Answer :

**Given:** ∠RPQ=30°

tangents PQ and PR are drawn from an external point P.

**To find:** ∠RSQ

**Theorem Used:**

1.) The length of two tangents drawn from an external point are equal.

2.) Tangent to a circle at a point is perpendicular to the radius through the point of contact.

**Explanation:**

As P is external point and PR and PQ are tangents,

By theorem (1) stated above,

PQ = PR

As angles opposite to equal sides are equal.

⇒ ∠RQP = ∠QRP

So PQR is an isosceles triangle.

By angle sum property,

∠RQP+∠QRP+∠RPQ=180°

⇒ 2∠RQP+30° =180°

⇒ 2∠RQP=150°

⇒∠RQP=∠QRP=75°

From the theorem (2) stated above,

∠OQP = ∠ORP = 90°

We know sum of angles of a quadrilateral is equal to 360°.

∠OQP + ∠ORP + ∠RPQ + ∠QOR = 360°

⇒ 90° + 90° + 30° + ∠QOR = 360°

⇒ 210° + ∠QOR = 360°

⇒ ∠QOR = 150°

As we know angle subtended by an arc at any point on the circle is half the angle subtended at the centre by the same arc,

Also ∠QSR = ∠SQU (alternate angles)

∠SQU = 75°

Angle on a straight line is 180°,

So,

∠SQU + ∠RQS + ∠PQR = 180°

⇒ 75° + ∠RQS + 75° = 180°

⇒ 150° + ∠RQS = 180°

⇒ ∠RQS = 30°

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