10.4 Other Angle
Relationships in Circles
Learning Target
• I can use theorems about tangents,
chords and secants to solve
unknown measure of arcs and
angles .
Review on inscribe angles
• You know that
measure of an
angle inscribed in
a circle is half the
measure of its
intercepted arc.
This is true even if
one side of the
angle is tangent to
the circle.
A
n
D
x
C
B

mADB = ½m AB
Angle x = ½ n
Theorem 10.12
• 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


x
1
2
A
m1= ½m AB
m2= ½m BCA
n
Angle x = ½ n
Ex. 1: Finding Angle and Arc
Measures
• Line m is tangent to
the circle. Find the
measure of the red
angle or arc.

• Solution:
m1= ½ AB
m1= ½ (150°)
m1= 75°
m
B
1
A
150°
Angle x = ½ n
Ex. 1: Finding Angle and Arc
Measures
S
• Line m is tangent to
the circle. Find the
measure of the red
angle or arc.
130°

• Solution:
m RSP = 2(130°)
m RSP = 260°
P
Angle x = ½ n
R
Ex. 2: Finding an Angle Measure
• In the diagram below, BC
C
A
is tangent to the circle.
Find mCBD
(9x + 20)°
• Solution:
5x°
mCBD = ½ m DAB
5x = ½(9x + 20)
10x = 9x +20
D
x = 20
 mCBD = 5(20°) =
100°
Angle x = ½ n

B
Lines Intersecting Inside or
Outside a Circle
• If two lines intersect a circle, there
are three (3) places where the lines
can intersect.
on the circle
Inside the circle
Outside the circle
Lines Intersecting
• You know how to find angle and arc
measures when lines intersect on the
circle.
• You can use the following theorems
to find the measures when the lines
intersect
INSIDE or OUTSIDE the circle.
D
Theorem 10.13
n
x
• If two chords intersect
2
in the interior of a
f
circle, then the
B
measure of each angle
is one half the sum of
the measures of the
m1 = ½ m CD + m AB
arcs intercepted by the
angle and its vertical
m2 = ½ m BC + m AD
angle.
Angle x = ½ ( f + n)
A
1
 
 
C
Theorem 10.14
B
A
• If a tangent and a
1
secant, two tangents x
n
or two secants
intercept in the
f
EXTERIOR of a circle,
then the measure of
the angle formed is
one half the difference
m1 = ½ m( BC - m AC )
of the measures of the
intercepted arcs.
Angle x = ½ ( f - n )
C
 
Theorem 10.14
P
• If a tangent and a
2
x n
f
secant, two tangents
or two secants
Q
R
intercept in the
EXTERIOR of a circle,
then the measure of
the angle formed is
one half the difference
Angle x = ½ ( f - n )
of the measures of the
intercepted arcs.
PQR
m2 = ½ m(
 
- m PR )
Theorem 10.14
X
W
• If a tangent and a
secant, two tangents 3 x n
f
Z
or two secants
intercept in the
EXTERIOR of a circle,
Y
then the measure of
the angle formed is
one half the difference
Angle x = ½ ( f - n )
of the measures of the
intercepted arcs.
 
m3 = ½ m( XY - mWZ )
Ex. 3: Finding the Measure of an
Angle Formed by Two Chords
P
106°
• Find the value of x
S
Angle x = ½ (f +n)
Q
x°
R
 
• Solution:
x° = ½ (mPS +m RQ
x° = ½ (106° + 174°)
x = 140
174°
Apply Theorem 10.13
Substitute values
Simplify
E
Ex. 4: Using Theorem 10.14
Angle x = ½ (f – n)
• Find the value of x
Solution:
 
mGHF = ½ m(EDG - m
GF
72° = ½ (200° - x°)
144 = 200 - x°
- 56 = -x
56 = x
200°
D
F
x°
G
) Apply Theorem 10.14
H
72°
Substitute values.
Multiply each side by 2.
Subtract 200 from both sides.
Divide by -1 to eliminate negatives.
Ex. 4: Using Theorem 10.14
 
M
Because MN and MLN make a
whole circle, m MLN =360°-92°=268°
L
• Find the value of x
Solution:
92°
x°
P
N
 
mGHF = ½ m(MLN - m MN ) Apply Theorem 10.14
= ½ (268 - 92)
= ½ (176)
= 88
Substitute values.
Subtract
Multiply
Angle x = ½ ( f – n )
Ex. 5: Describing the View
from Mount Rainier
• You are on top of
Mount Rainier on a
clear day. You are
about 2.73 miles
above sea level.
Find the measure
of the arc CD that
represents the part
of Earth you can
see.

Ex. 5: Describing the View
from Mount Rainier
• You are on top of
Mount Rainier on a
clear day. You are
about 2.73 miles
above sea level.
Find the measure
of the arc CD that
represents the part
of Earth you can
see.

Ex. 5: Describing the View
from Mount Rainier
• BC and BD are
tangent to the Earth.
You can solve right
∆BCA to see that
mCBA  87.9°. So,
mCBD  175.8°. Let
m CD = x° using Trig
Ratios


CD
175.8  ½[(360 – x) – x] Apply Theorem 10.14.
Simplify.
175.8  ½(360 – 2x)
Distributive Property.
175.8  180 – x
Solve for x.
x  4.2
From the peak, you can see an arc about
4°.
Reminders:
• Pair-Share:
Work on page. 624-625 #2-35
• Refer to the summary sheet “Angles
Related to Circles” to identify what
formula to use.
• Quiz on Friday about Trigonometry
and Circles.