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

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

3. EXAMPLE 3 Standardized Test Practice SOLUTION Notice that JKM and JLM intercept the same arc, and so JKM JLM by Theorem 10.8. Also, KJL and KML 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.

4. GUIDED PRACTICE for Examples 1, 2 and 3 Find the measure of the red arc or angle. 1. SOLUTION 1 2 1 2 M G = mHF = (90 o ) = 45 o a.

5. GUIDED PRACTICE for Examples 1, 2 and 3 Find the measure of the red arc or angle. 2. SOLUTION mTV = 2m U = 2 38 o = 76 o. b.

6. GUIDED PRACTICE for Examples 1, 2 and 3 Find the measure of the red arc or angle. 3. SOLUTION ZYNZXN 72° Notice that ZYN and ZXN intercept the same arc, and so ZYN by Theorem 10.8. Also, KJL and KML intercept the same arc, so they must also be congruent. ZXN