In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If ∠PQB = 55°.

(a) We know that,

Angle between the radius and tangent at the point of contact is 90°.

∴ ∠PBA = ∠PBQ = 90°

In △ PBQ,

By angle sum property of triangle,

⇒ ∠PBQ + ∠BQP + ∠QPB = 180°

⇒ 90° + 55° + ∠QPB = 180°

⇒ ∠QPB = 180° - 90° - 55° = 35°.

From figure,

⇒ ∠SPB = ∠QPB = 35°

We know that,

Angles in same segment are equal.

⇒ ∠SRB (x°) = ∠SPB = 35°

⇒ x° = 35°.

We know that,

The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.

∴ ∠SOB = 2∠SRB

⇒ y° = 2x° = 2 × 35° = 70°.

From figure,

⇒ z° = x° = 35° (Angles in alternate segment are equal)

Hence, x° = 35°, y° = 70° and z° = 35°.

(b) From figure,

⇒ OB = OR (Radius of the same circle)

⇒ ∠OBR = ∠ORB (Angles opposite to equal sides are equal)

⇒ ∠OBR = x° = 35°

∴ ∠OBR = ∠OPS

The above angles are alternate angles.

∴ RB // PS.

Hence, proved that RB // PS.