Question

mate a doubt, dont spam​ please

Answer 1

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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°

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Answer 2

Answer:

inbox me plz here is ur 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°

PLz MARK ME AS BRAINLIEST