As we know that the tangents drawn from an external point to a circle are equal.
Therefore, PQ = PR
Also, from the figure, PQR is an isosceles triangle, because PQ = PR
Therefore, ∠RQP =∠QRP (Because the corresponding angles of the equal sides of the isosceles triangle are equal)And, from the angle sum property of a triangle,
∠RQP + ∠QRP + ∠RPQ = 180°
∠RQP + ∠RQP + ∠RPQ = 180°
2∠RQP +∠RPQ = 180°
2∠RQP +30o = 180°
2∠RQP = 180° - 30°
2∠RQP = 150°
∠RQP = 150°/2
Therefore, ∠RQP = 75°
SR || QP and QR is a transversal
∵ ∠SRQ = ∠ PQR …[Alternate interior angle]
∴ ∠SRQ = 75°
⇒ ∠ORP = 90°…[Tangent is Perpendicular to the radius through the point of contact]
∠ORP = ∠ORQ + ∠QRP
⇒ 90° = ∠ORQ + 75°
⇒ ∠ORQ = 15°
Similarly, ∠ RQO = 15°
In Δ QOR,
∠QOR + ∠QRO + ∠OQR = 180°
⇒ ∠QOR + 15° + 15° = 180°
⇒ ∠QOR = 150°
⇒ ∠QSR = ∠QOR/2
⇒ ∠QSR = 150°/2 = 75°
In ΔRSQ,
∠RSQ + ∠QRS + ∠RQS = 180°
⇒ 75° + 75° + ∠RQS = 180°
∠RQS = 30º