Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 6 - Section 6.1 - Circles and Related Segments and Angles - Exercises - Page 286: 13e

Answer

$m\angle ABO=18^{\circ}$

Work Step by Step

We know that the $m\angle A=36^{\circ}$. If we draw in the radius $\overline{OA}$ we will create an isosceles triangle; $\triangle BOA$, with congruent base angles of $\angle OBA$ and $\angle BAO$. The measure of central angle $BOA$ is equal to the measure of its intercepted arc, which is $144^{\circ}$ (we found that in part b). $180-144=36\div2=18$ $m\angle ABO=18^{\circ}$
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