In the figure (ii) given below, O is the center of the circle and PT is the tangent to the circle at P. Given ∠QPT = 30°, calculate

(i) ∠PRQ

(ii) ∠POQ.

From figure,

OP ⊥ PT (∵ tangent at a point and radius through the point are perpendicular to each other.)

∴ ∠OPT = 90°.

Given, ∠QPT = 30°.

From figure,

∠OPT = 90°
∠OPQ + ∠QPT = 90°
∠OPQ + 30° = 90°
∠OPQ = 90° - 30° = 60°.

In △OPQ,

OP = OQ (∵ both are equal to radius of the circle.)

So, the triangle is isosceles. So,

∠OQP = OPQ = 60°.

Since sum of angles in a triangle = 180°.

In △OPQ,

⇒ ∠OQP + ∠OPQ + ∠POQ = 180°
⇒ 60 + 60 + ∠POQ = 180°
⇒ 120 + ∠POQ = 180°
⇒ ∠POQ = 180° - 120°
⇒ ∠POQ = 60°

Reflex ∠POQ = 360° - ∠POQ = 360° - 60° = 300°.

Arc PQ subtends Reflex ∠POQ at the centre and ∠PRQ at the remaining part of the circle.

∴ Reflex ∠POQ = 2∠PRQ (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle)

300° = 2° × ∠PRQ
∠PRQ = $\dfrac{300°}{2}$ = 150°.

(i) Hence, the value of ∠PRQ = 150°.

(ii) Hence, the value of ∠POQ = 60°.