12.3 Inscribed Angles

1. 12.3 Inscribed Angles

2. A B C Theorem 12-9: The measure of an inscribed angles is half the measure of its intercepted arc. mB=1/2mAC (

3. B 1000 900 C A 600 1100 D Example” Find the measure of A ( mA=1/2mBCD mA=1/2(900+600) mA=1/2(1500) mA=750

4. B 1000 900 C A 600 1100 D Example” Find the measure of D ( mD=1/2mABC mD=1/2(1000+900) mD=1/2(1900) mD=950

5. Corollaries #1 Two inscribed angles that intercept the same arc are congruent. mBmC B C

6. B Corollaries #2 An angle inscribed in a semicircle is a right angle mB=900

7. C B D A Corollaries #3 The opposite angles of a quadrilateral inscribed in a circle are supplementary. mA+mC=1800 mB+mD =1800

8. O b0 320 a Example” Find the measure of a and b. A is inscribed in a semi-circle,  a is a right angle

9. O b0 320 a Example” Find the measure of a and b. a=900 The sum of the angles of a triangle is 1800, the other angle is 1800-900-320=580 580

10. O b0 320 a Example” Find the measure of a and b. a=900 580=1/2b 2 2 580 1160 =b

11. B D C Theorem 12-10: The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc. mC=1/2mBDC (

12. B T R A 1260 S U Example: RS and TU are diameters of A. RB is tangent to A at point R. Find mBRT and mTRS.

13. B T R A 1260 S U mBRT ) mBRT=1/2m RT ) ) ) mRT=mURT-mUR ) ) mRT=1800-1260 mRT=540 mBRT=1/2(540) mBRT=270

14. B T R A 1260 S U mTRS mBRS=mBRT+mTRS 900=270+mTRS 630=mTRS 270

15. Here comes the assignment