3.   In Fig. 6.41, if AB || DE, \angle BAC = 35° and \angle CDE = 53°, find \angle DCE.

                    

Answers (1)


We have, 
AB || DE, \angle BAC = 35° and \angle CDE = 53°

AE is a transversal so, \angle BAC = \angle AED = 35^0

Now, In \Delta CDE,
\angleCDE + \angleDEC + \angleECD = 180^0 (By angle sum property)
Therefore, \angleECD = 180^0-53^0-35^0
                 \angle ECD = 92^0

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