In the following figure, if angles ∠ABC = 95°, ∠FED = 115° (not to scale). Then the angle ∠APC is equal to:

Answer: Option B

(External angle of ΔAFP)
∠ABC = 95°, ∠FED = 115°
In, $$\square$$ FABC,
∠AFC = 180° - 95° = 85°
In $$\square$$ FEDA,
∠FAD = 180° - 115° = 65°
So, ∠APC = 85° + 65° = 150°