In the figure (i) given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.

From figure,

XT = YT (∵ tangents from an external point to a circle are of equal length.)

So, △XTY is an isosceles triangle with

∠YXT = ∠XYT = a.

Since sum of angles in a triangle = 180°.

In △XTY,

⇒ ∠YXT + ∠XYT + ∠XTY = 180°
⇒ a + a + 80° = 180°
⇒ 2a + 80° = 180°
⇒ 2a = 180° - 80°
⇒ 2a = 100°
⇒ a = 50°.

From figure,

OX = OZ = radius of the circle

So, △OXZ is an isosceles triangle with

∠OXZ = ∠OZX = b.

Since sum of angles in a triangle = 180°.

In △OXZ,

⇒ ∠OXZ + ∠OZX + ∠XOZ = 180°
⇒ b + b + 140° = 180°
⇒ 2b + 140° = 180°
⇒ 2b = 180° - 140°
⇒ 2b = 40°
⇒ b = 20°.

From figure,

OX ⊥ XT (∵ tangent at a point and radius through the point are perpendicular to each other.)

∴ ∠OXT = 90°

∠OXY + ∠YXT = 90°
∠OXY + 50° = 90°
∠OXY = 90° - 50°
∠OXY = 40°.

From figure,

∠ZXY = ∠OXZ + ∠OXY = 20° + 40° = 60°.

Hence, the value of ∠ZXY = 60°.