Let length of side of Pentagon be $x$ then $CE=\frac{(\sqrt{5} +1)x}{2}$ and $\angle ACE=30^\circ, \angle ACB=\angle BAC=42^\circ$ Take point $K\in CE$ such that $\angle BKC=\angle BCK =72^\circ$ also observe $B$ is circumcenter of $\triangle ACK$ and $AK=EK\implies \angle AEK=\angle KAE=66^\circ$ so $\boxed{\angle E=102^\circ}$.