Question

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then $\angle POA$ will be

• A. 40°
• B. 50°
• C. 70°
• D. 80°

Hint:

When tangents are inclined at angle $\theta$, the angle between the centre lines equals $180° - \theta$.

Answer 1

The Correct Option is A

Step 1: Recall property of tangents.
Tangents drawn from an external point are equal in length and make equal angles with the line joining the centre and the external point.
Step 2: Relationship between angles.
If $\angle APB = 80°$, then $\angle AOB = 180° - 80° = 100°$ because the quadrilateral $OAPB$ is cyclic (tangent-secant theorem). But the question asks for $\angle POA$. Since $\triangle OAP$ is isosceles ($OA = OP$), \[ \angle POA = \frac{1}{2} (180° - 100°) = 40°. \] Step 3: Conclusion.
Therefore, $\angle POA = 40°$.