Chapter 8 circular measure

Additional Mathematics Module Form 4
Chapter 8- Circular Measure SMK Agama Arau, Perlis
Page | 94
CHAPTER 8- CIRCULAR MEASURE
8.1 RADIAN
1. In lower secondary, we have learned the unit for angle is degree. In this chapter, we will learn one
more unit for angle that is radian.
P
r
O 1 radian r
r
Q
2. When the value of the angle 1 radian, then the length of the arc is equal to the length of the radius.
3. From this information, we can deduce that:
r
rrad
π2360
1
=
360
2
1 ×=
r
r
rad
π
°= 3602 radπ
4. °= 180radπ
π
°
=
180
1 rad
°= 3.57 or '
1857°
5. °= 3602 radπ
radian
180
1
π
=°
8.1.1 Converting Measurements in degree to radian
Example 1:
Convert °120 to radians
Solution:
radian
180
1
π
=°

Page | 95
radian
180
120120
π
×=°
radian
3
2π
= or 2.0947 radian
Example 2:
Convert '
36112° to radians
Solution:
''
3611236112 +°=°
°+°= )
60
36
(112
°=
°+°=
6.112
6.0112
radian
180
1
π
=°
radian
180
6.1126.112
π
×=°
= 1.965 rad
8.1.2 Converting Measurements in radian to degree
Example 1:
Convert rad
6
π
to degree
Solution:
π
°
=
180
1 rad
π
ππ °
×=
180
66
rad
°= 30
Example 2:
Convert rad36.1 to degree
Solution:
π
°
=
180
1 rad
'601 =°
Tips

Page | 96
π
°
×=
180
36.136.1 rad
°=
π
244
°= 92.77
EXERCISE 8.1
1. Convert each of the following values to degrees and the nearest minute. ( )142.3=π
(a) 0.37 rad
(b) 2.04 rad
(c) 1.19 rad
2. Convert each of the following values to radians, giving your answer correct to 4 significant figures.
( )142.3=π
(a) '
9248°
(b) '
22304°
(c) '
1446°
8.2 LENGTH OF ARC OF A CIRCLE
P S
Q
turnwholeaofanglecircleaofnceCircumfere
S θ
=
If the unit of angle is degrees,
The angle of a whole turn is °360 .
We know that °= 3602 radπ
If the unit of the angle is in radians,
The angle of a whole turn is π2 radian.
θ
O
rS
r
S
π
π
θ
θ
π
2
2
3602
×=
°
=
θ is in degrees.

Page | 97
Hence,
π
θ
2
=
circleaofnceCircumfere
S
π
θπ
π
θ
π
2
2
22
r
S
r
S
=
=
Example 1:
P 7.68cm
Q
The diagram above shows a circle with a sector POQ and radius 6 cm. Given the length of minor arc PQ is
7.68 cm. Find the value ofθ , in radians.
Solution:
The formula of length of arc if angle in radians is
θrS =
Given r = 6 cm and S= 7.68 cm,
θ668.7 =
θ = 1.28 rad
Example 2:
Given a circle with centre O and radius 5 cm. Find the length of arc PQR if the angle POR∠ is 1.2 rad.
Solution:
θrS =
Given r = 5 cm and θ = 1.2 rad,
)2.1)(5(=S
=S 6 cm
θrS = θ is in radians.
θ
O

Page | 98
EXERCISE 8.2
1. In the diagram 3, the perimeter of sector OPQ is 32 cm.
P
O Q
Diagram 3
(a) Express r in terms ofθ .
(b) Find the value of r if 2.1=θ rad.
2. The angle subtended at the centre by an arc ABC with a radius 4.2 cm is 1.4 radians. Find the length of
arc AB.
3. The length of a minor arc of a circle is π2 cm. The angle subtended at the centre of by the major arc is
°240 . Find the radius of the circle.
4. Diagram 4 shows two arcs AB and CD with a common centre O. It is given that BD= AC= 3cm
.
Diagram 4
If the perimeter of the shaded region ABCD is 12 cm, find the length of radius OB.
8.3 AREA OF SECTOR OF A CIRCLE
A
B
turnwholeaofanglecircleofArea
A θ
=
θ
θ
O
rad
3
1
O B D
C
A

Page | 99
If the unit of angle is degrees,
The angle of a whole turn is °360 .
We know that °= 3602 radπ
If the unit of the angle is in radians,
The angle of a whole turn is π2 radian.
Hence,
Besides that,
θ2
2
1
rA =
θ..
2
1
rr=
Both formula can be used depends on the information given in the question.
Area Of Triangle
B
c a
h
A b C
Area = hAC ××
2
1
2
2
2
360
rA
r
A
π
π
θ
θ
π
×=
°
=
θ is in degrees.
θ
π
θπ
π
θ
π
2
2
2
2
1
2
2
rA
r
A
r
A
=
=
=
θ is in radians.
θrS =
rSA
2
1
=
1
The formula for area of triangle that is
heightbase ××
2
1
can only be used in
situation where there is right angle triangle.
In the situations that the triangle is not a
right-angled triangle, we cannot use the
formula.

Page | 100
AB
h
SinA =
=h AB Sin A
Substitute into ,
Area of segments
The formula is
Example 1:
P 7.68cm
Q
The diagram above shows a circle with a sector POQ and radius 6 cm. Given the length of minor arc PQ is
7.68 cm. Find the value ofθ , in radians. Hence, find the area of the shaded region.
Solution:
θrS =
Given r = 6 cm and S= 7.68 cm,
θ668.7 =
θ = 1.28 rad
O
θ
2
2 1
AABACArea sin
2
1
×××=
BBCABArea sin
2
1
×××=
CBCACArea sin
2
1
×××=
This formula can be used to find the area of
all types of triangle as long as there is
enough information given. The sine of an
angle is multiplied by the length of line that
joining to form the angle. For example, sine
A is multiply by AB and AC that are the lines
that joining to form the angle A.
Use the formula for the area of triangle.
θθ sin
2
1
2
1 22
rrA −=
)sin(
2
1 2
θθ −= rA
or
From the formula area of triangle CBCAC sin
2
1
××× ,
we can deduce that the area of triangle in circle is
θsin
2
1 2
r because the two lines are having the same
length.

Page | 101
Area of shaded region )28.1)(sin6(
2
1
)28.1)(6(
2
1 22
−=
2
8.5
24.1704.23
cm=
−=
Example 2:
The diagram shows a sector POQ with centre O and radius 16 cm. Point R lies on OP such that OR: OP =
5: 8. Given that °=∠ 90ORQ
P
R
O Q
Calculate
(a) the value of x, in radians.
(b) the area of shaded region, in 2
cm
Solution:
Given
8
5
=
OR
OP
and cmOR 16= .
Substitute cmOR 16= into
8
5
=
OR
OP
.
16
8
5
8
5
16
×=
=
OP
OP
cm10=
a) Given that °=∠ 90ORQ
16
10
cos =x
)
16
10
(cos 1−
=x
rad8957.0=
x
cm16
When to solve question involving
radians using calculator, make sure
the calculator is set in radians
mode.
Tips…
Use the formula for the area
of triangle.

Page | 102
b) Area of shaded region )8957.0)(sin6)(10(
2
1
)8957.0)(16(
2
1 2
−=
2
1979.52
4517.626496.114
cm=
−=
EXERCISE 8.3
1. In the diagram 4, the area of sector OPQ is 2
6.21 cm .
P
O Q
Diagram 4
(a) Express r in terms ofθ .
(b) Find the value of r if 2.1=θ rad.
2. The angle subtended at the centre by an arc ABC with a radius 3.6 cm is 1.15 radians. Find the area of
arc AB.
3. The area of a sector of a circle is π12 cm2
. The angle subtended at the centre of by the sector is °270 .
Find the radius of the circle.
8.4 SOLVING PROBLEMS INVOLVING RADIANS
Example:
The diagram above shows a semicircle ABCDE with centre P and rhombus PBQD.
Calculate:
(a) the radius of semicircle ABCDE
(b) the angle of θ in radian
(c) the area of sector PBCD
(d) the area of shaded region
θ
Use the formula for the area
of triangle.
A
B
C
)6,8(Q
)4,8(P
)4,12(E
D
θ

Page | 103
Solution:
(a) 222
)44()812( −+−=PE
16=PE
4±=PE
0>radius
Radius = 4 unit
(b)
4
1
cos =∠PBX






=∠ −
4
1
cos 1
PBX
rad1181.1=
23181.1 ×=θ
rad6362.2=
(c) Area of sector PBCD )6362.2)(4(
2
1 2
=
2
0896.21 unit=
(d) Area of shaded region 





−= )6362.2)(sin4)(4(
2
1
20896.21
2
3442.13
7464.70986.21
cm=
−=
CHAPTER REVIEW EXERCISE
1. Diagram below shows a sector OPQ with centre O. and radius 4 cm.
Given that the perimeter of the sector is 20 cm, find the angle of the sector in radians.
Q
P
O
PE is the radius of the semicircle
Length cannot be a negative number
X is the mid-point between line BD and
AQ. Line XP divides the angle θ into two.
Find one of the angles by using
trigonometry. Then multiply by two to
find the angleθ .
Divide the rhombus into two triangles. Find
the area of one triangle. Then multiply by
two to find the area of the rhombus.

Page | 104
2. The perimeter and area of a sector of a circle are 19 cm and 22.5 cm2
respectively. Calculate the
possible values of the radius of the sector and the angle of the sector.
3. A sector has an area of L cm2
and a radius of 4.5 cm. Given that angle subtended at the centre of the
circle is °135 , find the value of L. ( )142.3=π
4. The angle subtended at the centre of a circle by a minor arc is °40 . Find the ratio of length of the
minor arc PQ to the length of the major arc PQ. ( )142.3=π
5. In diagram below, O is the centre of the arcs PQ and RS. The perimeter of the figure is 30 cm.
(a ) Express α in terms of β
(b) Find the values α and β in radians, if the area of sector OPQ is 10 cm2
6. The diagram below shows a sector OPQ constructed by bending a wire of length 40 cm.
Calculate
(a) the value of r
(b) the area of sector OPQ
Q
P O
R
S
cm5
β
α
cm8
°42
rO
Q
P

Chapter 8 circular measure

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