Geometry Section 10.4

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

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. 1 2

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.

EXAMPLE 4 Use a circumscribed circle Photography Your camera has a 90o field of vision and you want to photograph the front of a statue. You move to a spot where the statue is the only thing captured in your picture, as shown. You want to change your position. Where else can you stand so that the statue is perfectly framed in this way?

EXAMPLE 4 Use a circumscribed circle SOLUTION From Theorem 10.9, you know that if a right triangle is inscribed in a circle, then the hypotenuse of the triangle is a diameter of the circle. So, draw the circle that has the front of the statue as a diameter. The statue fits perfectly within your camera’s 90o field of vision from any point on the semicircle in front of the statue.

EXAMPLE 5 Use Theorem 10.10 Find the value of each variable. a. SOLUTION PQRS is inscribed in a circle, so opposite angles are supplementary. a. m P + m R = 180o m Q + m S = 180o 75o + yo = 180o 80o + xo = 180o y = 105 x = 100

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

Assignment 2 of 2 #13-25 odd, pages 676 & 677