Hunting For Angles!

Using the law of cosines on $\triangle{AOC} \implies \overline{AC} = 2r\sin(\dfrac{\theta}{2})$ and $30^{\circ} = \dfrac{1}{2}m(\widehat{ARC} ) \implies$

$m(\widehat{ARC}) = 60^{\circ} \implies \theta = 60^{\circ}$ $\implies \overline{AC} = r \implies \triangle{AOC}$ is an equilateral triangle

$\implies m\angle{ACE} = 40^{\circ}$ $\implies m\angle{C} = 80^{\circ}$ and $m\angle{A} = 70^{\circ}$ in $\triangle{ABC}$

In $\triangle{ACE} \:\ m\angle{AEC} = 70^{\circ} \implies \triangle{ACE}$ is an isosceles triangle with $\overline{AC} = \overline{CE} = r$

$\implies \triangle{CEO}$ is an isosceles triangle with $m\angle{ECO} = 20^{\circ} \implies 20^{\circ} + 2x = 180 \implies$

$x = \boxed{80^{\circ}}$ .