Angles in Circles h 1

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Angles in Circles Angles in Circles tives: B Grade Use the angle properties of a circle. Grade Prove the angle properties of a circle.

description

Angles in Circles h 1

Transcript of Angles in Circles h 1

Page 1: Angles in Circles h 1

Angles in CirclesAngles in Circles

Objectives:B Grade Use the angle properties of a circle.

A Grade Prove the angle properties of a circle.

Page 2: Angles in Circles h 1

Angles in CirclesAngles in Circles

Parts of a circle

Radius

diameter

circumference

Major Sector

Major arc

Minor arc

Minor segment

Major segment

Minor Sector

A line drawn at right angles tothe radius at the circumference is

called theTangent

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Angles in CirclesAngles in Circles

Key words:

Subtend: an angle subtended by an arc is one whose two rays pass through the end points of the arc

arctwo rays

angle subtended by the arc

end points of the arc

arcend points of the arctwo rays

angle subtended

Supplementary: two angles are supplementary if they add upto 180o

Cyclic Quadrilateral: a quadrilateral whose 4 vertices lie on thecircumference of a circle

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Angles in CirclesAngles in Circles

The angle subtended in a semicircle is a right angle

Theorem 1:

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Angles in CirclesAngles in CirclesNow do these:

a70o b45o

c 130ox 2x

a = 180-(90+70)a = 20o

b = 180-(90+45)b = 45o

180-130 = 50o

c = 180-(90+50)c = 40o

3x = 180-90 x = 30o

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Angles in CirclesAngles in Circles

The angle subtended by an arc at the centre of a circle is twice thatat the circumference

Theorem 2:

This can also appear like

a

2a

arc

a2a

arc

2a

a

or

Page 7: Angles in Circles h 1

Angles in CirclesAngles in Circles

d = 80 ÷ 2d = 40o

e = 72 × 2e = 144o

f = 78 ÷ 2f = 39o

67 ÷ 2 = 33.5o

g = 180-33.5g = 147.5

Now do these:

d

80o

f78o

72o

e

g67o

96o

h

h = 96 × 2h = 192o

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Angles in CirclesAngles in Circles

The opposite angles in a cyclic quadrilateral are supplementary(add up to 180o)

Theorem 3:

a

b

c

d

a + d = 180o

b + c = 180o

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Angles in CirclesAngles in Circles

i = 180 – 83 = 97o

j = 180 – 115 = 65o

k = 180 – 123 = 57o

l = m = 180 ÷ 2 = 90o

115o

i

j

Now do these:

83o

123o

k

l m

Because the quadrilateral is a kite

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Angles in CirclesAngles in Circles

Angles subtended by the same arc (or chord) are equalTheorem 4:

same arcsame angle

same arc same angle

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n = 15o

p = 43o

q = 37o

r = 54o

s = 180 – (37 + 54) = 89o

Angles in CirclesAngles in CirclesNow do these:

15o

p

43o

n

r

54o

q

37o

s

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Angles in CirclesAngles in CirclesSummary

The angle subtended in a semicircle is a right angle The angle subtended by an arc at the centre of a circle is twice that at the circumference

a

2a

arcThe opposite angles in a cyclic quadrilateral are Supplementary (add up to 180o)

a

b

c

d

a + d = 180o

b + c = 180o

Angles subtended by the same arc (or chord) are equal

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More complex problemsAngles in CirclesAngles in Circles

64o

a

b

c

d

e65o

f28o

a = 90o

The angle subtended in a semicircle is a right angle

Angle AED is supplementary to angle ACDb = 180 – 64 = 116o

A

B

C

D

E

O

Cyclic quadrilateral ACDE

Angle ABD is supplementary to angle AEDb = 180 – 116 = 64o

Cyclic quadrilateral ABDE

The angle subtended by an arc at the centre of a circle is twice that at the circumference

d = 130o

The angle subtended by an arc at the centre of a circle is twice that at the circumference

orAngles subtended by the same arc (or chord) are equal

e = 65o

Opposite angles are equal, therefore triangles FGX andHXI are congruent.

F

GH

I

X

f = 28o

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a70o b45oc 130o

x 2x

Worksheet 1Angles in CirclesAngles in Circles

a = b = c = x =

d

80o

f78o

72o

e g67o

96o

h

d = e = f = g =

h =

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Worksheet 2Angles in CirclesAngles in Circles

115o

i

j

83o

123o

k

l m

15o

p

43o

n

r

54o

q

37o

s

i =

j = k=

l =

m =

n =

p =

q =

r =

s =

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Worksheet 3Angles in CirclesAngles in Circles

64o

a

b

c

d

e65o

f28o

A

B

C

D

E

O

F

H

I

a =

b =

c =

d =

e =

f =