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1 2. x = ( 80 + 70 ) Substitute. 1 2. x = mDEF Inscribed Angle Theorem. 1 2. x = ( mDE + mEF ) Arc Addition Postulate. Because EFG is the intercepted arc of D , you need to find mFG in order to find mEFG. Inscribed Angles. LESSON 12-3. Additional Examples.
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1 2 x = (80 + 70) Substitute. 1 2 x = mDEFInscribed Angle Theorem 1 2 x = (mDE + mEF) Arc Addition Postulate Because EFG is the intercepted arc of D, you need to find mFG in order to find mEFG. Inscribed Angles LESSON 12-3 Additional Examples Find the values of x and y. x = 75 Simplify.
The arc measure of a circle is 360°, so mFG = 360 – 70 – 80 – 90 = 120. 1 2 y = mEFGInscribed Angle Theorem 1 2 y = (mEF + mFG) Arc Addition Postulate 1 2 y = (70 + 120) Substitute. Inscribed Angles LESSON 12-3 Additional Examples (continued) y = 95 Simplify. Quick Check
. The sum of the measures of the three angles of the triangle inscribed in O is 180. Inscribed Angles LESSON 12-3 Additional Examples Find the values of a and b. By Corollary 2 to the Inscribed Angle Theorem, an angle inscribed in a semicircle is a right angle, so a = 90. Therefore, the angle whose intercepted arc has measure b must have measure 180 – 90 – 32, or 58. Because the inscribed angle has half the measure of the intercepted arc, the intercepted arc has twice the measure of the inscribed angle, so b = 2(58) = 116. Quick Check
. . 1 2 m BRT = mRTThe measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc (Theorem 12-10). mRT = mURT – mURArc Addition Postulate 1 2 m BRT = (180 – 126) Substitute 180 for m and 126 for mUR. m BRT = 27 Simplify. Inscribed Angles LESSON 12-3 Additional Examples RS and TU are diameters of A. RB is tangent to A at point R. Find m BRT and m TRS.
Use the properties of tangents to find m TRS. m BRS = 90 A tangent is perpendicular to the radius of a circle at its point of tangency. m BRS = m BRT + m TRSAngle Addition Postulate 90 = 27 + m TRSSubstitute. 63 = m TRSSolve. Inscribed Angles LESSON 12-3 Additional Examples (continued) Quick Check