In ΔPQR, PS is the bisector of ∠P and PT ⊥ OR, then ∠TPS is equal to:

A. ∠Q + ∠R

B. 90° + $$\frac{1}{2}$$ ∠Q

C. 90° - $$\frac{1}{2}$$ ∠R

D. $$\frac{1}{2}$$ (∠Q - ∠R)

Answer: Option D

Solution(By Examveda Team)

∠1 + ∠2 = ∠3 [PS is bisector.] - - - - - - (1)
∠Q = 90° - ∠1
∠R = 90° -∠2 - ∠3
So,
∠Q - ∠R = (90° - ∠1) - (90° - ∠2 - ∠3)
∠Q - ∠R = ∠2 + ∠3 - ∠1
∠Q - ∠R = ∠2 + (∠1 + ∠2) -∠1[using equation 1]
∠Q - ∠R = 2∠2
$$\frac{1}{2}$$ × (∠Q - ∠R) = ∠TPS