Question

please give question 8 ans​

Answer 1

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°