In the diagram, angle 200o is subtended at the centre of the circle. Find the value of x
30o
50o
80o
100o
\[\begin{align} & \left| OA \right|=\left| OB \right|\text{ }\!\!\{\!\!\text{ radius of the circle }\!\!\}\!\!\text{ } \\ & \angle OAB=\angle OBA\text{ }\!\!\{\!\!\text{ Base }\angle \text{s of Isso }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle OAB+\angle OBA+\angle AOB={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of }\angle \text{s in }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \text{3}{{\text{0}}^{\circ }}+{{30}^{\circ }}+\angle AOB={{180}^{\circ }} \\ & \angle AOB={{120}^{\circ }} \\ & \angle BOC=\angle AOC-\angle AOB \\ & \angle BOC={{200}^{\circ }}-{{120}^{\circ }}={{80}^{\circ }} \\ & \left| OB \right|=\left| OC \right|\text{ }\!\!\{\!\!\text{ radius of circle }\!\!\}\!\!\text{ } \\ & \angle OBC=\angle OCB=x\text{ }\!\!\{\!\!\text{ base }\angle s\text{ of Isso }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & x+x+\angle BOC={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of angles in triangle }\!\!\}\!\!\text{ } \\ & x+x+{{80}^{\circ }}={{180}^{\circ }} \\ & x=50\circ \\\end{align}\]