Question

In the given figure, find
$\angle AEB$

Answer 1

Sum of angles of $\Delta AEB$

$\begin{array}{c}\angle A+\angle AEB+\angle EBA={180}^0\\ {90}^0+\angle AEB+{20}^0={180}^0\\ \angle AEB={180}^0-{90}^0-{20}^0\\ {180}^0-{110}^0\\ {70}^0\end{array}$