Question

In the figure, if $AB\parallel DE,\angle BAC={35}^o$ and $\angle CDE={53}^o$, find $\angle DCE$.

Answer 1

Given, $AB\parallel DE,\angle BAC={35}^{\circ },\angle CDE={53}^{\circ }$
According to the question,
$\angle BAC=\angle CED$ (Alternate interior angles.)
$\therefore \angle CED={35}^{\circ }$
Now, $\angle DCE+\angle CED+\angle CDE={180}^{\circ }$ (Sum of the interior angles of the triangle.)
$\Rightarrow \angle DCE+{35}^{\circ }+{53}^{\circ }={180}^{\circ }$
$\Rightarrow \angle DCE+{88}^{\circ }={180}^{\circ }$
$\Rightarrow \angle DCE={92}^{\circ }$