Elementary Geometry for College Students (7th Edition)

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 6 - Section 6.1 - Circles and Related Segments and Angles - Exercises - Page 285: 11h


$m\angle 5=50^{\circ}$

Work Step by Step

In part d we determined that the $m\angle 1=80^{\circ}$. All radii of a circle are congruent, therefore $\overline{BQ}=\overline{AQ}$, creating an isosceles triangle. The two base angles of an isosceles triangle are congruent. So if we subtract the vertex angle ($\angle 1$) from $180^{\circ}$ and then divide the resulting number by two (two base angles), we will find the measure of the missing angle. $180-80=100\div2=50$ $m\angle 5=50^{\circ}$
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