In the given diagram, AB is a tangent to the circle at P. If ∠PQR = 112° and PS = SR, find:

(a) PQRS is a cyclic quadrilateral.

We know that,

The sum of opposite angles of a cyclic quadrilateral is 180°.

$$\begin{aligned}\Rightarrow \angle PQR + \angle PSR = 180° \\[1em] \Rightarrow 112° + \angle PSR = 180° \\[1em] \Rightarrow \angle PSR = 180° - 112° = 68°.\end{aligned}$$

Hence, ∠PSR = 68°.

(b) We know that,

By the alternate segment theorem, the angle between the tangent and a chord equals the angle subtended by the chord in the alternate segment.

∠BPR is the angle between tangent PB and chord PR, and its alternate segment contains Q.

∴ ∠BPR = ∠PQR = 112°.

Hence, ∠BPR = 112°.

(c) Given, PS = SR.

Equal chords subtend equal arcs.

∴ arc PS = arc SR.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended by it on the circumference.

∠PQR is the inscribed angle standing on arc PSR.

$\Rightarrow \text{arc PSR} = 2 \times \angle PQR = 2 \times 112° = 224°.$

Since arc PS = arc SR,

$\Rightarrow \text{arc PS} = \frac{224°}{2} = 112°.$

∠PTS is the inscribed angle standing on arc PS.

$\Rightarrow \angle PTS = \frac{1}{2} \times \text{arc PS} = \frac{1}{2} \times 112° = 56°.$

Hence, ∠PTS = 56°.