Almost Isoceles

The trick here is constructing equilateral $\triangle ABF$ .

$m\angle FBC + m\angle BCD = 15° + 165° = 180°$ and $FB = BC = CD$

⇒ $BCDF$ is a rhombus

⇒ $\overline{BC} \parallel \overline{DF}$ and $\triangle AFD$ is isoceles

⇒ $m\angle E = m\angle ADF = 90° - \frac{1}{2} m\angle AFD = 90° - \frac{1}{2} (360° - 60° - 165°) = \fbox{22.5°}$

Connect $AC$ and observe that $\triangle ABC$ is isosceles, which implies that $\angle BCA=\angle BAC =52.5^\circ$ . Applying the cosine rule to $\triangle ABC$ , we find that $AC=\sqrt{2-2\cos{75^\circ}}$

Now, note that $\angle ACD = 160^\circ-52.5^\circ=112.5^\circ$ . Applying the cosine rule to $\triangle ACD$ , we find that $AD = \sqrt{AC^2+1-2AC\cos{112.5^\circ}}=1.84776...$

Apply the sine rule to $\triangle ACD$ . $\frac{AD}{\sin{122.5^\circ}}=\frac{1}{\angle CAD}\Rightarrow \angle CAD = 30^\circ$

Therefore, $\angle E = 180^\circ-112.5^\circ-30^\circ-15^\circ=12.5^\circ$