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Trapezoid $EFGH$ is inscribed in a circle, with $EF \parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees.

 

 Aug 4, 2020
 #1
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220o since arc(HE) and arc(GF) are both 35o and arc(HG) is 70o so arc EPF is (360-35-35-70)o=220o and we have that arc(EPF) is \(\boxed{220^\circ}.\)

 Aug 4, 2020

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