In the given figure, PT touches the circle with center O at point R. Diameter SQ is produced to meet the tangent TR at P.

Given ∠SPR = x° and ∠QRP = y°;

prove that :

(i) ∠ORS = y°

(ii) write an expression connecting x and y.

(i) From figure,

⇒ ∠QRP = ∠OSR = y° [Angles in alternate segment are equal]

⇒ OS = OR (Radius of same circle)

As, angles opposite to equal sides are equal,

∴ ∠ORS = ∠OSR = y°.

Hence, proved that ∠ORS = y°.

(ii) From figure,

∠ORP = 90° [As, tangent to a point and radius from that point are perpendicular to each other.]

⇒ ∠ORQ = ∠ORP - ∠QRP = 90° - y° ………..(1)

OQ = OR (Radius of same circle)

As, angles opposite to equal sides are equal,

∴ ∠OQR = ∠ORQ = 90° - y°

In △PQR,

⇒ ∠OQR = ∠QPR + ∠QRP (As exterior angle in a trinagle is equal to the sum of two opposite interior angles.)

⇒ 90° - y° = x° + y°

⇒ x° + 2y° = 90°.

Hence, x + 2y = 90°.