Transcript Slide 1

Warm-Up
A circle and an angle are drawn in the same
plane. Find all possible ways in which the
circle and angle intersect at two points.
12.5 Other Angle
Relationships in Circles
Geometry
Mr. Calise
Objectives/Assignment
• 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
• 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
C
D
B

mADB = ½m AB
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.


m1= ½m AB
m2= ½m BCA
B
C
1
2
A
Ex. 1: 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= ½ AB
m1= ½ (150°)
m1= 75°
B
1
A
150°
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
R
Ex. 2: Finding an Angle Measure
• In the diagram below, BC
is tangent to the circle.
Find mCBD
(9x + 20)°
• Solution:
mCBD = ½ m DAB
5x = ½(9x + 20)
10x = 9x +20
D
x = 20
 mCBD = 5(20°) =
100°

C
A
5x°
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
Summary
• When an angle has its vertex on the circle
the angle is equal to ½ the measure of its
intercepted arc.
• When the angle has a vertex inside the
circle the angle is equal to ½ the sum of
its two intercepted arcs.
• When the angle has a vertex outside of
the circle then the angle is equal to ½ the
difference of the two intercepted arcs.
D
Theorem 10.13
• If two chords intersect
in the interior of a
circle, then the
measure of each angle
is one half the sum of
the measures of the
arcs intercepted by the
angle and its vertical
angle.
1
A
C
2
B
 
 
m1 = ½ m CD + m
AB
m2 = ½ m
AD
BC
+m
Theorem 10.14
B
A
• If a tangent and a
secant, two tangents
or two secants
intercept in the
EXTERIOR of a circle,
then the measure of
the angle formed is
one half the difference
of the measures of the
intercepted arcs.
1
C
 
m1 = ½ m( BC - m AC )
Theorem 10.14
• If a tangent and a
secant, two tangents
or two secants
intercept in the
EXTERIOR of a circle,
then the measure of
the angle formed is
one half the difference
of the measures of the
intercepted arcs.
P
2
R
Q
 
m2 = ½ m( P QR - m PR )
Theorem 10.14
X
• If a tangent and a
secant, two tangents 3
or two secants
intercept in the
EXTERIOR of a circle,
then the measure of
the angle formed is
one half the difference
of the measures of the
intercepted arcs.
W
Z
Y
 
m3 = ½ m( XY
- mWZ )
Ex. 3: Finding the Measure of an
Angle Formed by Two Chords 106°
P
• Find the value of x
S
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
200°
• Find the value of x
Solution:
D
F
x°
 
G
mGHF = ½ m(EDG - m GF ) Apply Theorem 10.14
72° = ½ (200° - x°)
144 = 200 - x°
- 56 = -x
56 = x
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:
N
 
mGHF = ½ m(MLN - m MN ) Apply Theorem 10.14
= ½ (268 - 92)
= ½ (176)
= 88
92°
Substitute values.
Subtract
Multiply
x°
P
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°.
Homework
• Page 617
• #’s 1 – 6, 16, 17