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Chapter 10 – Circles

Chapter 10 – Circles. Section 10.3 – Inscribed Angles. Unit Goal. Use inscribed angles to solve problems. is an inscribed angle. is the intercepted arc. Basic Definitions. INSCRIBED ANGLE – an angle whose vertex is on the circle

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Chapter 10 – Circles

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  1. Chapter 10 – Circles Section 10.3 – Inscribed Angles

  2. Unit Goal • Use inscribed angles to solve problems.

  3. is an inscribed angle. is the intercepted arc. Basic Definitions • INSCRIBED ANGLE – an angle whose vertex is on the circle • INTERCEPTED ARC – the arc whose endpoints are are on the inscribed angle

  4. What Is the Measure ofan Inscribed Circle? • What is the measure of ? • What is the measure of ?

  5. Theorem 10.8Measure of an Inscribed Angle • The measure of an inscribed angle is ½ of its intercepted arc.

  6. 20º Example • Find the measure of the angle or arc:

  7. 50º Example • Find the measure of the angle or arc:

  8. 60º Example

  9. 60º Theorem 10.9 • If two inscribed angles of a circle intercept the same arc, then the angles are congruent.

  10. Properties of Inscribed Polygons • If all the vertices of a polygon lie on a circle, the polygon is INSCRIBED in the circles and the circle is CIRCUMSCRIBED about the polygon

  11. Theorems About Inscribed Polygons • Theorem 10.10 • If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle. Conversely, if one side of an inscribed triangle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the diameter is the right angle • <B is a right angle iff segment AC is a diameter of the circle

  12. Theorem 10.11 • A quadrilateral can be inscribed in a circle iff its opposite angles are supplementary • D, E, F, and G lie on some circle C iff m<D + m<F = 180° AND m<E + m<G = 180°

  13. Example • In the diagram, ABCD is inscribed in circle P. Find the measure of each angle. • ABCD is inscribed in a circle, so opposite angles are supplementary • 3x + 3y = 180 and 5x+ 2y = 180 • 3x + 3y = 180 (solve for x) • - 3y -3y • 3x = -3y + 180 • 3 3 • x = -y + 60 • Substitute Substitute this into the second equation 5x + 2y = 180 5 (-y + 60) + 2y = 180 -5y + 300 + 2y = 180 -3y = -120 y = 40 x = -y + 60 x = -40 + 60 = 20

  14. Example (cont.) • x = 20, y = 40 • m<A = 2y, m<B = 3x, m<C = 5x, m<D = 3y • m<A = 80° • m<B = 60° • m<C = 100° • m<D = 120°

  15. HW Assignment p. 616-617 (4 – 28 even)

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