jmc

In $△ABC$ shown below, $$\begin{array}{rl}\angle BAC& =180^{\circ }-\angle ABC-\angle ACB\\ & =180^{\circ }-60^{\circ }-90^{\circ }\\ & =30^{\circ }.\end{array}$$ Since $\angle ADC$ is a straight angle, $$\begin{array}{rl}\angle ADE& =180^{\circ }-\angle CDE\\ & =180^{\circ }-48^{\circ }\\ & =132^{\circ }.\end{array}$$ In $△AED,$ $$\begin{array}{rl}\angle AED& =180^{\circ }-\angle ADE-\angle EAD\\ & =180^{\circ }-132^{\circ }-30^{\circ }\\ & =18^{\circ }.\end{array}$$ Since $\angle AEB$ is a straight angle, $$\begin{array}{rl}\angle DEB& =180^{\circ }-\angle AED\\ & =180^{\circ }-18^{\circ }\\ & =162^{\circ }.\end{array}$$ Thus, the value of $x$ is $\boxed{162}.$