If tangents PA and PB from a point P to a circle with centre

Solution:

Let us first put the given data in the form of a diagram.

Given data is as follows:

$\angle A P B=80^{\circ}$

We have to find $\angle A O B$.

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

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

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

Also, we know that sum of all angles of a quadrilateral will always be equal to 360°. Therefore in quadrilateral AOBP, we have

$\angle O B A+\angle O A B+\angle A P B+\angle A O B=360^{\circ}$

$90^{\circ}+90^{\circ}+80^{\circ}+\angle A O B=360^{\circ}$

 

$260^{\circ}+\angle A O B=360^{\circ}$

$\angle A O B=100^{\circ}$

Now consider ΔPOA and ΔPOB. We have,

PO is the common side for both the triangles.

PA = PB (Length of tangents drawn from the same external point will be equal)

OA = OB(Radii of the same circle)

Therefore, from SSS postulate of congruent triangles,

Therefore,

We have found that

$\angle A O B=100^{\circ}$

That is,

$\angle P O A+\angle P O B=100^{\circ}$

 

$2 \angle P O A=100^{\circ}$

$\angle P O A=50^{\circ}$

Therefore, the correct answer is choice (a).