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 48 - 4x degrees, where x > 0 find arc EPF, in degrees.
Since EF ll GH.......then EH = GF
Then
x^2 - 2x = 48 - 4x
x^2 + 2x - 48 = 0
(x + 8) ( x - 6) = 0
The second factor set to 0 ad solved for x gives x = 6
So
EF = (6)^2 - 2(6) = 24° = GH
Then arc EPF = 360 - 2*24 - 70 = 242°
