Answer:
[tex]x=80^\text{o}[/tex]
Step-by-step explanation:
[tex]1.\ \angle\text{ADC}=\frac{1}{2}\times\angle \text{AOC}\ \ \ [\text{Inscribed angle is half of central angle on same arc.}]\\\\\text{or, }\angle\text{ADC}=\frac{1}{2}\times160^\text{o}\\\\\therefore\ \angle \text{ADC}=80^\text{o}[/tex]
[tex]\text{2. }\angle \text{CBY}=\angle \text{ADC}\ \ \ [\text{Exterior angle of cyclic quadrilateral is equal to the}\\\text{opposite interior angle.}]\\\therefore\ x=80^\text{o}[/tex]
