Modul 7 Circular Measure

CIRCULAR MEASURE 1. RADIAN 1.1 Converting Measurements in Radians to Degrees and Vice Versa S r where S is the length of the arc AB, r is the radius of the circle, and θ is the angle subtended at the centre O by the arc AB.

Definition:  rad 

A S r O

o

For the full circle, the circumference is 2πr and the angle is 360 . 2r Therefore,   . r  2 rad Hence, 2π rad = 360o => π rad = 180o

 r

B

For each of the following measurement in degree, change it to radian. Example: 25.8o (a) 30o = A Since 180o = π rad π 25.8  25.8  180  0.4504 rad

o

25.8 O

B

[0.5237 rad] (b) 45o =

(c) 81.5o =

[0.7855 rad]

[1.423 rad]

o

o

(d) 122.8 =

(e) 154 20’ =

[2.144 rad]

[2.694 rad]

(f) 225o 36’ =

(g) -50o =

[3.938 rad] Circular Measure

[-0.02773 rad]

1

For each of the following measurement in radian, change it to degree. Example: 0.75 rad (a) 0.6 rad = Q Since π rad = 180o P 180 0.75 rad  0.75  0.75 rad π  42.97 O Always use π = 3.142 [34.37o] (b) 1.54 rad =

(c)

18 rad  5

[88.22o]

[206.24o] (e) 1.2π rad =

3 (d) 2 rad  7

[216o]

[139.13o] (f)

π rad  3

(g) -0.25 rad =

[-14.32o]

[60o] (i) (1 + π) rad =

1 (h) - 2 rad  3

[237.29o]

[-133.67o] Circular Measure

2

2. ARC LENGTH OF A CIRCLE S => S = rθ r For each of the following, find the length of the arc S given the values of r and θ. Example: (a) r = 20 cm and θ = 0.6 rad P r = 10 cm and θ = 1.5 rad 10 cm Since S = rθ = 10 × 1.5 S 1.5 rad O = 15 cm

2.1 Arc Length, Radius and Angle



Q

[12 cm]

 (c) r = 30 cm and θ = rad 6

(b) r = 15 cm and θ = 1.4 rad

[21 cm] (d) r = 10 cm and θ = 60

[15.71 cm] (e) r = 1.5 m and θ = 82.8o

o

[10.47 cm]

[2.168 m]

4.5 cm

For each of the following, calculate the length of the radius r given the values of S and θ. Example: (a) S = 9 cm and θ = 1.5 rad A S = 4.5 cm and θ = 1.5 rad Since S = rθ r 4.5 = 1.5r O 1.5 rad 4 .5 r 1 .5 = 3 cm B (b) S = 4.5 cm and θ = 1.8 rad

(c) S = 9 m and θ = 30

[2.5 cm]

[17.19 m]

 (e) S = 10 cm and θ = rad 8

(d) S = 15 cm and θ = 32.8o

[26.20 cm] Circular Measure

[6 cm]

o

[25.46 cm]

3

10 cm

5

cm

For each of the following, find the angle θ in radian given the values of r and S. Example: (a) r = 9 cm and S = 36 cm A r = 5 cm and S = 10 cm Since S = rθ 10 = 5θ O  10 θ= 5 = 2 rad B (b) r = 7 cm and S = 5.6 cm

[4 rad]

(c) r = 20 cm and S = 0.7 m

[0.8 rad]

[3.5 rad]

(d) r = 1.2 cm and S = 20 cm

(e) r = 3.5 cm and S = 112 cm

2 [16 cm] 3

[32 cm]

20

For each of the following, find the angle θ in degree given the values of r and S. Example: (a) r = 10 cm and S = 6 cm A r = 8 cm and S = 20 cm Since S = rθ 20 = 8θ  20 O θ= rad 8 180 = 2 .5  B  = 143.22o (b) r = 5 cm and S = 5.4 cm (c) r = 5.6 cm and S = 12 cm cm

8c

m

[57.29o]

[122.76o]

(d) r = 7.15 cm and S = 0.28 m

(e) r = 5.6 cm and S = 3π cm

[224.34o] Circular Measure

[34.37o]

[96.43o]

4

3. AREA OF SECTOR 3.1 Area of Sector, Radius and Angle

P r

Area of a circle, A = πr2

  r 2 360  rad   r 2 2 rad 1 A = r2θ, θ must be in radian. 2

Area for a sector OPQ of angle θo, A  Converting to radian,

O

 r Q

3c

m

For each of the following, find the area of the sector A given the values of r and θ. Example: (a) r = 5 cm and θ = 0.8 rad r = 3 cm and θ = 1.8 rad P 1 Since A  r 2 2 1 O 1.8 rad   3 2  1 .8 2 = 8.1 cm2 3c

m

Q

[10 cm2]

(b) r = 2.11 cm and θ = 1.78 rad

(c) r = 2.7 cm and θ = 1

2 rad 3

[3.962 cm2] (d) r = 12 mm and θ =

[6.075 cm2]

4 rad 5

(e) r = 1.5 m and θ = 122.8o

[2.411 m2]

[180.98 mm2] (f) r =

(g) r = 13.2 cm and θ = 67.2o

3 m and θ = 85o 5

[102.19 cm2]

[0.2671 m2] Circular Measure

5

For each of the following, find the length of the radius r given the values of θ and A. Example: θ = 1.2 rad and A = 15 cm2 (a) θ = 3 rad and A = 6 cm2 1 Since A  r 2 2 1 15   r 2  1.2 2 O 15 2 2 r  r r 1 .2 1.2 rad = 25 P Q r = 5 cm [2 cm] (b) θ = 1.4 rad and A = 118.3 cm2

(c) θ =

 rad and A = 3π m2 6

[6 m]

[13 cm] (e) θ = 102o and A = 215 cm2

3 (d) θ = 1 rad and A = 78.25 cm2 7

[15.54 cm]

[109.55 cm] (f) θ = 65 and A = 78 cm o

(g) θ = 102 and A = 1215 cm

2

o

[11.73 cm] Circular Measure

2

[36.94 cm]

6

5 cm

For each of the following, find the angle θ in radian given the values of r and A. (a) r = 2.4 cm and A = 1.44 cm2 3 Example: r = 5 cm and A = 9 cm2 8 1 Since A  r 2 2 3 1 9   52  O 5 cm B 8 2  75  2  _ 3 8  25 9 8 cm2 3 A  rad 4 [0.5 rad] 2

2

(b) r = 8 cm and A = 64 cm

(c) r = 0.6 m and A = 3.24 m

[2 rad]

[18 rad]

4 cm

For each of the following, find the angle θ in degree given the values of r and A. Example: r = 4 cm and A = 16 cm2 (a) r = 15 cm and A = 50π cm 1 Since A  r 2 2 A 4c 1 2 m 16   4   2 O 16  2   16 cm2 16 = 2 rad 180 = 2 B 3.142 o = 114.58 [80o] (b) r = 8 cm and A = 48 cm2

(c) r = 3.2 m and A = 5 m2

[85.93o] Circular Measure

[55.95o]

7

2.2 Perimeter of Segments of a Circle Example: For the shaded segment in the given diagram beside, find its perimeter. The perimeter of the shaded segment is the total length of the arc AB and the chord AB. SAB = rθ = 10 × 1.2 = 12 cm To find the length of the chord AB, consider the triangle OAB in the diagram beside. A 1 AB m 2  sin 0.6 rad 0c = 1 10 ad r 6 . 1 180 O 00 AB  10  sin(0.6  ) .6 ra 2 3.142 10 d = cm AB = 2 × 5.646 = 11.292 cm B  perimeter of shaded segment = 12 + 11.292 = 23.292 cm

A 10

O

cm

1.2 rad 10 c

m

B

Note: The length of the chord  AB = 2r sin 2

3.2 Area of Segments of a Circle Example: With reference to the sector in the same diagram beside, find the area of the shaded segment.

A 10

cm

The area of the shaded segment is the difference between the area of the sector OAB and the triangle OAB. O 1.2 rad 1 2 10 Area of sector OAB = r θ cm 2 1   10 2 1.2 B 2 2 = 60 cm To find the area of the triangle OAB, please refer to the diagram beside. A AC is perpendicular to the radius OB. cm AC 10 Note:  sin 1.2 rad AO Area of triangle OAB O 1.2 rad 180   1 AC  AO sin 1.2   = r 2 sin  3.142   2 10 C c m 1 Area of OAB   OB  AC 2 B 1   OB  OA sin 68.75 2 1   10  10  0.9320 2 Area of segment  46.60 cm 2 1 1  r 2  r 2 sin  2 2 Area of shaded segment = 60 – 46.60 2 1 = 13.40 cm  r 2   sin   2 Circular Measure

8

For each of the following diagrams, find (i) the perimeter and (ii) the area of the shaded segment. (b) (a)

A

P 6

0.5 rad 5c

cm

m

6 cm

O

m

d 0.8 ra

5c

Q

O B

[Perimeter = 9.472 cm, Area = 1.489 cm2]

[Perimeter = 4.974 cm, Area = 0.2579 cm2] (c)

(d) P

4.2 cm

O 3.8 cm

4 .2 c

3.8 cm

m

1 rad O

A

60o

Q

B

[Perimeter = 10.19 cm, Area = 1.145 cm2] Circular Measure

[Perimeter = 8.599 cm, Area = 4.827 cm2]

9