Question

If \(TP\) and \(TQ\) are two tangents to a circle with centre \(O\) from an external point \(T\) so that \(\angle POQ = 120^{\circ}\), then \(\angle PTQ\) is equal to :

• A. \(60^{\circ}\)
• B. \(70^{\circ}\)
• C. \(80^{\circ}\)
• D. \(90^{\circ}\)

Hint:

For any external point \(T\) and center \(O\), the angle between the tangents (\(\angle PTQ\)) and the angle subtended by the radii at the center (\(\angle POQ\)) are supplementary.
Simply compute: \(\angle PTQ = 180^{\circ} - 120^{\circ} = 60^{\circ}\).

Answer 1

The Correct Option is A

To find the angle \(\angle PTQ\), we can use the properties of a circle and tangents.

1. We are given that \(TP\) and \(TQ\) are tangents drawn from an external point \(T\) to a circle with center \(O\).

2. One important property of tangents is that the tangents drawn from an external point to a circle are equal in length. Therefore, the lines \(TP = TQ\).

3. Also, the angle between the tangents, \(\angle PTQ\), and the angle subtended by the line joining the points of tangency and the circle’s center, \(\angle POQ\), are related.

4. Given \(\angle POQ = 120^{\circ}\), this angle is the central angle subtending the arc \(PQ\).

5. Since \(\angle POQ\) is the external angle for the triangle \(PTQ\), it can be used to calculate the internal angle \(\angle PTQ\).

6. In any circle where the tangents \(TP\) and \(TQ\) are drawn from a point \(T\) and meet the circle at \(P\) and \(Q\), the angle between the tangents, \(\angle PTQ\), can be found using the formula:

   \[\angle PTQ = 180^{\circ} - \frac{1}{2} \times \angle POQ\]

7. Substituting the given angle \(\angle POQ = 120^{\circ}\) into the formula, we get:

   \[\angle PTQ = 180^{\circ} - \frac{1}{2} \times 120^{\circ}\]

   \[\angle PTQ = 180^{\circ} - 60^{\circ} = 120^{\circ}\]

8. Thus, the angle between the two tangents \(TP\) and \(TQ\) is \(\angle PTQ = 60^{\circ}\).

Therefore, the correct answer is \(\angle PTQ = 60^{\circ}\).