In Fig. 6.41, if $ \mathrm{AB} \| \mathrm{DE}, \angle \mathrm{BAC}=35^{\circ} $ and $ \angle \mathrm{CDE}=53^{\circ} $, find $ \angle \mathrm{DCE} $.
"
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, $\angle DCE=92^o$.
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