9.6 Secants, Tangents and Angle Measures

Geometry

Objectives

• Use angles formed by tangents and chords to solve problems in geometry.

• Use angles formed by lines that intersect a circle to solve problems.

Using Tangents and Chords

• Measure of an angle inscribed in a circle is half the measure of its intercepted arc.

C

B

A

D

m ADB = ½m AB

Theorem 9.11• If a tangent and a

chord intersect at a point on a circle, then the measure of each angle formed is one half the measure of its intercepted arc. 2

1

A

B

C

m1= ½m AB

m2= ½m ABC

Finding Angle and Arc Measures

• Line m is tangent to the circle. Find the measure of the red angle or arc.

• Solution:

m1= ½ AB

m1= ½ (150°)

m1= 75°

m

1

B

A

150°

Finding Angle and Arc Measures

• Line m is tangent to the circle. Find the measure of the red angle or arc.

• Solution:

m RSP = 2(130°)

m RSP = 260°

130°RP

S

Finding an Angle Measure is tangent to the

circle. Find m CBD• Solution:

m CBD = ½ m DAB

5x = ½(9x + 20)

10x = 9x +20

x = 20

mCBD = 5(20°) = 100°

(9x + 20)°

BC

B

C

A

5x°

D

2

1

B

A C

D

m1 = ½ ( m CD + m AB)

m2 = ½ ( m BC+ m AD)

Finding the Measure of an Angle Formed by Two Chords

• Find the value of x• Solution:

x° = ½ (m QR +m PS)

x° = ½ (106° + 174°)

x = 140R

S

P

Q

174°

106°

x°

Using Theorem 9.13

• Find the value of x

72° = ½ (200° - x°)

144 = 200 - x°

- 56 = -x

56 = x

m GHF = ½ (m EGD - m GF )F

GH

E

D

200°

x°

72°

P

N

M

L

Using Theorem 9.13

• Find the value of x

= ½ (268 - 92)

= ½ (176)

= 88

m GHF = ½ (m MLN - m MN)

x°92°

Because MN and MLN make a whole circle, m MLN =360°-92°=268°

Practice

Practice

m1 = ½ ( 40 + 52) =46

m2 = ½ ( 134) = 67

m3 = ½ ( 100 – 70) = 15

Practice

100 = ½ ( 130 + x)

200 = 130+ x

X = 70

50 = ½ ( (360 – x) -x)

100 = 360- 2x

260 = 2x

X = 130

20 = ½ ( 70 – x)

40 = 70-x

X = 30

CD = CQD = 120

E = ½ ( AD -BC)

25 = ½ (x -30)

50 = x – 30

X = 80

AB = 360-30 – 120 – 80 =

AB = 130

QDC = (180- 120) / 2 = 30

360 = 140+ 2y + y +2y

360= 140 +5y

220 = 5y

Y = 44

Y = 44

2 * 44 = 88

Y = 44

2 * 44 = 88

BCD = ½( AE – BD)

BCD = ½( 140-44)

BCD = 48

ÐA = FB = 50

BCA = ½ * FB = 25

ABC = 180- 50 -25 = 105

ÐGBC =180-105 =75

ÐFHE = ½( 35 + 50)ÐFHE = 42.5

X = 50CFD = ½*50 = 25

360 = 4x – 50 +x + x + 25+ x – 15 + 50360=7x +10350 = 7x X = 50