jmc

Since $\angle ACE$ is a straight angle, $$\angle ACB=180^{\circ }-105^{\circ }=75^{\circ }.$$In $△ABC,$ $$\begin{array}{rl}\angle BAC& =180^{\circ }-\angle ABC-\angle ACB\\ & =180^{\circ }-75^{\circ }-75^{\circ }\\ & =30^{\circ }.\end{array}$$Since $AB$ is parallel to $DC,$ we have $$\angle ACD=\angle BAC=30^{\circ }$$due to alternate angles. In $△ADC,$ $$\begin{array}{rl}\angle DAC& =180^{\circ }-\angle ADC-\angle ACD\\ & =180^{\circ }-115^{\circ }-30^{\circ }\\ & =35^{\circ }.\end{array}$$Thus, the value of $x$ is $\boxed{35}.$