we are learning to: - enhance mathematical basic skills knowledge. (which plt skills?) -accurately...

We are learning to:- Enhance Mathematical basic skills knowledge. (Which PLT skills?) -Accurately calculate missing angles using circle theorems. (Grades B/A).

Always aim high!LESSON OBJECTIVES

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Write in standard form: 2.61

STARTER


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EXTENSION

TASK1)

2) 3)

5)

4)

6)

Solve:4x + 3y = 105x - 2y = 24

Distance = 615milesSpeed = 82mphTime = ……….minutes

Write the bearing.

Find the sum of one exterior angle in an regular nonagon and one exterior angle in a regular heptagon?

Area shaded black?

5mWork out the perimeter

11cm

6cm

N

3cm

= 2.61 x 100

= 61582 = 7.5hours

450

Area of circle – Area of triangleΠ x r x r -

6 x 32

b x h2

= 3.14 x 5.5 x 5.5 -

= 95.0 - 9 =86cm2

095°

8x + 6y = 2015x - 6y = 72+23x = 92

x = 4 16 + 3y = 10

3y = -6

y = -2

5m

Π x d4 + r + r

= 3.14 x 104 + 5 + 5

=17.85m

Nonagon → 3609 =40°

Heptagon → 3607 =51.4°


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using?PLT SkillsCIRCLE THEOREMSINTRODUCTION

oDIAMETER

oRADIUS

RA

DIU

S

xo

xo

yo

yo

2xo

xo

o

xo2xo

o

xoyo

xo

yo

Angle at centre is twice the angle at the circumference.

bo

xo

yo

aoa + b = 180°

x + y = 180°


Angle made by a diameter is 90°. Opposite angles in a cyclic

quadrilateral add up to 180°

Angles subtended by an arc or a chord in the same

segment are equal. Angle between a tangent and a radius make 90°

Alternate segment theorem.


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EXAMPLES

CIRCLE THEOREMS

ao

30°

o

bo86°

o31°

co

do

42°

eo

fo

123°

82°

30°

o

g°

i°

j°

70o

h°

k°

38° l°

m°

45o

n°

p°

60o

q°

1) 2) 3) 4)

5) 6) 7)

60° 43°

42°

31°

57°

98°

90°

90°

90°

20°

60°

38°

38°45°

60°75°

Angles in a triangle add to 180°

Angles on a straight line add to 180°


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TASK 1 – ANGLE AT THE CENTRE (GRADE B)CIRCLE THEOREMS

1) 2) 3) 4)

5) 6) 7) 8)

.

.O . O .O .O

.O . .OO

g

O

h

a

140°b

55°

84°

c22°

d

226°

e

96°

f43°

11°

Find the angles marked in each of these circles with centre O.

70°

110° 42°

44°

113° 192°

47°86°

11°158°

79°


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TASK 2 – ANGLE MADE BY DIAMETER (GRADE B)CIRCLE THEOREMS

1) 2) 3) 4)

5) 6) 7) 8)

.

.O .O .O

.O . .O OO


.O30° a

b42°c

26°

18°

d

ef

g

80°

h

60°

48° 64°

72°

45°45° 60°

30°30°

50°

40°

40°


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TASK 3 – CYCLIC QUADRILATERALS (GRADE B)CIRCLE THEOREMS

1) 2) 3) 4)

5) 6) 7) 8)

.

.O .O .O

. .OO O


.Oa

b

110°92° c d

102° 109° e f

g 80°

103°

h

i

j

.O88°

o p

q

108°

87°k

l

121°

76°

m

n

119°

r

s

t

88°

70°71° 78°

100° 100°

80° 77°

103°

77°

72°

108°

93°

59°

121°

104°

92°

88°

92°

88°61°

119°

61°

119°


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TASK 4 – ANGLE MADE BY TANGENT AND RADIUS (GRADE B)CIRCLE THEOREMS

1) 2) 3)

4) 5) 6)

.

.O .O

.O

.O O


.O55°

ab

c

41°

25°

d

60°

4x

x

3x

x

ef

68°

35°49°

65° 120°

4x + x + 90 = 1805x + 90 = 180

5x = 90

x = 18°

18°

72°O

3x + x + 90 + 90 = 3604x + 180 = 360

4x = 180 x = 45°

45°

135°22°

136°


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TASK 5 – ANGLES IN THE SAME SEGMENT (GRADE B)CIRCLE THEOREMS

1) 2) 3)

4) 5) 6)

.

.O

.O

.O .O

O


.Oa

b31°

13°

c

d

9°56° 87°

42°

e

44° f 62°103°

g72°

h

13°31°

56°

9°51°

51°

44°

44° 62°

15°18° 18°


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TASK 6 – ALTERNATE SEGMENT THEOREM (GRADE A)CIRCLE THEOREMS

1) 2) 3)

4) 5) 6)

.

.O .O

.O .O O


.Oa82°

b

c 73°d

ef

36°

85°

47° 59°g

h

i

72° j

kl

m

np

qr

49°

114°

s

t

73°

82°

25°

85°59°

36°

74°

59°

74°

72°

72°36°

72°

49°82°

82°

49°

66°t°

s = t

57°

57°


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TASK 7 – MIXTURE OF THEOREMS (GRADE B)CIRCLE THEOREMS

1) 2) 3)

4) 5) 6)


.Oa

120°

.O

50°

b

.Oc

d

113°

106°

.O37° e

.Of g

6°

37°

.O

hi

j

32°

88°

60°

40°74°

67°

53°

37°6°

88°60°

32°


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7) 8) 9)

10) 11) 12)


.O

a31°

.O

b

.O

121°

76°

c

d

.O

7x

2x

.O

49°

98°

e .O

np

qr

34°

62° 45°59°

121°104°

7x + 2x + 90 = 1809x + 90 = 180

9x = 90 x = 10°

20°

70°

98°

33°

33°

34°112°

112°

34°

115°20°


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EXTENSION 1 – MIXTURE OF THEOREMS (GRADE A)CIRCLE THEOREMS

1) 2) 3)

4) 5) 6)

.

.O

.O

.O .O O


.O130°

160°a

b cd

18°

33°

ef48° 17°

g h26°

73°

ij

36°

74°

k

230°

115°20°

55°55°

129°51°

18°111°

115°

230°

130°

25°

26°

128°64° 116°

73°

34°

146°17°

106°38°


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using?PLT SkillsCIRCLE THEOREMS

7) 8) 9)

10) 11) 12)

.

.O .O

.O .O

O


.a

112°

b

c

d

80°

124° 56°e

f

g

35°

84°

h

36° 72°i

EXTENSION 1 – MIXTURE OF THEOREMS (GRADE A)

56°

124°

56°

100°

80°

160°10°

56°

68°

35°55°

84°

6°

72°

108°36°


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EXTENSION 2 – EXAM QUESTIONS (GRADES B-A)CIRCLE THEOREMS

1) 2)

3) 4)

.O

.O .O


.O

Prove that triangle QSR is

isosceles (3 marks)

Q

S R

80°

40°

Work out the value of each letter

(3 marks)

40°pq


(4 marks)

40°

62°

x

y


(2 marks)

65°

x

y

50°50°

40°

Therefore, QSR is

isosceles140°70°

28°

28°

124°

140°96°42°

65°

25°


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5) 6)

7) 8)

.O

.O .O


.O

Work out the value of the letter

(2 marks)


(4 marks)


(2 marks)

Work out the value of the letter(2 marks)

116°

a

74°58°

b

c

40°x

y

30°x

EXTENSION 2 – EXAM QUESTIONS (GRADES B-A)

32°

32°

32°

116°32°

32°

74°42°

140°

40°

30°

60°

60°

X = 120°


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9) 10)

.O


.O

Work out the value of the letter (5 marks)

Show that x = 15° and work out the missing letter (6 marks)

32°a

3x

2x9x

b


32° 32°Alternate

angles are equal

32° 116°

64°

9x + 3x = 18012x = 180

x = 15°

45°

30°135°15°

15°120°

MINI-PLENARY 1 – CIRCLE THEOREMS CATCHPHRASE


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MINI-PLENARY 2 – TICK OR TRASH

http://www.tes.co.uk/teaching-resource/Circle-Theorems-Tick-or-Trash-angles-6328139/

Click on the link and download the worksheet. Answers are provided.

DISCOVERY


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LINK BACK TO OBJECTIVES

- Accurately calculate missing angles using circle theorems.

What grade are we

working at?

PLENARY ACTIVITY 1

T

HS

ER

MA

T

USB S


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A

F

K

P

U

B

G

L

Q

V

C

H

M

R

W

D

I

N

S

X

E

J

O

T

Y

Clear Board

Find the value of aFind the value of a

55°55°

.Oa

110°

Points to the Green TeamPoints to the Red Team


132°132°

.Oa

66°


Find the value of eFind the value of e

44°44°

.O

88°e


Find the value of dFind the value of d

36°36°

.O

18°d



104°104°

.Oa

52°


Find the value of a Find the value of a

50°50°

.O40° a



53°53°

.O37°

d


Find the value of eFind the value of e

45°45°

.Oe


Find the value of a Find the value of a

25°25°

.O65° a



70°70°

.O20°

d


Find the values of a and bFind the values of a and b

a = 85° and b = 70°a = 85° and b = 70°

.Oa

b

110°95°



15°15°

.O 75°

a



a = 75° and b = 60°a = 75° and b = 60°

.Oa

b

120°105°



42°42°

.O 48°

a



a = 65° and b = 50°a = 65° and b = 50°

.Oa

b

130°115°


Find the values of a and b Find the values of a and b

a = 15° and b = 33° a = 15° and b = 33°

.O

a

b33°

15°


Find the value of fFind the value of f

56°56°

.O

56° f



a = 19° and b = 42° a = 19° and b = 42°

.O

a

b42°

19°


Find the value of fFind the value of f

61°61°

.O

61° f



a = 22° and b = 56° a = 22° and b = 56°

.O

a

b56°

22°


Find the values of d and eFind the values of d and e

d = 75° and e = 26°d = 75° and e = 26°

.O

de

26°

75°



a = 71° and b = 85°a = 71° and b = 85°

.Oa85°

b

71°



d = 69° and e = 19°d = 69° and e = 19°

.O

de

19°

69°



a = 75° and b = 64°a = 75° and b = 64°

.Oa64°

b

75°



d = 86° and e = 15°d = 86° and e = 15°

.O

de

15°

86°


PLENARY ACTIVITY 2 – GRADE A* EXAM STYLE QUESTION


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.

P

A

B

T

x

O

PT is a tangent to a circle with centre O. AB are points on the circumference. Angle PBA is x.

(a) Write down the value of angle AOP (b) Calculate the angle OPA in terms of x.

(c) Prove that the angle APT is equal to the angle PBA

2x

2xThe triangle OPA is isosceles since sides OP

and OA are both radii.Angles in a triangle add up to 180° and the

base angles are equal.

OPA = 180° - 2x2

OPA = 90° - x

90 - xx

Angle APT = Angle PBA because of the alternate segment theorem

Draw your brain

What have you learnt?

In your brain, write or draw everything you can remember about calculating missing angles using circle theorems. It can be a skill or a reflection, or something else that might be prominent in your brain.

What grade are we

working at?

Where are we in our

journey?

.

How well do you understand the task?

I don’t understand

I nearly understand

I fully understand

Creative Entrepreneur

Responsible Citizen

Independent Learner

Positive Thinker

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Reflective Learner

Enterprise Skills Which ones are you using?

SELF ASSESSMENT

Plenary Activity

www.mistrymaths.co.uk

SELF ASSESSMENT

Plenary Activity

On your post it notes…

Think about how you can improve your

work.

WWW (What Went Well)

EBI (Even Better If)

Creative Entrepreneur

Responsible Citizen

Independent Learner

Positive Thinker

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Reflective Learner

Enterprise Skills Which ones are you using?

www.mistrymaths.co.uk


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using?PLT SkillsCIRCLE THEOREMSINTRODUCTION

oDIAMETER

oRADIUS

RA

DIU

S

xo

xo

yo

yo

2xo

xo

o

xo2xo

o

xoyo

xo

yo


bo

xo

yo

aoa + b = 180°

x + y = 180°


Angle made by a diameter is 90°. Opposite angles in a cyclic

quadrilateral add up to 180°

Angles subtended by an arc or a chord in the same

segment are equal. Angle between a tangent and a radius make 90°

Alternate segment theorem.


Self Manager


Creative Thinker

Team Worker


using?PLT Skills

TASK 1 – ANGLE AT THE CENTRE (GRADE B)CIRCLE THEOREMS

1) 2) 3) 4)

5) 6) 7) 8)

.

.O . O .O .O

.O . .OO

g

O

h

a

140°b

55°

84°

c22°

d

226°

e

96°

f43°

11°



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TASK 2 – ANGLE MADE BY DIAMETER (GRADE B)CIRCLE THEOREMS

1) 2) 3) 4)

5) 6) 7) 8)

.

.O .O .O

.O . .O OO


.O30° a

b42°c

26°

18°

d

ef

g

80°

h


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TASK 3 – CYCLIC QUADRILATERALS (GRADE B)CIRCLE THEOREMS

1) 2) 3) 4)

5) 6) 7) 8)

.

.O .O .O

. .OO O


.Oa

b

110°92° c d

102° 109° e f

g 80°

103°

h

i

j

.O88°

o p

q

108°

87°k

l

121°

76°

m

n

119°

r

s

t


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TASK 4 – ANGLE MADE BY TANGENT AND RADIUS (GRADE B)CIRCLE THEOREMS

1) 2) 3)

4) 5) 6)

.

.O .O

.O

.O O


.O55°

ab

c

41°

25°

d

60°

4x

x

3x

x

ef

68°


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TASK 5 – ANGLES IN THE SAME SEGMENT (GRADE B)CIRCLE THEOREMS

1) 2) 3)

4) 5) 6)

.

.O

.O

.O .O

O


.Oa

b31°

13°

c

d

9°56° 87°

42°

e

44° f 62°103°

g72°

h


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TASK 6 – ALTERNATE SEGMENT THEOREM (GRADE A)CIRCLE THEOREMS

1) 2) 3)

4) 5) 6)

.

.O .O

.O .O O


.Oa82°

b

c 73°d

ef

36°

85°

47°59°

g

h

i

72° j

kl

m

np

qr

49°

114°

s

t


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1) 2) 3)

4) 5) 6)


.Oa

120°

.O

50°

b

.Oc

d

113°

106°

.O37° e

.Of g

6°

37°

.O

hi

j

32°

88°


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7) 8) 9)

10) 11) 12)


.O

a31°

.O

b

.O

121°

76°

c

d

.O

7x

2x

.O

49°

98°

e .O

np

qr

34°


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1) 2) 3)

4) 5) 6)

.

.O

.O

.O .O O


.O130°

160°a

b cd

18°

33°

ef48° 17°

g h26°

73°

ij

36°

74°

k



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7) 8) 9)

10) 11) 12)

.

.O .O

.O .O

O


.Oa

112°

b

c

d

80°

124° 56°e

f

g

35°

84°

h

36° 72°i



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1) 2)

3) 4)

.O

.O .O


.O

Prove that triangle QSR is

isosceles (3 marks)

Q

S R

80°

40°


(3 marks)

40°pq


(4 marks)

40°

62°

x

y


(2 marks)

65°

x

y



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5) 6)

7) 8)

.O

.O .O


.O

Work out the value of the letter

(2 marks)


(4 marks)


(2 marks)

Work out the value of the letter(2 marks)

116°

a

74°58°

b

c

40°x

y

30°x



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9) 10)

.O


.O

Work out the value of the letter (5 marks)

Show that x = 15° and work out the missing letter (6 marks)

32°a

3x

2x9x

b


PLENARY ACTIVITY 2 – GRADE A* EXAM STYLE QUESTION


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.

P

A

B

T

x

O

PT is a tangent to a circle with centre O. AB are points on the circumference. Angle PBA is x.

(a) Write down the value of angle AOP (b) Calculate the angle OPA in terms of x.

(c) Prove that the angle APT is equal to the angle PBA

we are learning to: - enhance mathematical basic skills knowledge. (which plt skills?) -accurately...

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