Question

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

Answer 1

$\angle BAE=\angle CED$

$\therefore \theta ={35}^{\circ }$ (Here $AB\parallel DE$, Here AE acts as transverse)

in $\Delta CDE\Rightarrow 53+35+\angle C=180$

$88+\angle C={180}^{\circ }$

$\angle C={92}^{\circ }$

$\therefore DCE={92}^{\circ }$