In the diagram $PR\parallel SV\parallel WY,\text{ }TX\parallel QY,\text{ }\angle TXW={{60}^{\circ }}$ find $\angle TQU$

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$\begin{align} & \angle QTU=\angle PQT={{48}^{\circ }}\text{ }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \angle XTU=\angle TXW={{60}^{\circ }}\text{ }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \angle TUQ=\angle XTU={{60}^{\circ }}\text{ }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \text{Consider }\vartriangle QTU \\ & \angle QTU+\angle TUQ+\angle TQU={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ sum of }\angle s\text{ in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & {{48}^{\circ }}+{{60}^{\circ }}+\angle TQU={{180}^{\circ }} \\ & \angle TQU={{180}^{\circ }}-{{108}^{\circ }}={{72}^{\circ }} \\\end{align}$
