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here is the question

Trapezoid HGFE 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. here is the pic

I found a list of wrong answers, 205, 210, 240 

if someone could please give me the correct answer that would be great

thx in advance 

 Jun 16, 2020
 #1
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Since GH || EF, arc(EH) = arc(GF)   --->   x2 - 2x  =  56 - 3x

                                                            x2 + x - 56  =  0

                                                         (x + 8)(x - 7)  =  0

Either  x = -8 (not allowed) or  x = 7.

 

arc(EH)  =  x2 - 2x  =  (7)2 - 2(7)  =  49 - 14  =  35

arc(GF)  =  56 - 3x  -  56 - 3(7)  =  56 - 21  =  35

arc(GH)  =  70

 

arc(EPF)  =  360 - arc(EH) - arc(GH) - arc(GF)  =  360 - 35 - 70 - 35  =  220

 Jun 16, 2020
 #2
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:o wow that was great!!! Thx for ur explantion and answer!!!

Guest Jun 16, 2020

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