Mate a doubt, dont spam please

1.	Mate a doubt, dont spam please
Answer» $\huge\underline{\overline{\mid{\bold{\orange{ANSWER-}}\mid}}}$ 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

Answer»

$\huge\underline{\overline{\mid{\bold{\orange{ANSWER-}}\mid}}}$

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)

>>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°

Discussion