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 May 18, 2020
 #1
avatar+26675 
+3

ECB = 180 - 55 = 125

Draw a line from F to A   so you have two right angles in a pentagon

Interior angles of a pentagon =  (n-2) 180 = 540

then   CEF = 540 - 90 -90 -120 - 125 = 115 o

 May 18, 2020
 #2
avatar+927 
0

First, I'll just disregard the fact that this is an AoPS question.

 

 BCD is a straight line, so we have BCE = 180 . We draw a line CG, which is parallel to AB and EF.

AB || CG, so

\(\angle ECG = \angle BCE - \angle BCG = 125^\circ - 60^\circ = 65^\circ\)

CG and EG are parallel so

 

\(\angle CEF = 180^\circ - \angle ECG = \boxed{115^\circ}.\)

 May 18, 2020
 #3
avatar+111455 
+3

Since AB and EF are parallel.....drop a perpendicular  from AB to EF

 

Then  both of the angles formed by this perpendicular  = 90°

 

And angle ECD is supplemental to angle  BCE....so angle BCE  = 180 - 55   = 125°

 

And we  have  an irregular pentagon......and the sum of the interior  angless =540°

 

So  angle CEF  can be found as

 

CEF  +  90  +  90  +  125  + 120    = 540

 

CEF  + 425  = 540

 

CEF  = 540  - 425  =  115°

 

 

 

cool cool cool

 May 18, 2020

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