1. 1. The measure of the interior angles of a quadrilateral are 80º, 100º, 55º, and 5xº. Find the value of x.
ANSWER 25
2. Two supplementary angles have measures 6xº and 12xº. Find each angle measure.
ANSWER
60º; 120º
3. Solve 3x = ( 4x + 12).
1 2
ANSWER 6
4. Solve 80 = ( 360 – 2x).
1 2
ANSWER
100
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2. Apply angle relationships in circles. Use inscribed angles of circles.
Target
Apply angle relationships in circles.
You will…
Use inscribed angles of circles.

3. Vocabulary
inscribed angle – an angle whose vertex is on the circle and whose sides contain chords of the circle
intercepted arc – the arc that lies in the interior of an inscribed angle and has endpoints on the angle
Theorem 10.7 –
Theorem 10.8 – inscribed angles that intercept the same arc are congruent.

4. Vocabulary
inscribed polygon – a polygon whose vertices all lie on a circle; the circle is called a circumscribed circle
Theorem 10.9 – an inscribed angle is a right angle if and only if it intercepts a semicircle
Theorem – a quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.

5. 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. –

6. 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

7. 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. GUIDED PRACTICE
for Examples 1, 2 and 3
Find the measure of the red arc or angle. 1. 2. 3.
SOLUTIONS
m G = mHF
= (90o)
= 45o 1 2
mTV = 2m U
= o
= 76o
ZYN
ZXN
72°
m ZXN =

9. 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?

10. 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. GUIDED PRACTICE
for Example 4 4.
What If ? In Example 4, explain how to find locations if you want to frame the front and left side of the statue in your picture.
SOLUTION
Make the diameter of your circle the diagonal of the
rectangular base.

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

13. EXAMPLE 5
Use Theorem 10.10
Find the value of each variable.
SOLUTION
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

14. GUIDED PRACTICE
for Example 5
Find the value of each variable. 5. 6.
SOLUTIONS
x = 98
c = 62
y = 112
x = 10