Question

In Fig. 6.31, if PQ II ST. Z PQR = 110° and ZRST= 130°, find ZQRS. [Hint : Draw a line parallel to ST through point R.)
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Answer 1

Interior angles on the same side of the transversal:The pair of interior angles on the same side of the transversal are called consecutive interior angles or allied angles or co interior angles.

If a transversal intersects two Parallel Lines then each pair of interior angles on the same side of the  transversal is supplementary.

If a transversal intersects two lines such that a pair of alternate interior angles is equal then the two lines are parallel.

SOLUTION :

Given :PQ || ST, ∠PQR = 110° and ∠RST = 130°

Construction:A line XY parallel to PQ and ST is drawn.

∠PQR + ∠QRX = 180° (Angles on the same side of transversal.)

110° + ∠QRX = 180°

∠QRX = 180° - 110°

∠QRX = 70°

Also,∠RST + ∠SRY = 180° (Angles on the same side of transversal.)

130° + ∠SRY = 180°

∠SRY = 50°

Now,∠QRX +∠SRY + ∠QRS = 180°

70° + 50° + ∠QRS = 180°

∠QRS = 60°

Hence, ∠QRS = 60°

Answer 2

Answer:

Let us draw a parallel line XY to PQ || ST and passing through point R.

Sum of interior angle on the same side of the transversal is always = 180°

So that

∠ PQR + ∠ QRX = 180°

Given that ∠ PQR= 110°

110° + ∠QRX = 180°

∠QRX = 180° -110°

∠QRX = 70°

Sum of interior angle on the same side of the transversal is always = 180°

∠RST + ∠SRY = 180° (Co-interior angles on the same side of transversal SR)

Also

130° + ∠SRY = 180°

∠SRY = 50°

XY is a straight line. Use property of linear pair we get

∠QRX + ∠QRS + ∠SRY = 180°

70° + ∠QRS + 50° = 180°

∠QRS = 180° − 120°

= 60°