ABCDE is part of a regular polygon which has an exterior angle of $20^\circ$. CDLM is a square. Find

1. Find the interior angle of the regular polygon at $C$.

Exterior angle $=20^\circ$

Interior angle of regular polygon:
$180^\circ-20^\circ=160^\circ$
So $\angle BCD=160^\circ$.

2. Use the square angle at $C$ to find $x$.

Since $CDLM$ is a square,
$\angle DCM=90^\circ$
Around point $C$ on the inside shown,
$x+90^\circ=160^\circ$
$x=70^\circ$

1. Find the angle at $B$ in triangle $BCM$.

From part (a), $\angle BCM = x = 70^\circ$.

Since $BC = CD$ as sides of the regular polygon, and $CM = CD$ as sides of the square,
$BC = CM.$
So triangle $BCM$ is isosceles.

Hence
$\angle CBM = \angle BMC = \frac{180^\circ - 70^\circ}{2} = 55^\circ.$

2. Use the interior angle of the regular polygon at $B$ to find $y$.

The interior angle of the regular polygon is
$180^\circ - 20^\circ = 160^\circ.$

At $B$,
$y + \angle CBM = 160^\circ.$

So
$y = 160^\circ - 55^\circ = 105^\circ.$