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

YT and XT are tangents to the circle.

∴ ∠OYT = 90° and ∠OXT = 90°.

In quadrilateral OYTX,

⇒ ∠XOY + ∠OYT + ∠OXT + ∠XTY = 360°

⇒ ∠XOY + 90° + 90° + 80° = 360°

⇒ ∠XOY = 360° - 260° = 100°.

From figure,

⇒ ∠XOZ + ∠YOZ + ∠XOY = 360°

⇒ 140° + ∠YOZ + 100° = 360°

⇒ ∠YOZ = 360° - 240° = 120°.

We know that,

When two angles are subtended by the same arc, the angle at the centre of a circle is twice the angle at the circumference.

∴ ∠YOZ = 2∠ZXY

⇒ ∠ZXY = $\dfrac{1}{2}$∠YOZ = $\dfrac{1}{2} \times 120°$ = 60°.

Hence, ∠ZXY = 60°.