# Document ```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.
A
C
D
B
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.
B
C
1
2
A
m1= &frac12;m AB
m2= &frac12;m ABC
Finding Angle and Arc
Measures
m
• Line m is tangent to
the circle. Find the
measure of the red
angle or arc.
• Solution:
m1= &frac12; AB
m1= &frac12; (150&deg;)
m1= 75&deg;
B
1
A
150&deg;
Finding Angle and Arc
Measures
S
• Line m is tangent to
the circle. Find the
measure of the red
angle or arc.
130&deg;
P
• Solution:
m RSP = 2(130&deg;)
m RSP = 260&deg;
R
Finding an Angle Measure
BC is tangent to the
circle. Find m CBD (9x + 20)&deg;
• Solution:
m CBD = &frac12; m DAB
5x = &frac12;(9x + 20)
10x = 9x +20
x = 20
D
mCBD = 5(20&deg;) = 100&deg;
C
A
5x&deg;
B
D
m1 = &frac12; ( m CD + m AB)
1
A
2
B
C
m2 = &frac12; ( m BC+ m AD)
Finding the Measure of an Angle
Formed by Two Chords
106&deg;
P
• Find the value of x
• Solution:
x&deg; = &frac12; (m QR +m PS)
x&deg; = &frac12; (106&deg; + 174&deg;)
x = 140
S
Q
x&deg;
R
174&deg;
E
Using Theorem 9.13
200&deg;
• Find the value of x
D
F
m GHF = &frac12; (m EGD - m GF )
72&deg; = &frac12; (200&deg; - x&deg;)
144 = 200 - x&deg;
- 56 = -x
56 = x
x&deg;
G
H
72&deg;
Using Theorem 9.13
M
Because MN and MLN make a whole
circle, m MLN =360&deg;-92&deg;=268&deg;
L
92&deg;
• Find the value of x
m GHF = &frac12; (m MLN - m MN)
= &frac12; (268 - 92)
= &frac12; (176)
= 88
N
x&deg;
P
Practice
Practice
m1 = &frac12; ( 40 + 52) =46
m2 = &frac12; ( 134) = 67
m3 = &frac12; ( 100 – 70) = 15
Practice
100 = &frac12; ( 130 + x)
200 = 130+ x
X = 70
50 = &frac12; ( (360 – x) -x)
100 = 360- 2x
260 = 2x
X = 130
20 = &frac12; ( 70 – x)
40 = 70-x
X = 30
CD = CQD = 120
 E = &frac12; ( AD -BC)
25 = &frac12; (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 = &frac12;( AE – BD)
BCD = &frac12;( 140-44)
BCD = 48
A = FB = 50
BCA = &frac12; * FB = 25
ABC = 180- 50 -25 = 105
GBC =180-105 =75
360 = 4x – 50 +x + x + 25+ x – 15 + 50
360=7x +10
350 = 7x
X = 50
FHE = &frac12;( 35 + 50)
FHE = 42.5
X = 50
CFD = &frac12;*50 = 25
```