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Published byRandolph McCarthy Modified over 9 years ago
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Theorem 12-9: The measure of an inscribed angles is half the measure of its intercepted arc. m B= 1 / 2 mAC ( B A C
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Example” Find the measure of A A B C D 90 0 110 0 60 0 100 0 m A= 1 / 2 mBCD ( m A= 1 / 2 (90 0 +60 0 ) m A= 1 / 2 (150 0 ) m A=75 0
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Example” Find the measure of D A B C D 90 0 110 0 60 0 100 0 m D= 1 / 2 mABC ( m D= 1 / 2 (100 0 +90 0 ) m D= 1 / 2 (190 0 ) m D=95 0
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Corollaries #1 Two inscribed angles that intercept the same arc are congruent. m B m C B C
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Corollaries #2 An angle inscribed in a semicircle is a right angle m B=90 0 B
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Corollaries #3 The opposite angles of a quadrilateral inscribed in a circle are supplementary. m A+m C=180 0 m B+m D =180 0 B D A C
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Example” Find the measure of a and b. a b0b0 O 32 0 A is inscribed in a semi- circle, a is a right angle
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Example” Find the measure of a and b. a b0b0 O 32 0 a=90 0 The sum of the angles of a triangle is 180 0, the other angle is 180 0 -90 0 -32 0 =58 0 58 0
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Example” Find the measure of a and b. a b0b0 O 32 0 a=90 0 58 0 = 1 / 2 b 58 0 2 2 116 0 =b
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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 / 2 mBDC ( B D C
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Example: RS and TU are diameters of A. RB is tangent to A at point R. Find m BRT and m TRS. B R U S A 126 0 T
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m BRT B R U S A 126 0 T m BRT= 1 / 2 m RT ) mRT=mURT-mUR ) ) ) mRT=180 0 -126 0 ) mRT=54 0 ) m BRT= 1 / 2 (54 0 ) m BRT=27 0
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m TRS B R U S A 126 0 T m BRS=mBRT+m TRS 27 0 90 0 =27 0 +m TRS 63 0 =m TRS
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