Multiple choice

If $TP$ and $TQ$ are two tangents to a circle with centre $O$, so that $\angle POQ = {120}^{o}$ then $\angle PTQ$ is equal to

  1. ${120}^{o}$
  2. ${30}^{o}$
  3. ${60}^{o}$
  4. ${90}^{o}$
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C Correct answer
Explanation

In the quadrilateral formed by the center O, the two points of tangency P and Q, and the intersection T, the angles at P and Q are 90 degrees. The sum of angles is 360. Angle PTQ = 360 - 90 - 90 - 120 = 60 degrees.

AI explanation

Since the tangent at any point on a circle is perpendicular to the radius at that point, both triangle OPT and triangle OQT are right-angled triangles. The quadrilateral OPTQ therefore has three right angles (two at the points of tangency and one at O if it were a rectangle, but here angle POQ is 120 degrees). The sum of angles in quadrilateral OPTQ is 360 degrees, so angle PTQ equals 360 degrees minus 90 degrees minus 90 degrees minus 120 degrees. This calculation results in angle PTQ being 60 degrees.

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