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Apply Other Angle Relationships in Circles. Sunday, September 14, 2014. Essential Question: How do we find the measures of angles inside or outside a circle?. Lesson 6.5. M2 Unit 3: Day 5. Find the value of x. 1. ANSWER. ANSWER. 38. 56. Daily Homework Quiz. Daily Homework Quiz. 2.

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Apply other angle relationships in circles

Apply Other Angle Relationships in Circles

Sunday, September 14, 2014

Essential Question:

How do we find the measures of angles inside or outside a circle?

Lesson 6.5

M2 Unit 3: Day 5


Find the value of x .

1.

ANSWER

ANSWER

38

56

Daily Homework Quiz

Daily Homework Quiz

2.


ANSWER

ANSWER

44

x = 54; y = 20

Daily Homework Quiz

Find the value of x .

3.

4.


ANSWER

x = 5; y = 10

Daily Homework Quiz

Find the value of each variable.

5.


Solve for x.

1. 2x = 84 + 32

1 2

58

ANSWER

2. x = ( 360 – 120) = 0

120

ANSWER

Warm Ups


3.180 – x = (( 2x + 4) + 28).

1 2

82

ANSWER

45º

ANSWER

Warm Ups

Solve for x.

4. One-half of the measure of an angle plus its supplement is equal to the measure of the angle. Find the measure of the angle.


Theorem 6 13
Theorem 6.13

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.


=

a.

m1

12

(130o)

(125o)

2

=

b.

m KJL

Find angle and arc measures

EXAMPLE

Line mis tangent to the circle. Find the measure of the red angle or arc.

SOLUTION

= 65o

= 250o


=

m1

12

(210o)

GUIDED PRACTICE

Guided Practice

Find the indicated measure.

SOLUTION

= 105o


m TSR

(98o)

2

=

GUIDED PRACTICE

Guided Practice

Find the indicated measure.

SOLUTION

= 196o


m XY

(80o)

2

=

GUIDED PRACTICE

Guided Practice

Find the indicated measure.

SOLUTION

= 160o


Theorem 6 14 angle inside the circle
Theorem 6.14 Angle Inside the Circle

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 arc intercepted by the angle and its vertical angle.


The chords JLand KMintersect inside the circle.

(mJM + mLK)

xo

=

12

12

xo

(130o + 156o)

=

xo

= 143

EXAMPLE 2

EXAMPLE

Find the value of x.

SOLUTION

Use Theorem 6.14.

Substitute.

Simplify.


The chords ACand CDintersect inside the circle.

=

12

12

78o

(yo + 95o)

=

61

= y

Guided Practice

4. Find the value of the variable.

SOLUTION

(mAB + mCD)

78°

Use Theorem 6.14

Substitute.

Simplify.


Theorem 6 15 angle outside the circle
Theorem 6.15Angle Outside the Circle


Theorem 6 15 secant and tangent
Theorem 6.15 Secant and Tangent


Theorem 6 15 two tangents
Theorem 6.15Two Tangents


Theorem 6 15 two secants
Theorem 6.15 Two Secants


The tangent CDand the secant CBintersect outside the circle.

(mAD – mBD)

m BCD

=

12

12

xo

(178o – 76o)

=

x

= 51

EXAMPLE

Find an angle measure outside a circle

Find the value of x.

SOLUTION

Use Theorem 10.13.

Substitute.

Simplify.


EXAMPLE

Find an angle measure outside a circle

Find the value of x.

= 40

= 63


SOLUTION

The tangent JFand the secant JGintersect outside the circle.

(mFG – mKH)

m FJG

=

12

12

30o

(ao – 44o)

=

a

= 104

GUIDED PRACTICE

Guided Practice

Find the value of the variable.

5.

Use Theorem 10.13.

Substitute.

Simplify.


Because QTand QRare tangents,

Also,TS SR and CA CA . So, QTS QRS by the Hypotenuse-Leg Congruence Theorem, and

(mTUR – mTR)

m TQR

QR RS and QT TS

12

12

TQS RQS.Solve rightQTS to find thatm RQS

73.7°.

=

73.7o

[(xo) –(360 –x)o]

xo

253.7

GUIDED PRACTICE

Guided Practice

6. Find the value of the variable.

SOLUTION

Use Theorem 10.13.

Substitute.

Solve for x.


Guided Practice

7. Find the value of x.

50° = 83° – x

x = 33°


The Northern Lights are bright flashes of colored light between 50 and 200 miles above Earth. Suppose a flash occurs 150 miles above Earth. What is the measure of arc BD, the portion of Earth from which the flash is visible? (Earth’s radius is approximately 4000 miles.)

EXAMPLE 4

Solve a real-world problem

SCIENCE


Because between CBand CDare tangents,

Also,BC DC and CA CA . So, ABC ADC by the Hypotenuse-Leg Congruence Theorem, and

BCA DCA.Solve rightCBA to find thatm BCA

74.5°.

(mDEB – mBD)

m BCD

=

CB AB and CD AD

12

12

149o

[(360o – xo) –xo]

xo

31

The measure of the arc from which the flash is visible is about 31o.

ANSWER

EXAMPLE 4

Solve a real-world problem

SOLUTION

Use Theorem 10.13.

Substitute.

Solve for x.


Homework
Homework between

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