In the figure (i) given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.

From figure,

In △SRT,

SR ⊥ ST (∵ tangent is perpendicular to radius from that point.)

so, ∠TSR = 90°

Since, sum of angles in a triangle = 180°

⇒ ∠TSR + ∠SRT + ∠STR = 180°
⇒ 90° + 65° + x = 180°
⇒ x + 155° = 180°
⇒ x = 25°.

SQ subtends ∠SOQ at the centre and ∠STQ on point D.

∴ ∠SOQ = 2∠STQ (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle.)

y = 2x = 2 × 25° = 50°.

In △OSP,

Since, sum of angles in a triangle = 180

⇒ ∠OSP + ∠SOP + ∠SPO = 180°
⇒ 90° + y + z = 180°
⇒ 90° + 50° + z = 180°
⇒ z + 140° = 180°
⇒ z = 180° - 140° = 40°.

Hence, the value of x = 25°, y = 50° and z = 40°.