In ΔABC, 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 ∠EPD = 110° and the bisectors of ∠A and ∠B meet at point Q, then ∠AQB = ?

∠EPD = 110°
∠PEC = ∠PDC = 90°
∴ ∠DCE = 180° - 110° = 70°
AQ and BQ angle bisector of ∠A and ∠B respectively.
$$\eqalign{ & \angle {\text{AQB}} = {90^ \circ } + \frac{{\angle {\text{DCE}}}}{2} \cr & = {90^ \circ } + \frac{{{{70}^ \circ }}}{2} \cr & = {125^ \circ } \cr} $$