Question

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

Answer 1

$\angle DCE=\angle EAB$ (Alternate angles as $AB||DE$ and $AE$ is transversal)

$\angle DEC={35}^{\circ }$

In $△DCE$, we have

$\angle DCE+\angle CED+\angle CED={180}^{\circ }$ (Angle sum property)

$\angle DCE+{35}^{\circ }+{53}^{\circ }={180}^{\circ }\angle DCE={180}^{\circ }-{88}^{\circ }={92}^{\circ }$