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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.

 

 Mar 8, 2020
 #1
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Hint:

 

EFGH is an isosceles trapezoid, because a pair of sides are parrelel.

 

That means Angle F and Angle E are congruent. Furthermore, we can conclude that arc EH and GF are congruent.

 

x2 - 2x = 56 - 3x

 

Solve for X, solve for arcs.

 Mar 8, 2020

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