In ΔPQR, ∠P = 120°, PS ⊥ QR at S and PQ + QS = SR, then the measured value of ∠PTQ is ?

20° 40° 60° 90°

40°

Draw a line PT such that QS = ST Step 1: In ΔQPT QS = ST and PS ⊥ QT the ΔQPT is isosceles triangle then PQ = PT Step 2: As PQ = PT so ∠Q = ∠T = 2θ and PT = TR ⇒ ∠TPR = ∠TRP = θ  In ΔPQR ∠P + ∠Q + ∠R = 180° (∠P = 120°) 3θ = 60° θ = 20° ∠PTQ = 2 × 20° = 40°