Question

in Fig. 6.31. ir PQ 1 ST. POR 110 and
RST= 130 find QRS.
Hint : Draw a line parallel to ST through
point R)​

Answer 1

Step-by-step explanation:

Let us draw a line XY

parallel to ST and passing through point R.

∠PQR + ∠QRX

= 180º (Co-interior angles on the same side of transversal QR)

⇒ 110º +

∠QRX = 180º

⇒ ∠QRX =

70º

Also,

∠RST + ∠SRY

= 180º (Co-interior angles on the same side of transversal SR)

130º + ∠SRY

= 180º

∠SRY = 50º

XY is a straight line.

RQ and RS stand on it.

∴ ∠QRX +

∠QRS + ∠SRY

= 180º

70º + ∠QRS

+ 50º = 180º

∠QRS = 180º

− 120º = 60º

Answer 2

Answer:

Solution: 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°

Step-by-step explanation:

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