# Chapter 6 - Section 6.1 - Circles and Related Segments and Angles - Exercises - Page 275: 12f

Yes. $m\angle EHG=\frac{1}{2}($measure of arc $EG +$ measure of arc $DF)$

#### Work Step by Step

$m\angle EHG=\frac{1}{2}($measure of arc $EG +$ measure of arc $DF)$ Substituting in known values from previous parts... $93=\frac{1}{2}(EG+104)$ Multiply both sides by 2 (to clear the fraction)... $186=EG+104$ Subtracting 104 from both sides... $82=EG$ So the question now is...does the measure of arc $EG=82^{\circ}$? Yes. The measure of a central angle is equal to the measure of its intercepted arc. We are given that the $m\angle EOG=82^{\circ}$, therefore the measure of its intercepted arc (arc $EG$) equals $82^{\circ}$.

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