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.

Guest Mar 8, 2020

#1**+2 **

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.

x^{2} - 2x = 56 - 3x

Solve for X, solve for arcs.

CalculatorUser Mar 8, 2020