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EXAMPLE 1

Find the indicated measure in P. a. m T. b. mQR. a. M T = mRS = (48 o ) = 24 o. mTQ = 2 m R = 2 50 o = 100 o . Because TQR is a semicircle,. b. mQR = 180 o mTQ = 180 o 100 o = 80 o . So, mQR = 80 o. –. –. 1. 1. 2. 2. EXAMPLE 1.

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EXAMPLE 1

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  1. 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 1 Use inscribed angles SOLUTION

  2. Find mRSand mSTR. What do you notice about STRand RUS? From Theorem 10.7,you know thatmRS = 2m RUS= 2 (31o) = 62o. Also, m STR = mRS = (62o) = 31o. So,STR RUS. 1 1 2 2 EXAMPLE 2 Find the measure of an intercepted arc SOLUTION

  3. a. m G = mHF = (90o) = 45o 1 1 2 2 EXAMPLE 3 Find the measure of the red arc or angle. 3. SOLUTION

  4. mTV = 2m U = 2 38o = 76o. b. EXAMPLE 4 Find the measure of the red arc or angle. 2. SOLUTION

  5. 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. SOLUTION 75o + yo = 180o 80o + xo = 180o y = 105 x = 100

  6. b. JKLMis inscribed in a circle, so opposite angles are supplementary. b. mK + m M = 180o m J + m L = 180o EXAMPLE 5 Use Theorem 10.10 Find the value of each variable. SOLUTION 4bo + 2bo = 180o 2ao + 2ao = 180o 6b = 180 4a = 180 b = 30 a = 45

  7. Assignment… Using your textbook: Pg. 207 #1-18

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