$\left. \begin{matrix}  x\hat{o}y+y\hat{o}z=90{}^\circ   \\ y\hat{o}z=22/5{}^\circ   \\ \end{matrix} \right\}\Rightarrow x\hat{o}y=90{}^\circ -22/5{}^\circ =67/5{}^\circ $   $\left. \begin{matrix} x\hat{o}u+x\hat{o}y=180{}^\circ   \\  x\hat{o}y=67/5{}^\circ   \\ \end{matrix} \right\}\Rightarrow x\hat{o}u=180{}^\circ -67/5{}^\circ =112/5{}^\circ $   $\frac{x\hat{o}u}{y\hat{o}y}=\frac{112/5{}^\circ }{22/5{}^\circ }=5$