1. Geometry
Section 10.4

3. 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 = o = 100o. Because TQR is a semicircle, b.
mQR = 180o mTQ = 180o 100o = 80o. So, mQR = 80o. –

4. 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

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

8. 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?

9. 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.

11. 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

12. 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

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