Question

Tangents PQ and PR are drawn from external points P to a circle with center O such that angle RPQ=30°. A chord RS is drawn parallel to the tangent PQ. Find angle RQS

Answer 1

For diagram pls refer to attachment




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 this helps you