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Question
In the given figure, PQ = QR and ∠RQP = 72°. CP and CQ are tangents. Determine ∠POQ.

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Solution
Given:
PQ = QR
∠RQP = 72°
CP and CQ are tangents to the circle.
O is the centre of the circle.
To find: ∠POQ.
Step 1: Use the triangle property
Since PQ = QR, triangle PQR is isosceles.
So, base angles are equal:
∠RPQ = ∠PRQ
Let each angle = x.
We know in triangle PQR:
x + x + 72° = 180°
2x = 108°
x = 54°
Thus,
∠RPQ = 54°
∠PRQ = 54°
Step 2: Use the tangent–radius theorem
The radius drawn to the point of tangency is perpendicular to the tangent.
So:
OQ ⟂ CQ
OP ⟂ CP
Thus, angles between PQ and OQ, and between PR and OP, relate directly to the triangle angles.
Relation between ∠PRQ and arc PQ
Step 3: Relation between inscribed and central angles.
Angle POQ is the central angle subtending arc PQ.
Angle PRQ (54°) is the inscribed angle subtending the same arc PQ.
Central angle = 2 × Inscribed angle
∠POQ = 2 × ∠PRQ
∠POQ = 2 × 54°
∠POQ = 108°
