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Core Mathematics 2 Radian Measures 1

Core Mathematics 2

Radian Measures

Edited by: K V Kumaran

Email: [email protected]

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Radian Measures

Radian measure, including use for arc length and area of sector.

Use of the formulae

s = rθ

A = 21 r2θ for a circle.


Measuring Angles in Radians

“A radian is the angle subtended at the centre of a circle by an arc length whose length is equal to

that of the radius of the circle”. This means a radian is the angle formed when the arc length and

the radius are the same.

The number of radians in a circle = length of circumference

Length of radius

=2r

r

= 2

=2r

r

= 2

360 = 2 rads 180 = rads 1 rad =

180

57·3

360 = 2 rads

180 = rads

1 rad = 180

57·3

Changing Degrees to Radians

Rule:-

Multiply by

180

Multiply by

180

Example 1. Convert 45° to radians

45

180=

45

180 =

4

45

180=

45

180

=

4

Example 2. Convert 75° to radians, give your answer to 2sf.

75

180 =

75

180

= 1·308996 = 1·3 (2sf)

75

180 =

75

180

= 1·308996

= 1·3 (2sf)

Changing Radians to Degrees

Rule:

Multiply by 180

Multiply by 180

Example 1. Convert 2

3 rads to degress

Example 1. Convert 2

3 rads to degress

2

3

180

= 120

2

3

180

= 120

Example 2. Covert 2.1c to degrees

2·1 180

= 120·3

2·1 180

= 120·3

Leave your answer in

terms of

unless asked

for more accuracy


Finding Arc Lengths

The length of an arc is always proportional to the angle at the centre of the arc and the radius of the

arc. So, if 2 arcs have the same radius but one has an angle twice the size of the other, it means one

arc length will be twice the size of the other.

Formula for finding an arc length: – l = r where l = arc length, r = radius = angle at the centre

Formula for finding an arc length: – l = r where l = arc length,

r = radius

= angle at the centre

Why?

l

2r=

2 ( 2r) l =

2r

2

l = r

l

2r=

2

( 2r) l =2r

2

l = r

Example 1. Find the length of the arc ABC.

l = r r = 5cm =2

3 l = 5

2

3 l =

10

3 or 10·47cm

l = r r = 5cm =2

3

l = 5 2

3

l =10

3 or 10·47cm

Length of arc = angle at the centre

Circumference total angle at centre

Ratio of sides = Ratio of angles


Example 2. Find the radius of the sector ABC

l = r r = 16cm =5

4 16 = r

5

4 16

4

5= r

64

5= r or r = 4·07cm

l = r r = 16cm =5

4

16 = r 5

4

16 4

5= r

64

5= r or r = 4·07cm

Example 3. An arc AB or a circle, with centre o and radius r cm, subtends an angle of θ

radians at O. The perimeter of the sector AOB is P cm.

Express r in terms of θ.

p = (2 x radius) + arc length

p = 2r + rθ

r = p / (2 + θ)


Finding the Area of a Sector

Formula for area of a sector is

A =1

2 r

2 where r = radius , = angle at centre

A =1

2 r

2 where r = radius , = angle at centre

Why?

A

r2

=

2 ( r

2) A =

2 r

2 A =

r2

2 or A =

1

2 r

2

A

r2

=

2

( r2) A =

2 r

2

A =r

2

2 or A =

1

2 r

2

NB. If the question gives the angle at the centre in degrees it must be changed to radians

Example 1. Find the area of the sector ABC, where ABC =

3 and r = 2cm,

give your answer in terms of

Example 1. Find the area of the sector ABC, where ABC =

3 and r = 2cm,

give your answer in terms of

A =1

2 r

2 where r = 2cm, =

3 A =

1

2 2 2

3 A =

2

3 cm

2

A =1

2 r

2 where r = 2cm, =

3

A =1

2 2 2

3

A =2

3 cm

2

Area of Sector = angle at the centre

Area of circle total angle at centre

Ratio of areas = Ratio of angles


Example 2. Find the area of the sector ABC, where ABC = 60 and r = 8cm, give your answer to 3sf

Example 2. Find the area of the sector ABC, where ABC = 60 and r = 8cm, give your answer to 3sf

60 = 60

180 =

3 A =

1

2 r

2 where =

3 , r = 8cm A =

1

2 8 8

.

3 A =

32

3 A = 33·5 cm

2 (2sf)

60 = 60

180

=

3

A =1

2 r

2 where =

3 , r = 8cm

A =1

2 8 8

.

3

A =32

3

A = 33·5 cm2 (2sf)


Finding the Area of a Segment

Formula for the area of a segment:-

A =1

2 r

2 –

1

2 r

2 Sin or A =

1

2 r

2( – sin ) where r = radius, = angle at centre

A =1

2 r

2 –

1

2 r

2 Sin

or A =1

2 r

2( – sin ) where r = radius, = angle at centre

Why?

Area of segment = Area of sector – area of a triangle

A =1

2 r

2 –

1

2 r

2 sin

as a and b = r ab = r2

A =1

2 r

2 –

1

2 r

2 sin

A =1

2 r

2 –

1

2 r

2 sin

as a and b = r ab = r2

A =1

2 r

2 –

1

2 r

2 sin

Example 1. Find the area of the shaded segment.

A =1

2 r

2( – sin) where r = 9cm and =

6

A =1

2 9

2

6 – sin

6

A = 40·5 (0·023598775) A = 0·95575 (5sf)

A =1

2 r

2( – sin) where r = 9cm and =

6

A =1

2 9

2

6 – sin

6

A = 40·5 (0·023598775)

A = 0·95575 (5sf)


Homework Questions 1 – Conversion between radians and degrees

1. Change the following angles from degrees into radians. Leave your answers as fractions

or multiples of

a) 135 ْ

b) 315 ْ

c) 27 ْ

2. Change the following angles from radians into degrees

a)

5

4

5

4

b)

3

8

3

8

c)

3

3

3. Convert the following angles from degrees into radians. Give your answer correct to 3 sf

a) 84 ْ

b) 72 ْ

c) 405 ْ

4. Convert the following angles from radians into degrees

a) 3.9 rads

b) 1.6 rads

c) 5.32 rads


Homework Questions 2 – Calculating Arc Lengths

1. Calculate the length of the arcs for the sectors given. Give your answers to 1dp

a) r = 6cm, =5

3

a) r = 6cm, =5

3

b) r = 13cm, =2

3

b) r = 13cm, =2

3

c) r = 7cm, = 3·8 rads

c) r = 7cm, = 3·8 rads

2. Calculate the length of the radius for each of these sectors if:- (give answers to 2dp)

a) l = 14mm =5

7

a) l = 14mm =5

7

b) l = 3.8cm =4

3

b) l = 3.8cm =4

3

c) l = 7.6cm = 5·2 rads

c) l = 7.6cm = 5·2 rads

3. Calculate the angle at the centre of the sector if:- (give your answer in degrees to 1dp)

a) l = 8cm, r = 5cm

b) l = 3.2m, r = 1.8m

c) l = 5.1cm, r = 11cm

4. Calculate the radius for each of the following sectors if:- (give answers to 2dp)

a) l = 10cm,

= 64

= 64 ْ

b) l = 25cm,

= 312

= 312 ْ

c) l = 13m,

= 240

= 240 ْ


Homework Questions 3 – Finding the Area of a Sector 1. Calculate the area of the sectors, given your answer correct to 3 sf

a) r = 4cm =5

2

a) r = 4cm =5

2

b) r = 8cm =9

4

b) r = 8cm =9

4

c) r = 12cm = 8·6 rads

c) r = 12cm = 8·6 rads

2. Calculate the area of these sectors, give your answers in terms of

a) r = 5cm = 72

a) r = 5cm = 72 ْ

b) r = 9m = 180

b) r = 9m = 180 ْ

c) r = 7cm = 60

c) r = 7cm = 60 ْ

3. Calculate the angle at the centre in degrees, given the area of a sector and the radius

a) A = 56 cm2, r = 9cm

b) A =

18

18 cm2, r = 12cm

c) A = 25.6 cm2, r = 13cm

4. Calculate angle at the centre in radians, given the area of a sector and the radius

a) A = 17 cm2, r = 3.1cm

b) A = 125 cm2, r = 19cm

c) A = 45 m2, r = 6.9m


Homework Questions 4 – Area of a Segment

1. Calculate the area of the segment, given your answer correct to 2 dp

a) r = 8m,

=5

4

=5

4

b) r = 9m,

=11

3

=11

3

c) r = 112cm,

= 17·4 rads

= 17·4 rads

d) r = 18.5cm,

=7

12

=7

12

e) r = 16.2m,

= 10·1 rads

= 10·1 rads ْ

f) r = 27.6cm,

=13

25

=13

25


Past Examination Questions

1. Figure 1

Figure 1 shows the triangle ABC, with AB = 8 cm, AC = 11 cm and BAC = 0.7 radians. The arc

BD, where D lies on AC, is an arc of a circle with centre A and radius 8 cm. The region R, shown

shaded in Figure 1, is bounded by the straight lines BC and CD and the arc BD.

Find

(a) the length of the arc BD,

(2)

(b) the perimeter of R, giving your answer to 3 significant figures,

(4)

(c) the area of R, giving your answer to 3 significant figures.

(5)

Q7,Jan 2005

2. In the triangle ABC, AB = 8 cm, AC = 7 cm, ABC = 0.5 radians and ACB = x radians.

(a) Use the sine rule to find the value of sin x, giving your answer to 3 decimal places.

(3)

Given that there are two possible values of x,

(b) find these values of x, giving your answers to 2 decimal places (3)

Q7,June 2005

B

C R

8 cm

0.7 rad

11 cm

D

A


O

6 m

5 m

A B

5 m

3. Figure 2

In Figure 2 OAB is a sector of a circle, radius 5 m. The chord AB is 6 m long.

(a) Show that cos BOA ˆ = 25

7.

(2)

(b) Hence find the angle BOA ˆ in radians, giving your answer to 3 decimal places.

(1)

(c) Calculate the area of the sector OAB.

(2)

(d) Hence calculate the shaded area.

(3)

Q5, Jan 2006


4. Figure 2

Figure 2 shows the cross-section ABCD of a small shed.

The straight line AB is vertical and has length 2.12 m.

The straight line AD is horizontal and has length 1.86 m.

The curve BC is an arc of a circle with centre A, and CD is a straight line.

Given that the size of BAC is 0.65 radians, find

(a) the length of the arc BC, in m, to 2 decimal places,

(2)

(b) the area of the sector BAC, in m2, to 2 decimal places,

(2)

(c) the size of CAD, in radians, to 2 decimal places,

(2)

(d) the area of the cross-section ABCD of the shed, in m2, to 2 decimal places.

(3)

Q8,May 2006

1.86 m

2.12 m

D A

C

B


6 m 6 m

63 m

S

Q

P R

5. Figure 2

Figure 2 shows a plan of a patio. The patio PQRS is in the shape of a sector of a circle with centre Q

and radius 6 m.

Given that the length of the straight line PR is 63 m,

(a) find the exact size of angle PQR in radians.

(3)

(b) Show that the area of the patio PQRS is 12 m2.

(2)

(c) Find the exact area of the triangle PQR.

(2)

(d) Find, in m2 to 1 decimal place, the area of the segment PRS.

(2)

(e) Find, in m to 1 decimal place, the perimeter of the patio PQRS.

Q9, Jan 2007


6.

Figure 1

Figure 1 shows the triangle ABC, with AB = 6 cm, BC = 4 cm and CA = 5 cm.

(a) Show that cos A = 43 .

(3)

(b) Hence, or otherwise, find the exact value of sin A.

(2)

Q4, May 2007

B

C

A 4 cm

6 cm

5 cm


7. Figure 1

Figure 1 shows 3 yachts A, B and C which are assumed to be in the same horizontal plane. Yacht B

is 500 m due north of yacht A and yacht C is 700 m from A. The bearing of C from A is 015.

(a) Calculate the distance between yacht B and yacht C, in metres to 3 significant figures.

(3)

The bearing of yacht C from yacht B is , as shown in Figure 1.

(b) Calculate the value of .

(4)

Q6,Jan 2008

N

A

15

B

500 m 700 m

C


8.

Figure 1

Figure 1 shows ABC, a sector of a circle with centre A and radius 7 cm.

Given that the size of BAC is exactly 0.8 radians, find

(a) the length of the arc BC,

(2)

(b) the area of the sector ABC.

(2)

The point D is the mid-point of AC. The region R, shown shaded in Figure 1, is bounded by CD, DB

and the arc BC.

Find

(c) the perimeter of R, giving your answer to 3 significant figures,

(4)

(d) the area of R, giving your answer to 3 significant figures.

(4)

Q7,June 2008


9.

Figure 3

The shape BCD shown in Figure 3 is a design for a logo.

The straight lines DB and DC are equal in length. The curve BC is an arc of a circle with centre A

and radius 6 cm. The size of BAC is 2.2 radians and AD = 4 cm.

Find

(a) the area of the sector BAC, in cm2,

(2)

(b) the size of DAC, in radians to 3 significant figures,

(2)

(c) the complete area of the logo design, to the nearest cm2.

(4)

Q7,Jan 2009


10.

Figure 1

An emblem, as shown in Figure 1, consists of a triangle ABC joined to a sector CBD of a circle with

radius 4 cm and centre B. The points A, B and D lie on a straight line with AB = 5 cm and BD = 4 cm.

Angle BAC = 0.6 radians and AC is the longest side of the triangle ABC.

(a) Show that angle ABC = 1.76 radians, correct to three significant figures.

(4)

(b) Find the area of the emblem.

(3)

Q4, Jan 2010

11.

Figure 1

Figure 1 shows the sector OAB of a circle with centre O, radius 9 cm and angle 0.7 radians.

(a) Find the length of the arc AB.

(2)

(b) Find the area of the sector OAB.

(2)

The line AC shown in Figure 1 is perpendicular to OA, and OBC is a straight line.

(c) Find the length of AC, giving your answer to 2 decimal places.

(2)

The region H is bounded by the arc AB and the lines AC and CB.

(d) Find the area of H, giving your answer to 2 decimal places.

(3)

Q6, Jun 2010

5 cm 4 cm

4 cm

0.6 rad

B D A

C


12.

Figure 1

The shape shown in Figure 1 is a pattern for a pendant. It consists of a sector OAB of a circle centre

O, of radius 6 cm, and angle AOB = 3

. The circle C, inside the sector, touches the two straight edges,

OA and OB, and the arc AB as shown.

Find

(a) the area of the sector OAB,

(2)

(b) the radius of the circle C.

(3)

The region outside the circle C and inside the sector OAB is shown shaded in Figure 1.

(c) Find the area of the shaded region.

(2)

Q5, May 2011


13.

Figure 2

Figure 2 shows ABC, a sector of a circle of radius 6 cm with centre A. Given that the size of angle

BAC is 0.95 radians, find

(a) the length of the arc BC,

(2)

(b) the area of the sector ABC.

(2)

The point D lies on the line AC and is such that AD = BD. The region R, shown shaded in Figure 2,

is bounded by the lines CD, DB and the arc BC.

(c) Show that the length of AD is 5.16 cm to 3 significant figures.

(2)

Find

(d) the perimeter of R,

(2)

(e) the area of R, giving your answer to 2 significant figures.

(4)

Q7, Jan 2012


14.

Figure 2

The triangle XYZ in Figure 1 has XY = 6 cm, YZ = 9 cm, ZX = 4 cm and angle ZXY = .

The point W lies on the line XY.

The circular arc ZW, in Figure 1 is a major arc of the circle with centre X and radius 4 cm.

(a) Show that, to 3 significant figures, = 2.22 radians.

(2)

(b) Find the area, in cm2, of the major sector XZWX.

(3)

The region enclosed by the major arc ZW of the circle and the lines WY and YZ is shown shaded in

Figure 1.

Calculate

(c) the area of this shaded region,

(3)

(d) the perimeter ZWYZ of this shaded region.

(4)

Q7, Jan 2013


15.

Figure 2

Figure 2 shows a plan view of a garden.

The plan of the garden ABCDEA consists of a triangle ABE joined to a sector BCDE of a circle with

radius 12 m and centre B.

The points A, B and C lie on a straight line with AB = 23 m and BC = 12 m.

Given that the size of angle ABE is exactly 0.64 radians, find

(a) the area of the garden, giving your answer in m2, to 1 decimal place,

(4)

(b) the perimeter of the garden, giving your answer in metres, to 1 decimal place.

(5)

Q5 ,May 2013


16.

Figure 2

Figure 2 shows the design for a triangular garden ABC where AB = 7 m, AC = 13 m and

BC = 10 m.

Given that angle BAC = θ radians,

(a) show that, to 3 decimal places, θ = 0.865

(3)

The point D lies on AC such that BD is an arc of the circle centre A, radius 7 m.

The shaded region S is bounded by the arc BD and the lines BC and DC. The shaded region S will

be sown with grass seed, to make a lawned area.

Given that 50 g of grass seed are needed for each square metre of lawn,

(b) find the amount of grass seed needed, giving your answer to the nearest 10 g.

(7)

Q8, May 2013_R

17.


Figure 2

The shape ABCDEA, as shown in Figure 2, consists of a right-angled triangle EAB and a triangle

DBC joined to a sector BDE of a circle with radius 5 cm and centre B.

The points A, B and C lie on a straight line with BC = 7.5 cm.

Angle EAB = 2

radians, angle EBD = 1.4 radians and CD = 6.1 cm.

(a) Find, in cm2, the area of the sector BDE.

(2)

(b) Find the size of the angle DBC, giving your answer in radians to 3 decimal places.

(2)

(c) Find, in cm2, the area of the shape ABCDEA, giving your answer to 3 significant figures.

(5)

Q5, May 2014

18.

Figure 2

Figure 2 shows the shape ABCDEA which consists of a right-angled triangle BCD joined to a sector

ABDEA of a circle with radius 7 cm and centre B.

A, B and C lie on a straight line with AB = 7 cm.

Given that the size of angle ABD is exactly 2.1 radians,

(a) find, in cm, the length of the arc DEA,

(2)

(b) find, in cm, the perimeter of the shape ABCDEA, giving your answer to 1 decimal place.

(4)

Q5, May 2014_R


19.

Figure 1

Figure 1 shows a sketch of a design for a scraper blade. The blade AOBCDA consists of an isosceles

triangle COD joined along its equal sides to sectors OBC and ODA of a circle with centre O and

radius 8 cm. Angles AOD and BOC are equal. AOB is a straight line and is parallel to the line DC.

DC has length 7 cm.

(a) Show that the angle COD is 0.906 radians, correct to 3 significant figures.

(2)

(b) Find the perimeter of AOBCDA, giving your answer to 3 significant figures.

(3)

(c) Find the area of AOBCDA, giving your answer to 3 significant figures.

(3)

Q4, May 2015

core mathematics 2 radian measureskumarmaths.weebly.com/.../radian_measures_pack.pdf · core...

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