In the figure, PA and PB are two tangents to the circle with centre O such that $\angle APB = 50^\circ$. Then, the measure of $\angle OAB$ is :

Step 1: Identify properties of tangents and radii

We know that the radius drawn to the tangent at the point of contact is perpendicular to the tangent. Therefore, the angle between the radius OA and tangent PA is $90^\circ$, and similarly, the angle between radius OB and tangent PB is $90^\circ$.

Step 2: Find $\angle$ AOB in quadrilateral OAPB

The sum of angles in a quadrilateral is $360^\circ$. In quadrilateral OAPB, we have $\angle OAP = 90^\circ$, $\angle OBP = 90^\circ$, and $\angle APB = 50^\circ$. We can find $\angle AOB$ by subtracting the sum of these three angles from $360^\circ$.

Step 3: Identify properties of triangle OAB

In triangle OAB, OA and OB are both radii of the same circle. Therefore, triangle OAB is an isosceles triangle, which means the angles opposite to the equal sides are also equal.

Step 4: Calculate $\angle$ OAB

Since $\triangle OAB$ is an isosceles triangle with $OA = OB$, the base angles $\angle OAB$ and $\angle OBA$ are equal. The sum of angles in a triangle is $180^\circ$. So, we can find $\angle OAB$ by subtracting $\angle AOB$ from $180^\circ$ and dividing the result by 2.

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