Elementary Geometry for College Students (7th Edition) Clone

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: 12e

Answer

$m\angle EHG=93^{\circ}$

Work Step by Step

We are given that the $m\angle EOG=82^{\circ}$. Radii $\overline{OE}=\overline{OG}$, creating an isosceles triangle. The sum of the interior angles of a triangle is $180^{\circ}$. Subtracting the vertex angle, and dividing the result by two will give us the measure of the base angles...more importantly, the $m\angle E$. $180-82=98\div2=49$ $m\angle E=49^{\circ}$ We learned from part d that the $m\angle DGE=38^{\circ}$. Using the sum of the interior angles for $\triangle GEH$... $180-49-38=93$ $m\angle EHG=93^{\circ}$
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