Question

pls tell this answer pls​

Answer 1

hope this helps u

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°

Answer 2

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°

$\red{ \angle {PQR} } \green { = 60\degree }$

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