Question

In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle OAB = 15^\circ \), then \( \angle APB \) equals :

• A. \( 30^\circ \)
• B. \( 15^\circ \)
• C. \( 45^\circ \)
• D. \( 10^\circ \)

Hint:

The angle between tangents is always double the angle between the chord and the tangent (\( \angle APB = 2 \times \angle OAB \)).

Answer 1

The Correct Option is A

To solve the problem of finding \( \angle APB \), we’ll follow these steps:

1. The tangents \( PA \) and \( PB \) from a point \( P \) external to the circle are equal, i.e., \( PA = PB \).
2. In the given circle, \( OA \) and \( OB \) are the radii, and thus, perpendicular to the tangents at the points \( A \) and \( B \) respectively. Therefore, \( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).
3. We know that \( \angle OAB = 15^\circ \), and since \( \angle OAP = 90^\circ \), the remaining angle in \( \triangle OAP \) at point \( A \) is \( \angle OPA \):
4. \[ \angle OPA = 90^\circ - \angle OAB = 90^\circ - 15^\circ = 75^\circ \]
5. Similarly, \( \angle OPB = 90^\circ - \angle OBA = 90^\circ - 15^\circ = 75^\circ \).
6. Now, considering \( \triangle APB \), we have:
   • \( \angle OPA = \angle OPB = 75^\circ \)
7. Using the angle sum property in \( \triangle APB \), we have:
   • \[ \angle APB + \angle OPA + \angle OPB = 180^\circ \]
   • Substituting the known angles: \[ \angle APB + 75^\circ + 75^\circ = 180^\circ \]
   • Simplifying: \[ \angle APB = 180^\circ - 150^\circ = 30^\circ \]

Therefore, the measure of \( \angle APB \) is \(30^\circ\).