In $$\triangle$$ ABC , the perpendiculars drawn from A, B and C meet the opposite sides at points D, E and F, respectively. AD, BE and CF intersect at point P. If $$\angle EPD = 110^{\circ}$$ and the bisectors of $$\angle$$ A and $$\angle$$ B meet at point Q, then $$\angle$$ AQB?

A

$$125^{\circ}$$

B

$$135^{\circ}$$

C

$$110^{\circ}$$

D

$$115^{\circ}$$