In the given figure, APB is a tangent to a circle with centre O at point P.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact.

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Therefore,

$\angle O P B=90^{\circ}$

That is,

$\angle O P Q+\angle Q P B=90^{\circ}$

It is given that,

$\angle Q P B=50^{\circ}$

Therefore, we have,

$\angle O P Q+50^{\circ}=90^{\circ}$

$\angle O P Q=40^{\circ}$

Now, consider . We have,

OP = OQ(Radii of the same circle)

Since angles opposite to equal sides will be equal, we have,

$\angle P Q O=\angle O P Q$

We have found that,

$\angle O P Q=40^{\circ}$

Therefore,

$\angle P Q O=40^{\circ}$

We know that sum of all angles of a triangle will be equal to . Therefore,

$\angle O P Q+\angle P Q O+\angle P O Q=180^{\circ}$

$40^{\circ}+40^{\circ}+\angle P O Q=180^{\circ}$

$\angle P O Q=100^{\circ}$

The correct answer is option (a).