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

M manish painkra

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,
$\angle$CDE + $\angle$DEC + $\angle$ECD = $180^0$ (By angle sum property)
Therefore, $\angle$ECD = $180^0-53^0-35^0$
$\angle ECD = 92^0$

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