EXAMPLE 1 Use inscribed angles

a. m T mQRb.

Find the indicated measure in P.

SOLUTION

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M T = mRS = (48o) = 24oa.

mQR = 180o mTQ = 180o 100o = 80o. So, mQR = 80o. – –

mTQ = 2m R = 2 50o = 100o. Because TQR is a semicircle,b.

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 (31o) = 62o.

Also, m STR = mRS = (62o) = 31o. So, STR RUS.12

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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.

GUIDED PRACTICE for Examples 1, 2 and 3

Find the measure of the red arc or angle.

1.

SOLUTION

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M G = mHF = (90o) = 45oa.

GUIDED PRACTICE for Examples 1, 2 and 3

Find the measure of the red arc or angle.

2.

SOLUTION

mTV = 2m U = 2 38o = 76o. b.

GUIDED PRACTICE for Examples 1, 2 and 3

Find the measure of the red arc or angle.

3.

SOLUTION

ZYN ZXN

ZXN 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