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 = ?

Answer: Option D

Solution

Answer: Option D

Solution:

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