Question

A regular pentagon $ABCDE$ and a square $ABMN$ are formed on opposite sides of $AB$. Find the value of $\angle BCM$

Answer 1

$\begin{array}{c}\angle ABN={90}^o\\ \angle ABC={108}^o\\ \angle MBC={360}^o-{90}^o-{108}^o={162}^o\\ \angle BMC+\angle MBC+\angle BCM={180}^o\\ \angle BMC+\angle BCM={180}^o-{162}^o={18}^o\\ \angle BMC={18}^o-\angle BCM\end{array}$
But, $\angle BCM=\angle BMC$
$\begin{array}{c}2\angle BCM={18}^o\\ \angle BCM=9^o\end{array}$