Question

In given figure, If $△EDC\sim △EBA,\angle BEC={115}^0$ and $\angle EDC={70}^0$ Find $\angle DEC,\angle DCE,\angle EAB,\angle AEB$ and $\angle EBA$.

Answer 1

Since BD is a line and EC is a ray on it.

$\therefore$ $\angle DEC+\angle BEC={180}^0$

$\Rightarrow$ $\angle DEC+{115}^0={180}^0$

$\Rightarrow$ $\angle DEC={180}^0-{115}^0={65}^0$

But, $\angle AEB=\angle DEC$ [Vertically opposite angles]

$\therefore$ $\angle AEB={65}^0$

In $△CDE$, we have

$\angle CDE+\angle DEC+\angle DCE={180}^0$

$\Rightarrow$ ${70}^0+{65}^0+\angle DCE={180}^0$

$\Rightarrow$ $\angle DCE={180}^0-{135}^0={45}^0$

It is given that $△EDC\sim △EBA$

$\therefore$ $\angle EBA=\angle EDC,\angle EAB=\angle ECD$

$\Rightarrow$ $\angle EBA={70}^0$ and $\angle EAB={45}^0$

Hence, $\angle DEC={65}^0,\angle DCE={45}^0,\angle EAB={45}^0,\angle AEB={65}^0$ and $\angle EBA={70}^0$.