Trapezoid HGFE is inscribed in a circle, with EF || GH. If arc GH is 70 degrees, arc EH is x^2 - 2x degrees, and arc FG is 65 - 3x degrees, where x > 0, find arc EPF, in degrees. Round your answer to the nearest tenth.
Since EF ll GH, then arcs EH and FG are equal .....so....
x^2 - 2x = 65 - 3x
x^2 + x - 65 = 0
x ≈ 7.58
So arc EH = arc FG = 65 - 3(7.58) = 42.26°
Then arc EPF =
360 - 2 * arc EH - arc GH
360 - 2(42.26) - 70 ≈ 205.5°
