1. Circles Unit 6: Lesson 3 Inscribed Angles Holt Geometry Texas ©2007

2. Objectives and Student Expectations
TEKS: G1A, G2B, G5B, G8C, G9C
The student will develop and awareness of the structure of a mathematical system.
The student will use patterns to make generalizations about angle relationships in circles.
The student will make conjectures about lines, angles, and circles, choosing a variety of approaches such as coordinate, transformational, and axiomatic.
The student will be expected to use Pythagorean Theorem.
The student will test conjectures about the properties and attributes of circles and the lines that intersect.

3. A central angle is an angle whose vertex is the center of a circle.
Remember!

4. An inscribed angle is an angle whose vertex is on a circle and whose sides contain chords of the circle.
An intercepted arc consists of endpoints that lie on the sides of an inscribed angle and all the points of the circle between them.

5. ______________

6. Example: 1 Find each measure. mPRU Inscribed  Thm.
Substitute 118 for mPU.

7. Example: 2 Find each measure. mSP Inscribed  Thm.
Substitute 27 for m SRP.
Multiply both sides by 2.

9. Example: 3
80
Find m GJL.
= 40⁰
Find m GHL.
= 40⁰

10. Example: 4
An art student turns in an abstract design for his art project.
Find mDFA.
mDFA = mDCF + mCDF
Ext  Thm.
Inscribed  Thm.
Substitute.
Simplify.
= 115°

11. Example: 5 Find mABD and mBC in the string art. Inscribed  Thm.
Substitute.
= 43
Inscribed  Thm.
Substitute.

13. Example: 6 Find a. WZY is a right angle
WZY is inscribed in a semicircle.
mWZY = 90
Def of rt. 
5a + 20 = 90 Substitute 5a + 20 for mWZY.
5a = 70 Subtract 20 from both sides.
a = 14 Divide both sides by 5.

15. Example: 7 Find the angle measures of GHJK.
Step 1 Find the value of b.
mG + mJ = 180
GHJK is inscribed in a .
3b b + 20 = 180
Substitute the given values.
9b + 45 = 180
Simplify.
9b = 135
Subtract 45 from both sides.
b = 15
Divide both sides by 9.

16. Example: 7 continued Step 2 Find the measure of each angle.
mG = 3(15) + 25 = 70
Substitute 15 for b
mJ = 6(15) + 20 = 110
in each expression.
mK = 10(15) – 69 = 81
mH + mK = 180
H and K are supp.
mH + 81 = 180
Substitute 81 for mK.
mH = 99
Subtract 81 from both sides