Download Lesson 10-6

Geometry
Chapter 10
Lesson 10-6
Example 1 Secant-Secant Angle
Find m2 if m MN = 60 and m LO = 50.
Method 1 Find m1.
1
m1 = 2(m MN + m LO ) Theorem 10.12
1
= 2(50 + 60) or 55
Substitution
m2 = 180 - m1
= 180 - 55 or 125
Method 2 Find the measures of the arcs intercepted by 2 and its vertical angle.
1
m2 = 2(m LM + m OTN )
Theorem 10.12
Find m LM and m OTN .
m LM + m OTN = 360 - (m MN + m LO )
= 360 - (50 + 60)
= 360 - 110 or 250
1
m2 = 2(m LM + m OTN )
Theorem 10.12
1
= 2(250) or 125
Simplify.
Example 2 Secant-Tangent Angle
Find mAGX if m GX = 68.
m XOG = 360 - m GX
= 360 - 68 or 292
1
mAGX = 2m XOG
1
= 2(292) or 146
Theorem 10.13
Simplify
Example 3 Secant-Secant Angle
Find a.
1
mF = 2(m BJ - m DH )
1
a = 2(135 - 30)
1
a = 2(105) or 52.5
Substitution
Simplify.
Theorem 10.14
Substitution
Simplify.
Geometry
Chapter 10
Example 4 Tangent-Tangent Angle
RECREATION Two sides of a fence built
around a circular patio are shown. The fence
touches the patio at points R and M and
m RM = 110. Find the measure of the angle
where the two sides of the fence meet, RZM.
To find mRZM., you need to know the measure of MAR .
m MAR = 360 - m RM
= 360 - 110
m RM = 110
= 250
1
mRZM = 2(m MAR - m RM )
1
= 2(250 - 110) or 70
Theorem 10.14
Simplify.
The measure of the angle where the two sides of the fence meet is 70.
Example 5 Secant-Tangent Angle
Find b.
MOD is a semicircle because MD is a diameter.
So, m MND = 180.
1
mC = 2(m MND - m MO )
1
54 = 2(180 - 3b)
108 = 180 - 3b
-72 = -3b
24 = b
Theorem 10.14
Substitution
Multiply each side by 2.
Subtract 180 from each side.
Divide each side by -3.