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Understanding Inscribed Angles and Arc Measures: Lesson 12-3 Examples

This lesson provides additional examples on inscribed angles and the relationships between angles and arcs in circles. Key concepts include the Inscribed Angle Theorem, the Arc Addition Postulate, and properties of angles formed by tangents and chords. We will find the measures of angles x, y, a, and b using these theorems. Learn how to solve for angle measures based on intercepted arcs and the relationships of diameters, tangents, and chords. Practice problems included to enhance understanding and application.

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Understanding Inscribed Angles and Arc Measures: Lesson 12-3 Examples

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  1. 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.

  2. 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

  3. . 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

  4. . . 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.

  5. 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

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