In Fig. if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

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°.