Question

If PA and PB are the tangents drawn from an external point P to a circle with centre at O and \(\angle APB = 80^\circ\) then \(\angle POA = \)

• A. \(40^\circ\)
• B. \(50^\circ\)
• C. \(80^\circ\)
• D. \(60^\circ\)

Hint:

In problems involving tangents from an external point P to a circle with center O, the triangles \(\triangle OAP\) and \(\triangle OBP\) are always congruent right-angled triangles.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem uses properties of tangents from an external point to a circle. Specifically, we use the fact that the line joining the center of the circle to the external point bisects the angle between the tangents, and the radius is perpendicular to the tangent at the point of contact.

Step 2: Key Formula or Approach:
1. The line segment PO bisects \(\angle APB\). So, \(\angle OPA = \frac{1}{2} \angle APB\).
2. The radius is perpendicular to the tangent at the point of contact. So, \(\angle OAP = 90^\circ\).
3. The sum of angles in a triangle is \(180^\circ\). In \(\triangle OAP\), \(\angle POA + \angle OAP + \angle APO = 180^\circ\).

Step 3: Detailed Explanation:
We are given that \(\angle APB = 80^\circ\).
The line PO bisects this angle, so:
\[ \angle APO = \frac{1}{2} \angle APB = \frac{1}{2} \times 80^\circ = 40^\circ \] Now, consider the triangle \(\triangle OAP\). We know that the radius OA is perpendicular to the tangent PA at the point of contact A.
Therefore, \(\angle OAP = 90^\circ\).
The sum of angles in \(\triangle OAP\) is \(180^\circ\).
\[ \angle POA + \angle OAP + \angle APO = 180^\circ \] Substitute the known values:
\[ \angle POA + 90^\circ + 40^\circ = 180^\circ \] \[ \angle POA + 130^\circ = 180^\circ \] \[ \angle POA = 180^\circ - 130^\circ = 50^\circ \]

Step 4: Final Answer:
The measure of \(\angle POA\) is \(50^\circ\). This matches option (B).