Apply Other Angle Relationships in Circles

1. Apply Other Angle Relationships in Circles Monday, November 17, 2014 Essential Question: How do we find the measures of angles inside or outside a circle? Lesson 6.5 M2 Unit 3: Day 5

2. Find the value of x . 1. ANSWER ANSWER 38 56 Daily Homework Quiz Daily Homework Quiz 2.

3. ANSWER ANSWER 44 x = 54; y = 20 Daily Homework Quiz Find the value of x . 3. 4.

4. ANSWER x = 5; y = 10 Daily Homework Quiz Find the value of each variable. 5.

5. Solve for x. 1. 2x = 84 + 32 1 2 58 ANSWER 2. x = ( 360 – 120) = 0 120 ANSWER Warm Ups

6. 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.

7. 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.

8. = 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

9. = m1 12 (210o) GUIDED PRACTICE Guided Practice Find the indicated measure. SOLUTION = 105o

10. m TSR (98o) 2 = GUIDED PRACTICE Guided Practice Find the indicated measure. SOLUTION = 196o

11. m XY (80o) 2 = GUIDED PRACTICE Guided Practice Find the indicated measure. SOLUTION = 160o

12. 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.

13. 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.

14. 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.

15. Theorem 6.15Angle Outside the Circle

16. Theorem 6.15 Secant and Tangent

17. Theorem 6.15Two Tangents

18. Theorem 6.15 Two Secants

19. 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.

20. EXAMPLE Find an angle measure outside a circle Find the value of x. = 40 = 63

21. 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.

22. 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.

23. Guided Practice 7. Find the value of x. 50° = 83° – x x = 33°

24. 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

25. Because 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.

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