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In the diagram below \(\overline{AB}\) and \(\overline{EF}\) are parallel. If \(\angle ABC=120^\circ\) and \(\angle ECD=55^\circ\), then what is \(\angle CEF \) in degrees.

 

 

Thanks so much for your help!

 Apr 8, 2020
 #1
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Angle CEF = 180 - (180 - 120) = 120 degrees.

 Apr 8, 2020
 #2
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That's what I initially thought but it was incorrect

Ako180  Apr 8, 2020
 #3
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Extend CE and extend AB, let us call the point of intersection \(L\)

 - Because of vertical angles angle \(LCB\) is 55 degrees.

 - Angle \(CBL\) is 180 - 120 = 60 degrees.

 

Since all angles in a triangle add up to 180 degrees, angle \(CLB\) = 65 degrees. Because of alternate interior angles, the supplement of CLB is congruent to CEF.

 

Therefore, CEF = 180 - 65 = \(\boxed{115}\) degrees.

 Apr 8, 2020
 #4
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Thank you!! That makes a lot of sense

Ako180  Apr 8, 2020
 #5
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I went at it a liffle differently. 

 

Since AB & EF are parallel, draw a line between them that is perpendicular to both.

 

Now you have a pentagram.  The total of the interior angles of a pentogram is (5 – 2)(180) = 540

 

One of those right angles is                                              90 

The other right angle is                                                     90

You are given that angle ABC is                                      120

Angle ECD is 55 therefore angle ECB is (180 – 55)        125

 

The total of those four angles is                                       425 

 

Subtract the total of the four angles from the total

of the interior angles to get the fifth angle (540 – 425)     115 

.

 Apr 8, 2020
 #6
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Nice!  I like that strategy!

AnExtremelyLongName  Apr 13, 2020

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