Question

In Fig. if$ AB\parallel DE$, $ \angle BAC = 35°$ and $ \angle CDE = 53°$, find $ \angle DCE$.

Answer 1

Step 1: Find the value of $\angle CED$

Given, $\angle DCE$

$AE$ is a transversal between parallel lines $AB$ and $DE$

$\angle BAE$ and $\angle AED$ are alternate angles between parallel lines, as shown in the diagram.

$\Rightarrow \angle BAE=AED$

$\therefore \angle AED={35}^{\circ }$

Points $A,C,E$ are collinear

$\therefore \angle AED=\angle CED={35}^{\circ }$

Step 2: Find the value of $\angle DCE$

In $△CDE$ by the angle sum property of a triangle

$\angle CDE+\angle CED+\angle DCE={180}^{\circ }$

$\Rightarrow$ $53+35+\angle DCE=180$

$\Rightarrow$ $\angle DCE=180-88$

$\Rightarrow$ $\angle DCE={92}^{\circ }$

Hence, $\angle DCE$ is ${92}^{\circ }$