Section 10.3 – Inscribed Angles

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Presentation transcript:

Section 10.3 – Inscribed Angles Chapter 10 – Circles Section 10.3 – Inscribed Angles

Unit Goal Use inscribed angles to solve problems.

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

What Is the Measure of an Inscribed Circle?

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

Example Find the measure of the angle or arc: 20º

Example Find the measure of the angle or arc: 50º

Example 60º

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

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

Theorems About Inscribed Polygons 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

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°

Example Substitute this into the second equation 5x + 2y = 180 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

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°