In Figure 5.61 OP || RS, ∠OPQ = 110° and ∠QRS = 130° then PQR is equal to-

For the above figure let us draw a straight line through Q as follows:

It is given that OP || RS and ∠ OPQ = 110° and ∠ QRS = 130°

Sum of interior angles lying on the same side of the transversal is always 180°.

So by referring above figure:

∠ OPQ + ∠ PQX = 180°

110° + ∠ PQX = 180°

∠ PQX = 180 – 110

∠ PQX = 70°

Again Sum of interior angles lying on the same side of the transversal is always 180°.

So by referring above figure:

∠ QRS + ∠ RQY = 180°

130° + ∠ RQY = 180°

∠ RQY = 180 – 130

∠ RQY = 50°

Now we have assumed that XY is a straight line.

So ∠ PQR + ∠ RQY + ∠ PQX = 180° as they form a linear pair.

∠ PQR + 70° + 50° = 180°

∠ PQR + 120° = 180°

∠ PQR = 180 – 120

∠ PQR = 60°

So the correct answer is C.