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Answers
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Given, PQ || ST, ∠PQR = 110° and ∠RST = 130°
Draw a line AB parallel to ST through R.
Now, ST || AB and SR is a transversal.
So, ∠RST + ∠SRB = 180°[since, sum of the interior angles on the same side of the transversal is 180°]
⇒ 130° + ∠SRB = 180° ⇒ ∠SRB = 180°-130°
⇒ ∠SRB = 50° …(i)
Since, PQ || ST and AB || ST, so PQ || AB and then QR is a transversal.
So, ∠PQR + ∠QRA = 180° [since, sum of the interior angles on the same side of the transversal is 180°] ⇒ 110° + ∠QRA = 180° ⇒ ∠QRA = 180° -110°
⇒ ∠QRA=70° ..(ii)
Now, ARB is a line.
∴ ∠QRA + ∠QRS + ∠SRB = 180° [by linear pair axiom]
⇒ = 70° + ∠QRS + 50° = 180° ⇒ 120° + ∠QRS = 180°
⇒ => ∠QRS = 180° -120° ⇒ ∠QRS = 60°

Given OP//RS
Draw a line 'l' parallel to RS through Q .
From the figure
a + 110° =180°
and c + 130° = 180°
[Interior angles on the same side of the Transversal ]
Therefore.
a = 180° - 110° = 70°
b = 180° - 130° = 50°
a + b + c = 180° [ Angle at a point on a line ]
=> 70° + b + 50° = 180°
=> b + 120° = 180°
=> b = 180° - 120°
=> b = 60°
•••♪
