In the given figure, PQ and PR are tangents drawn from P to a circle with centre O. If ∠OPQ = 35°, then

Solution:

Consider and . We have,

PO is the common side for both the triangles.

OQ = OR(Radii of the same circle)

PQ = PR(Tangents from an external point will be equal)

Therefore, from SSS postulate of congruent triangles, we have,

$\triangle P O Q \cong \triangle P O R$

Therefore,

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

That is,

$\angle O P Q=\angle a$

It is given that,

$\angle O P Q=35^{\circ}$

Therefore,

$\angle a=35$

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

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

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

$\angle b+90^{\circ}+35^{\circ}=180^{\circ}$

$\angle b=55^{\circ}$

Therefore, the correct answer is option (b).