Inscribed Angles LESSON 12-3 Additional Examples

Find the values of

x

and

y

.

x

=

mDEF

2

x

1 = (

mDE

2 +

mEF

) Inscribed Angle Theorem Arc Addition Postulate

x

1 = ( 80 2 + 70 ) Substitute.

x

= 75 Simplify.

Because

EFG

is the intercepted arc of

D

, you need to find

mFG

in order to find

mEFG

.

HELP GEOMETRY

Inscribed Angles LESSON 12-3 Additional Examples (continued)

The arc measure of a circle is 360 °, so

mFG

= 360 – 70 – 80 – 90 = 120.

y

1 =

mEFG

2

y

1 = (

mEF

+

mFG

)

y

1 = ( 70 2 + 120 )

y

= 95 Inscribed Angle Theorem Arc Addition Postulate Substitute.

Simplify.

Quick Check HELP GEOMETRY

Inscribed Angles LESSON 12-3 Additional Examples

Find the values of

a

and

b

.

By Corollary 2 to the Inscribed Angle Theorem,

an angle inscribed in a semicircle is a right angle

, so

a

= 90.

The sum of the measures of the three angles of the triangle inscribed in

O

is 180. Therefore, the angle whose intercepted arc has measure

b

must have measure 180 – 90 – 32, or 58. Because the inscribed angle has

half

the measure of the intercepted arc, the intercepted arc has

twice

the measure of the inscribed angle, so

b

= 2(58) = 116.

HELP Quick Check GEOMETRY

Inscribed Angles LESSON 12-3 Additional Examples

RS

and

TU

are diameters of

A

.

RB

is tangent to

A

at point

R

. Find

m BRT

and

m TRS

.

m BRT

1 =

mRT

2 The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc (Theorem 12-10).

mRT

=

mURT

–

mUR

Arc Addition Postulate

m BRT

1 = (180 – 126) 2 Substitute 180 for

m

and 126 for

mUR

.

m BRT

= 27 Simplify.

HELP GEOMETRY

Inscribed Angles LESSON 12-3 Additional Examples (continued)

Use the properties of tangents to find

m TRS

.

m BRS

= 90 A tangent is perpendicular to the radius of a circle at its point of tangency.

m BRS

=

m BRT

+

m TRS

Angle Addition Postulate 90 = 27 +

m TRS

63 =

m TRS

Substitute.

Solve.

Quick Check HELP GEOMETRY