# In Fig. 6.41, if $\mathrm{AB} \| \mathrm{DE}, \angle \mathrm{BAC}=35^{\circ}$ and $\angle \mathrm{CDE}=53^{\circ}$, find $\angle \mathrm{DCE}$."

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Given:

$AB \parallel DE, \angle BAC=35^o$ and $\angle CDE=53^o$.

To do:

We have to find $\angle DCE$.

Solution:

We know that,

The lines $AB \parallel DE$

Therefore,

$AE$ becomes the transversal of $AB$ and $DE$.

Since the lines intersected by the transversal are parallel, alternate interior angles are equal.

This implies,

$\angle BAC=\angle AED$

Since the value of $\angle BAC=35^o$ we get,

$\angle AED=35^o$

In a similar way, in $\triangle CDE$ we get,

$\angle DCE+\angle CED+\angle CDE=180^o$  (since the sum of the interior angles of a triangle is $180^o$)

By substituting the values we get,

$\angle DCE+\angle CED+53^o=180^o$

Since $\angle BAC=35^o$ we also get, $\angle CED=35^o$  (alternate interior angles)

Therefore,

$\angle DCE+35^o+53^o=180^o$

$\angle DCE+88^o=180^o$

This implies,

$\angle DCE=180^o-88^o$

$\angle DCE=92^o$

Hence, $\angle DCE=92^o$.

Updated on 10-Oct-2022 13:40:46