Question

Fig. 6.30
4.
S
T
In Fig. 6.31, if PQ II ST. PQR = 110° and
ZRST=130°, find ZQRS.
[Hint : Draw a line parallel to ST through
point R.]
P
1309
1109
R
in 631​

Answer 1

Here is your answer:

60°

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°.