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# HALP ASAP

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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

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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