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Use Inscribed Angles and Polygons

Use Inscribed Angles and Polygons. Wednesday, September 17, 2014. How do we use inscribed angles of circles?. Lesson 6.4. M2 Unit 3: Day 4. Find the value of x in C . Explain. 1. ANSWER.

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Use Inscribed Angles and Polygons

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  1. Use Inscribed Angles and Polygons Wednesday, September 17, 2014 How do we use inscribed angles of circles? Lesson 6.4 M2 Unit 3: Day 4

  2. Find the value of x in C. Explain. . 1. ANSWER 6; If a diameter of a circle is to the chord, then the diameter bisects the chord and its arc. Daily Homework Quiz Daily Homework Quiz

  3. ~ Find the value of x in C. Explain. = . ANSWER 4; In the same circle, if two chords are equidistant from the center, then they are . Daily Homework Quiz Daily Homework Quiz 2.

  4. Determine whether RS is a diameter. ANSWER Yes. Sample answer: RS is the bisector of TU by Theorem 5.3. Then RS is a diameter of the circle by Theorem 10.4. Daily Homework Quiz Daily Homework Quiz 3.

  5. 25 ANSWER 60º; 120º ANSWER Warm Ups 1.The measure of the interior angles of a quadrilateralare 80º, 100º, 55º, and 5xº. Find the value of x. 2.Two supplementary angles have measures 6xº and12xº. Find each angle measure.

  6. 3.Solve 3x = ( 4x + 12). 1 2 1 2 6 ANSWER 4. Solve 80 = ( 360 – 2x). 100 ANSWER Warm Ups

  7. Inscribed Angle Inscribed Angle: An angle whose vertex lies on a circle and whose sides are chords of the circle (or one side tangent to the circle). A Examples: Inscribed Angle Intercepted Arc C B 3 1 2 4 Yes! No! No! Yes!

  8. Theorem 6.9 Inscribed Angle Theorem The measure of an inscribed angle is half the measure of its intercepted arc. A B O C

  9. Find the indicated measure inP. a. mT b. mQR a. M T = mRS = (48o) = 24o mTQ = 2m R = 2 50o = 100o. BecauseTQR is a semicircle, b. mQR = 180o mTQ = 180o 100o = 80o. So, mQR = 80o. – – 1 1 2 2 EXAMPLE Use inscribed angles SOLUTION

  10. a. M G = mHF = (90o) = 45o 1 1 2 2 GUIDED PRACTICE Guided Practice Find the measure of the red arc or angle. 1. SOLUTION

  11. mTV = 2m U = 2 38o = 76o. b. GUIDED PRACTICE Guided Practice Find the measure of the red arc or angle. 2. SOLUTION

  12. Theorem 6.10 If two inscribed angles of a circle intercept the same arc, then the angles are congruent. A D B C

  13. Find mRSand mSTR. What do you notice about STRand RUS? From Theorem 6.9,you know thatmRS = 2m RUS= 2 (31o) = 62o. Also, m STR = mRS = (62o) = 31o. So,STR RUS. 1 1 2 2 EXAMPLE 2 EXAMPLE Find the measure of an intercepted arc SOLUTION

  14. Notice thatJKM andJLM intercept the same arc, and soJKM JLM by Theorem 6.10 Also, KJL andKML intercept the same arc, so they must also be congruent. Only choice C contains both pairs of angles. So, by Theorem 10.8, the correct answer is C. EXAMPLE 3 Standardized Test Practice SOLUTION

  15. Notice thatZYN andZXN intercept the same arc, and soZYN byTheorem 6.10. Also, KJL andKML intercept the same arc, so they must also be congruent. ZXN ZYN ZXN ZXN 72° GUIDED PRACTICE Guided Practice Find the measure of the red arc or angle. 3. SOLUTION

  16. mLJM = mLKMmLJM and mLKM both intercept LM. Guided Practice 4. Find mLJM. 5b – 7 = 3b Substitute the given values. 2b – 7 = 0 Subtract 3b from both sides. 2b = 7 Add 7 to both sides. b = 3.5 Divide both sides by 2.

  17. mEDF = mEGFmEGF and mEDF both intercept EF. Guided Practice 5. Find mEDF. 2x + 3 = 75 – 2x Substitute the given values. 4x = 72 Add 2x and subtract 3 from both sides. x = 18 Divide both sides by 4. mEDF= 2(18) + 3 = 39°

  18. Theorem 6.11 Inscribed Right Angle Theorem An angle inscribed in a semicircle is a right angle. P 180 90 S R

  19. EXAMPLE Find a. 5a + 20 = 90 Substitute 5a + 20 for mWZY. 5a = 70 Subtract 20 from both sides. a = 14 Divide both sides by 5.

  20. Guided Practice 6. Find z. 8z – 6 = 90 Substitute. 8z = 96 Add 6 to both sides. z = 12 Divide both sides by 8.

  21. Theorem 6.12 Inscribed Quadrilateral Theorem If a quadrilateral is inscribed in a circle, then the opposite angles are supplementary. mDAB + mDCB = 180  mADC + mABC = 180 

  22. a. PQRS is inscribed in a circle, so opposite angles are supplementary. a. mQ + m S = 180o m P + m R = 180o EXAMPLE 5 Find the value of each variable. EXAMPLE SOLUTION 75o + yo = 180o 80o + xo = 180o y = 105 x = 100

  23. b. JKLMis inscribed in a circle, so opposite angles are supplementary. b. mK + m M = 180o m J + m L = 180o EXAMPLE 5 Find the value of each variable. EXAMPLE SOLUTION 4bo + 2bo = 180o 2ao + 2ao = 180o 6b = 180 4a = 180 b = 30 a = 45

  24. ABCD is inscribed in a circle, so opposite angles are supplementary. mD+ m B = 180o m A + m C = 180o GUIDED PRACTICE Guided Practice Find the value of each variable. 7. SOLUTION 82o + xo = 180o y°+ 68o = 180o y = 112 x = 98

  25. STUV is inscribed in a circle, so opposite angles are supplementary. mV+ m T = 180o m S + m U = 180o GUIDED PRACTICE Guided Practice Find the value of each variable. 8. SOLUTION c°+ (2c – 6 )o = 180o 8xo + 10xo = 180o (3c)o = 186o 18xo = 180o c = 62 x = 10

  26. Guided Practice 9. Find the angle measures of GHJK.

  27. Example 4 Continued Step 2 Find the measure of each angle.

  28. Find x and y. Guided Practice 110o J 90o Q 35o yo 70o xo L K 180o

  29. Homework: pg 207-208 (#1, 2, 5, 6, 8-18, 22)

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