Catch a Star

Sum of the the exterior angles of every polygon is 360. sum of the exterior angles of then interior pentagon is 360. If we extend each side of the pentagon in the same direction then the sum of all the exterior angles will be 360. And also if we extend the each side in opposite direction then again sum of eterior angles will be 360. Adding both sums 360+360=720 sum of all the outer triangles 5*180=900 THEN SUM OF THE GIVEN ANGLES WILL BE 900-720=180.

Let's mark intersection points between $A$ and $C$ as $P$ and $Q$ ( $BE \cap AC = P$ and $BD \cap AC = Q$ ). Then $\angle BPQ = \angle C + \angle E$ and $\angle BQP = \angle A + \angle D$

Therefore the sum of angles in a star equals to the sum of angles in a triangle:

$\angle A + \angle B + \angle C + \angle D + \angle E = \angle B + \angle P + \angle Q = \boxed{180}$