Question

In fig 6.41, if $AB||DE$, $\angle BAC={35}^o$ and $\angle CDE={53}^o$, find $\angle DCE$.

Answer 1

$\angle AED=\angle BAC$ (alternate interior angles)
$\Rightarrow \angle AED={35}^0$
$\Rightarrow \angle CED={35}^0$
$In\Delta DEC$
$\angle CDE+\angle DCE+\angle CED={180}^0$(by angle sum property)
${53}^0+\angle DCE+{35}^0={180}^0$
$\angle DCE+{88}^0={180}^0$
$\angle DCE={180}^0-{88}^0$
$\angle DCE={92}^0$