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Understanding Inscribed Angles Theorem and Calculating Angle Measures

Explore the Inscribed Angles Theorem, which states that the measure of an inscribed angle is half the measure of its intercepted arc. This is illustrated through examples involving angles A, B, C, and D with given measures. You'll learn how to find angles in various scenarios, including those formed by tangents and chords, and apply corollaries such as the congruence of angles intercepting the same arc. Perfect for geometry students seeking to enhance their understanding of circular angles and properties.

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Understanding Inscribed Angles Theorem and Calculating Angle Measures

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  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

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