In triangle ABC, ∠ABC = 90°. BP is drawn perpendicular to AC. If ∠BAP = 50°, then what is the value (in degrees) of ∠PBC?

Given : ∠ABC = ∠BPC = 90° and ∠BAP = 50°

To find : ∠PBC = $$\theta$$ = ?

Solution : In $$\triangle$$ ABC,

=> $$\angle$$ BAC + $$\angle$$ ABC + $$\angle$$ ACB = $$180^\circ$$

=> $$50^\circ+90^\circ$$ + $$\angle$$ ACB = $$180^\circ$$

=> $$\angle$$ ACB = $$180-140=40^\circ$$

Similarly, in $$\triangle$$ BPC

=> $$\angle$$ BPC + $$\angle$$ PBC + $$\angle$$ PCB = $$180^\circ$$

=> $$90^\circ+\theta$$ + $$40^\circ$$  = $$180^\circ$$

=> $$\theta$$  = $$180-130=50^\circ$$

=> Ans - (C)