In the diagram $\overline{PE},\text{ }\overline{QT},\text{ }\overline{RG}$ intersect at S and $PQ\parallel RG$. If \[\angle SPQ={{113}^{\circ }}\]and \[\angle RST={{22}^{\circ }}\], find \[\angle SPQ={{113}^{\circ }}\]\[\angle PS{{Q}^{\circ }}\]
22o
45o
67o
89o
$\begin{align} & \angle FSG=\angle SPQ={{113}^{\circ }}\text{ }\!\!\{\!\!\text{ corresponding angles }\!\!\}\!\!\text{ } \\ & \angle RST+\angle TSF+\angle FSG={{180}^{\circ }} \\ & {{22}^{\circ }}+x+{{113}^{\circ }}={{180}^{\circ }} \\ & x={{180}^{\circ }}-{{135}^{\circ }}={{45}^{\circ }} \\ & \angle PSQ=\angle TSF={{45}^{\circ }}\text{ }\!\!\{\!\!\text{ vertically opposite angles }\!\!\}\!\!\text{ } \\ & \text{Alternative method} \\ & \text{Produce }PQ\text{ to point }O \\ & \angle OPS={{180}^{\circ }}-\angle SPQ\text{ }\!\!\{\!\!\text{ Sum of }\angle s\text{ on a straight line }\!\!\}\!\!\text{ } \\ & \angle OPS180-{{113}^{\circ }}={{67}^{\circ }} \\ & \angle QSG=\angle OPS={{67}^{\circ }}\text{ }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \angle PSQ=\angle RST={{22}^{\circ }}\text{ }\!\!\{\!\!\text{ vertically opposite angles }\!\!\}\!\!\text{ } \\ & \angle PSQ=\angle QSG-\angle PSQ \\ & \angle PSQ={{67}^{\circ }}-{{22}^{\circ }}={{45}^{\circ }} \\\end{align}$