Question

In Fig , tangents PQ and PR are drawn to a circle such that ∠RPQ =30°. A chord RS is drawn parallel to the tangent PQ. Find the ∠RQS.

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]
∴ ∠PRQ = ∠PQR [Angles opposite to equal sides are equal in a Δ]
In ΔPQR
∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a Δ]
2∠RQP + 30° = 180°
2∠RQP = 150°
∠RQP = 75°
so ∠RQP = ∠QRP = 75°
∠RQP = ∠RSQ = 75°  [ By Alternate Segment Theorem]
Given, RS || PQ
∴ ∠RQP = ∠SRQ = 75°    [Alternate angles]
∠RSQ = ∠SRQ = 75°                                          
∴ 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°
∴ ∠RQS = 30°

Answer 2

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°