In the given circle, AB, BC and CD are the congruent arcs. The angle subtended at the centre by arc MN is 110° and the angle subtended by arc BC is θ. If MD and AN are diameters  of the given circle and $\theta ={\left(\frac{2x}{3}\right)}^{°}$ , then the value of x is: 

It can be observed from the given figure that ∠MON and ∠AOD are vertically opposite angles. ∴ ∠AOD = ∠MON = 110°

It is given that AB, BC, and CD are congruent arcs.

We know that congruent arcs of a circle subtend equal angles at the centre of the circle.

∴ ∠AOB = ∠BOC = ∠COD = θ

∠AOD = ∠AOB + ∠BOC + ∠COD

⇒ 110° = 3θ

$$\begin{gathered} \Rightarrow 3\times \frac{2x}{3}=110° \\ \Rightarrow x=55° \end{gathered}$$