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A triangle PQR is inscribed in a circle with its centre at O. A tangent PT is drawn at P such that \(\angle \text{QPT} = 36°\). What is \(\angle \text{POQ}\) equal to ?

This question was previously asked in
CDS 2 2026 Maths Question Paper (13-Sep-2026)
The correct answer is
72°

Circle Geometry: Tangent-Chord Theorem Application

The problem involves a triangle PQR inscribed in a circle with center O and a tangent PT drawn at point P. We are given the angle between the tangent PT and the chord PQ, which is ∠QPT = 36°. We need to find the angle subtended by the arc PQ at the center, ∠POQ.

Applying the Tangent-Chord Theorem

  • Tangent-Chord Theorem: This theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment subtended by the chord.

    In this case, the angle between tangent PT and chord PQ is ∠QPT. The angle in the alternate segment subtended by chord PQ is ∠PRQ.

    Therefore, according to the theorem: \(\angle \text{PRQ} = \angle \text{QPT}\) Given ∠QPT = 36°, we have: \(\angle \text{PRQ} = 36°\)

Relating Circumference Angle to Center Angle

  • Angle at Center Theorem: The angle subtended by an arc at the center of a circle is double the angle subtended by the same arc at any point on the circumference.

    The arc PQ subtends ∠POQ at the center O and subtends ∠PRQ at the circumference point R.

    Using the theorem: \(\angle \text{POQ} = 2 \times \angle \text{PRQ}\)

Calculating Angle POQ

  • Substitute the value of ∠PRQ found earlier:

    \(\angle \text{POQ} = 2 \times 36°\) \(\angle \text{POQ} = 72°\)

Thus, the angle ∠POQ is 72°.

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