In the diagram, O is the centre of the circle $\overline{PR}$is a tangent to the circle at Q and $\angle SOQ={{86}^{\circ }}$ Calculate the value of $\angle SQR$
43o
47o
54o
86o
\[\begin{align} & \left| PO \right|=\left| OS \right|\text{ }\!\!\{\!\!\text{ radius of a circle }\!\!\}\!\!\text{ } \\ & \angle OPS=\angle OSP\text{ }\!\!\{\!\!\text{ Base }\angle s\text{ of Iss}\text{. }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & x+x+{{86}^{\circ }}\text{=18}{{\text{0}}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of }\angle s\text{ in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & x={{47}^{\circ }} \\ & \angle SQR={{90}^{\circ }}-x={{90}^{\circ }}-{{47}^{\circ }}={{43}^{\circ }}\text{ }\!\!\{\!\!\text{ tangent to a circle }\!\!\}\!\!\text{ } \\\end{align}\]