In the following figure, PQ and PR are tangents to the circle, with center O. If ∠QPR = 60°, calculate :

(i) ∠QOR,

(ii) ∠OQR,

(iii) ∠QSR.

(i) We know that,

The tangent at any point of a circle and the radius through this point are perpendicular to each other.

In quadrilateral ORPQ,

∠OQP = ∠ORP = 90° [∵ The tangent at any point of a circle and the radius through this point are perpendicular to each other]

∠QPR = 60° [Given]

⇒ ∠OQP + ∠ORP + ∠QPR + ∠QOR = 360° [By angle sum property of quadrilateral]

⇒ 90° + 90° + 60° + ∠QOR = 360°

⇒ 240° + ∠QOR = 360°

⇒ ∠QOR = 120°.

Hence, ∠QOR = 120°.

(ii) Join QR.

In △QOR,

OQ = OR (Radii of same circle)

As, angles opposite to equal sides are equal.

∴ ∠OQR = ∠ORQ ……….(1)

⇒ ∠OQR + ∠ORQ + ∠QOR = 180°

⇒ ∠OQR + ∠ORQ + 120° = 180°

⇒ ∠OQR + ∠ORQ = 180° - 120°

⇒ ∠OQR + ∠ORQ = 60°

⇒ ∠OQR + ∠OQR = 60°

⇒ 2∠OQR = 60°

⇒ ∠OQR = $\dfrac{60°}{2}$

⇒ ∠OQR = 30°.

Hence, ∠OQR = 30°.

(iii) Arc RQ subtends ∠QOR at the center and ∠QSR at the remaining part of the circle.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended at the circumference.

⇒ ∠QOR = 2∠QSR

⇒ ∠QSR = $\dfrac{1}{2}$∠QOR = $\dfrac{1}{2} \times 120°$ = 60°.

Hence, ∠QSR = 60°.