+34 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!
+658 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.
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
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