Answer : (d) 80º
Given,
∠XTY = 80°
TX = TY (Tangents from the same external point are equal)
⇒ ∠TXY = ∠TYX
= \(\frac{1}{2}\) (180° – ∠XTY)
= \(\frac{1}{2}\) (180° – 80°) = 50°
OX ⊥ XT (radii ⊥ tangent at point of contact)
⇒ ∠OXT = 90° ⇒ ∠OXY = ∠OXT – ∠TXY = 90° – 50° = 40°
Also, OM ⊥ ZY
∴ In Δ XMY, ∠XYM = 180° – (∠XMY + ∠MXY)
= 180° – (90° + 40°) = 50°
Also, by alternate segment theorem,
∠XZY = ∠TXY = 50°
∴ In Δ XZY, ∠X = 180° – (∠XZY + ∠XYZ)
= 180° – (50° + 50°) = 80°.