It is given that PQ || ST,
Draw a line XY || ST,
So, XY || PQ, i.e, PQ || ST || XY
Since, PQ || XY & QR is the transversal.
We know that,
Sum of co-interior angles = 180°.
⇒ ∠PQR + ∠QRX = 180°
⇒ 110° + ∠QRX = 180°
⇒ ∠QRX = 180° - 110°
⇒ ∠QRX = 70°.
Also, ST || XY & SR is transversal
⇒ ∠SRY + ∠RST = 180° (Interior angles on the same side of the transversal are supplementary)
⇒ ∠SRY + 130° = 180°
⇒ ∠SRY = 180° - 130°
⇒ ∠SRY = 50°.
From figure,
⇒ ∠QRX + ∠QRS + ∠SRY = 180° (Linear pairs)
⇒ 70° + ∠QRS + 50° = 180°
⇒ 120° + ∠QRS = 180°
⇒ ∠QRS = 180° - 120°
⇒ ∠QRS = 60°
Hence, ∠QRS = 60°.