Question

In the figure given, $\angle A,\angle B$ and $\angle C$ are right angles. If and $\angle AEB={40}^{\circ }$ and $\angle BED=\angle BDE$, then find $\angle CDE$.

Answer 1

$\angle AFE={180}^o-({90}^o+{90}^o)$

$={50}^o$

$=\angle BFG$

$\therefore \angle BGF={180}^o-({90}^o+{50}^o)$

$={40}^o$

$=\angle CGD$

$\therefore \angle CDG={180}^o-({90}^o+{90}^o)={50}^o=\angle CDB$

Now $\angle BED+\angle BDE={180}^o-{90}^o={90}^o$

& $\angle BED=\angle BDE\Rightarrow \angle BDE=\frac{{90}^o}{2}={45}^o$

$\therefore \angle CDE=\angle CDG+\angle BDE={50}^o+{45}^o$

${95}^o$