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

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°