$$ABCDE$$ is a pentagon with $$\angle B = 90^\circ$$ and $$\angle E = 150^\circ$$. If $$\angle C + \angle D = 180^\circ$$ and $$\angle A + \angle D = 180^\circ$$, then the external angle $$\angle D$$ is

The interior angles of a pentagon add up to $$540^\circ$$, so $$\angle A + \angle C + \angle D = 540^\circ - 90^\circ - 150^\circ = 300^\circ$$. Substituting $$\angle A = 180^\circ - \angle D$$ and $$\angle C = 180^\circ - \angle D$$ gives $$360^\circ - \angle D = 300^\circ$$, so the interior angle $$\angle D = 60^\circ$$. The external angle at $$D$$ is therefore $$180^\circ - 60^\circ = 120^\circ$$.