Question

Solve it for angle RQS.
It is hard question.

Answer 1

✌️Given ∠RPQ=30° and PR and PQ are tangents drawn from P to the same circle.

 Hence
PR = PQ [Since tangents drawn from an external point to a circle are equal in length]

Therefore, ∠PRQ = ∠PQR [Angles opposite to equal sides are equal in a triangle] 

In ΔPQR
∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a triangle] 
2∠RQP + 30° = 180°
2∠RQP = 150°
∠RQP = 75° 

Hence,
∠RQP = ∠QRP = 75°
∠RQP = ∠RSQ = 75° [ By Alternate Segment Theorem] 
Given, RS || PQ
Therefore ∠RQP = ∠SRQ = 75° [Alternate angles] 
∠RSQ = ∠SRQ = 75°
Therefore QRS is also an isosceles triangle. [Since sides opposite to equal angles of a triangle are equal.]
∠RSQ + ∠SRQ + ∠RQS = 180° [Angle sum property of a triangle]
 75° + 75° + ∠RQS = 180° 
150° + ∠RQS = 180°
Therefore, ∠RQS = 30°

♥️HOPE IT HELPS U IF YEAH PLZ MARK IT AS BRAINLIEST❤️☺️

Answer 2

hey mate here is your answer ✌♥✌♥

by the ppty of tangent and circle

pq=pr

so by the ppty of isosceles triangle

angle PQR =prq

now by asp of triangle

prq + PQR +rpq=180

2rqp=150

rqp=75

now ppty of circle

rqp=rsq

rsq=75

now by alternate Angles

rsq=sqa ... a is a point at the end of tangent pq

sqa=75

now by Linear pair

sqa+sqr+rqp=180

sqr=180-150

sqr=30°

hope it will be helpful to you ✔✔✔

mark me brainliest ✌✌✌