CAT 2025 Slot 2 · Quant · Question 10

Let AB touch the circle at C.

AQ and AC are both tangents drawn from A, so triangles AQO and ACO are congruent and OA bisects \angle QOC.

By the same argument at B, OB bisects \angle COR.

Adding the halves, \angle AOB=\frac{1}{2}\angle QOC+\frac{1}{2}\angle COR=\frac{1}{2}\angle QOR.

So \angle QOR=2\times{50}^{∘}={100}^{∘}.

In quadrilateral PQOR, \angle PQO=\angle PRO={90}^{∘}, because a radius meets a tangent at right angles.

The four angles total {360}^{∘}, so \angle QPR={360}^{∘}-{90}^{∘}-{90}^{∘}-{100}^{∘}={80}^{∘}.

A lies on PQ and B lies on PR, so \angle APB is that same angle, {80}^{∘}.

Shortcut. Two standard results do the whole question: the third tangent always subtends half the central angle at O, and the angle at an external point is the supplement of the central angle. So \angle APB={180}^{∘}-2\times \angle AOB={180}^{∘}-{100}^{∘}={80}^{∘} in one line.