
= 1 − 2 sin 2 θ
≈ 1 − 2θ 2 b The curves intersect twice in the given
range, so the equation has two solutions.
19 a 3 sin θ = 2, 0 ⩽ θ ⩽ π
2 c 5 sin x = 3 cos x
sin θ =
3 3
tan x =
θ = 0.730, 2.41 5
x = 0.540, 3.68
b sin θ = −cos θ, −π ⩽ θ ⩽ π
tan θ = −1 π
21 a 4 sin θ − cos  − θ=
 4 sin θ − sin θ

π 3π 2 
θ= − ,
4 4 = 3sin θ

1 π
c tan θ + =2, 0 ⩽ θ ⩽ 2π b 4 sin θ − cos  − θ  = 1, 0 ⩽ θ ⩽ 2π
tan θ 2 
tan 2 θ + 1 =2 tan θ 3sin θ = 1
tan 2 θ − 2 tan θ + 1 =0 1
sin θ =
(tan θ − 1) 2 =0 3
θ = 0.340, 2.80
tan θ = 1
π 5π sin 2 x + 0.5 3π
θ= , 22 = 2, 0 < x <
4 4 1 − sin 2 x 2
sin 2 x + 0.5 =2 − 2 sin 2 x
d 2 sin 2 θ − sin θ − 1 =sin 2 θ , −π ⩽ θ ⩽ π
3sin 2 x = 1.5
sin 2 θ − sin θ − 1 =0
sin 2 x = 0.5
1± 5 Let X = 2 x
sin θ =
2
sin=X 0.5, 0 < X < 3π
sin θ = 1.618 (no solutions)
π 5π 13π 17 π
or sin θ = − 0.618 X = , , ,
6 6 6 6
⇒θ = −0.666, − 2.48
π 5π 13π 17 π
x= , , ,
12 12 12 12

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23 a Cosine can be negative, so do not reject 26 a 4 sin 2 x + 9 cos x − 6 =0
1 4(1 − cos 2 x) + 9 cos x − 6 =0
−
2
4 − 4 cos 2 x + 9 cos x − 6 =0
b Rearranged incorrectly, so incorrectly 4 cos 2 x − 9 cos x + 2 =0
square rooted.
b 4 cos 2 x − 9 cos x + 2 =0, 0 ⩽ θ ⩽ 4π
c 2
2 cos x = 1 (4 cos x − 1)(cos x − 2) =0
1
cos 2 x = cos x = 2 (no solution)
2
or cos x= 0.25 ⇒ x= 1.3, 5.0, 7.6,11.2
1
cos x = ±
2
27 a tan 2 x = 5 sin 2 x
3π π π 3π sin 2 x
x=− ,− , , = 5 sin 2 x
4 4 4 4 cos 2 x
sin 2 x = 5 sin 2 x cos 2 x
24 a Not all solutions have been calculated.
There will be four solutions in the given (1 − 5 cos 2 x) sin 2 x = 0
interval.
b sin 2 x(1 − 5 cos 2 x) =0, 0 ⩽ x ⩽ π
b 2 tan 2x = 5, 0 ⩽ x ⩽ 2π Let X = 2 x
Let X = 2 x sin X (1 − 5 cos X ) = 0, 0 ⩽ X ⩽ 2π
2 tan X = 5, 0 ⩽ X ⩽ 4π sin X = 0 ⇒ X = 0, π, 2π
tan X = 2.5
or cos X = 0.2 ⇒ X = 1.37, 4.91
X = 1.19, 4.33, 7.47,10.6
π
x = 0.595, 2.17, 3.74, 5.31 = x 0, 0.7, , 2.5, π
2

25 a 5 sin x= 1 + cos 2 x 28 a
2
5 sin x =+
1 2(1 − sin x)
2 sin 2 x + 5 sin x − 3 =0

b 2 sin2 x + 5 sin x − 3 = 0, 0 ⩽ x ⩽ 2π
(2 sin x − 1)(sin x + 3) =0
 3   π   4π 
π 5π b  0,  ,  , 0 ,  , 0
sin x = 0.5 ⇒ x = ,  2  3   3 
6 6
or sin x = −3 (no solution)
π
c cos  x +  = 0.65, 0 ⩽ x ⩽ 2π
 6
π
Let X= x +
6
π 13π
cos X = 0.65, ⩽x⩽
6 6
X = 0.863, 5.42
x = 0.34, 4.90

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  π θ 37 − 2 cos 2θ
c sin 4=
29 sin  3 x + = 0.45, 0 ≤ x ≤ π When is small:
 3
π π 10π LHS ≈ 4θ
Let X = 3 x + , ≤ X ≤
3 3 3  (2θ ) 2 
and RHS ≈ 37 − 2 1 − 
sin X = 0.45  2 
X = 2.67, 6.75,8.96 so 4θ = 37 − 2 + 4θ 2
x = 0.54,1.90 or 2.64 ( 2 d.p.) 4θ 2 − 4θ + 35 =
0
b 2 − 4ac < 0
Challenge So there are no solutions.

a 9 sin θ tan θ + 25 tan θ =

6
When θ is small:
LHS ≈ 9θ 2 + 25θ
so 9θ 2 + 25θ =
6
9θ 2 + 25θ − 6 =0
(9θ − 2)(θ + 3) =0
2
θ = or θ = −3
9

2
θ= is ‘small’, so this value is valid.
9
θ = −3 is not‘small’, so this value is not
valid. ‘Small’ in this context is ‘close to 0’.

b 2 tan θ + 3 =5 cos 4θ
When θ is small:
 (4θ ) 2 
LHS ≈ 2θ + 3 and RHS ≈ 5 1 − 
 2 
so 2θ + 3 = 5 − 40θ 2
40θ 2 + 2θ − 2 =0
20θ 2 + θ − 1 = 0
(4θ + 1)(5θ − 1) =0
− 1 ,θ =
θ= 1
4 5

Both values of θ could be considered

‘small’ in this case so both solutions are
valid.

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