In Fig. 7, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.

PR = PQ (Tangents drawn from an external point are equal)

⇒ ∠PQR = ∠ PRQ (Angles opposite to equal sides)

In ΔPQR,

∠PQR + ∠ PRQ + ∠ RPQ = 180° (Sum of all angles of a triangle is 180°)

⇒ ∠ PRQ + ∠ PRQ + 30° = 180°

⇒ 2∠PRQ = 150°

⇒

Given that, RS||PQ

So, ∠SRQ = ∠PQR = 75° (Alternate interior angle)

Join OR and OQ.

∠ ORP = 90° (tangent is perpendicular to the radius)

⇒ ∠ ORQ + ∠ QRP = 90°

⇒ ∠ ORQ + 75° = 90°

⇒ ∠ ORQ = 15°

In ΔORQ,

∠ OQR = ∠ ORQ = 15° (Angles opposite to equal sides as OR = OQ = radius)

∠ OQR + ∠ ORQ + ∠ ROQ = 180° (Sum of all angles of a triangle is 180°)

⇒ 15° + 15° + ∠ROQ = 180°

⇒ ∠ ROQ = 150°

Now, ∠ RSQ = ∠ROQ/2 (Angle in the major sector)

⇒

In Δ RSQ,

∠ RSQ + ∠ SRQ + ∠ RQS = 180° (Sum of all angles of a triangle is 180°)

⇒ 75° + 75° + ∠RQS = 180°

∴ ∠ RQS = 30°