mm2g3 students will understand properties of circles. mm2g3 b understand and use properties of...

MM2G3 Students will understand properties of circles.

MM2G3 b Understand and use properties of central, inscribed, and related angles. MM2G3 d Justify measurements and relationships in circles and geometric and algebraic properties.

Apply Other Angle Relationships in Circles

Essential Question:

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

M2 Unit 3: Day 5Lesson 6.5

Wednesday, April 19, 2023


MM2G3 b Understand and use properties of central, inscribed, and related angles. MM2G3 d Justify measurements and relationships in circles and geometric and algebraic properties.MM2G3 b Understand and use properties of central, inscribed, and related angles. MM2G3 d Justify measurements and relationships in circles and geometric and algebraic properties.

MM2G3 Students will understand properties of circles.Daily Homework Quiz

Find the value of x .

1.

Daily Homework Quiz

2.




3.

Find the value of x .

Daily Homework Quiz

4.




5.

Find the value of each variable.

ANSWER x = 5; y = 10

Daily Homework Quiz



ANSWER 58

2. x = ( 360 – 120) = 01 2

1. 2x = 84 + 32

Solve for x.

Warm Ups



3. 180 – x = (( 2x + 4) + 28).1 2

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.

Solve for x.

ANSWER 45º

Warm Ups



Theorem 6.13If 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.

21

C

A

B

m 2 = 1

2 m BCA

m 1 = 1

2 m AB




EXAMPLE Find angle and arc measures

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

SOLUTION

= 12 (130o) = 65o

= 2 (125o) = 250ob. m KJL

a. m 1



MM2G3 Students will understand properties of circles.GUIDED PRACTICE

Find the indicated measure.

SOLUTION

= 12 (210o) = 105om 1

Guided Practice





= 2 (98o) = 196om TSR

SOLUTION

Guided Practice





= 2 (80o) = 160om XY

SOLUTION

Guided Practice



Theorem 6.14 Angle Inside the CircleIf 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.

1

A

C

B

D

m 1 = 1

2( mCD + mAB )



MM2G3 Students will understand properties of circles.EXAMPLE 2

Find the value of x.

SOLUTION

The chords JL and KM intersect inside the circle.

Use Theorem 6.14.xo = 12

(mJM + mLK)

xo = 12

(130o + 156o) Substitute.

xo = 143 Simplify.

EXAMPLE




4. Find the value of the variable.

SOLUTION

The chords AC and CD intersect inside the circle.

Use Theorem 6.14

Substitute.

Simplify.

= 12

(yo + 95o) 78o

= y 61

78° (mAB + mCD)= 12

Guided Practice



Theorem 6.15Angle Outside the Circle



Theorem 6.15 Secant and Tangent

1

A

B

C

m 1 = 1

2( m BC - m AC )



Theorem 6.15Two Tangents

2

R

P

Q

m 2 = 1

2( m PQR - m PR )



Theorem 6.15 Two Secants

3

Z

W

X

Y

m 3 = 1

2( m XY - m WZ )




EXAMPLE Find an angle measure outside a circle


SOLUTION

Use Theorem 10.13.

Substitute.

Simplify.

The tangent CD and the secant CB intersect outside the circle.

= 12

(178o – 76o)xo

= 51 x

m BCD (mAD – mBD)= 12




= 40 = 63


EXAMPLE Find an angle measure outside a circle




Find the value of the variable.5.

SOLUTION

The tangent JF and the secant JG intersect outside the circle.

Use Theorem 10.13.

Substitute.

Simplify.

= 12

(ao – 44o) 30o

= 104 a

m FJG (mFG – mKH)= 12

Guided Practice




6. Find the value of the variable.

SOLUTION

Use Theorem 10.13.

Substitute.73.7o 12

[(xo) –(360 –x)o]

Solve for x.xo 253.7

= 12

m TQR (mTUR – mTR)

Because QT and QR are tangents, QR RS and QT TS

Also,TS SR and CA CA . So, QTS QRS by the Hypotenuse-Leg Congruence Theorem, and TQS RQS.Solve right QTS to find that m RQS 73.7°.

Guided Practice




7. Find the value of x.

50° = 83° – x

x = 33°

Guided Practice




Solve a real-world problemSCIENCE

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.)




Solve a real-world problemSOLUTION

Use Theorem 10.13.

Substitute.149o 12

[(360o – xo) –xo]

Solve for x.xo 31

= 12

m BCD (mDEB – mBD)

Because CB and CD are tangents, CB AB and CD AD

Also,BC DC and CA CA . So, ABC ADC by the Hypotenuse-Leg Congruence Theorem, and BCA DCA.Solve right CBA to find that m BCA 74.5°.

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




Homework

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