If $\angle A = 60^\circ$, $\angle E = 40^\circ$ and $\angle C = 30^\circ$, then $\angle BDC =$

$\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ$
The sum of the angles in a triangle is $180^\circ$. We can find $\angle ABE = 80^\circ$, so $\angle CBD = 180-80=100^\circ$.
\[\angle BDC = 180-100-30=\boxed{\text{(B)}\ 50^\circ}\]