jmc

There are several ways to proceed, and here is one. Since $△ABC$ and $△DEF$ are both isosceles, it should be easy to find that $\angle B=\angle C=80^{\circ }$ and $\angle E=\angle F=75^{\circ }.$ Now, connect $C$ and $E$:

Since the two triangular wedges have the same height, we see that $AD\parallel CE,$ thus $\angle DAC=\angle ACB=80^{\circ }.$ Likewise, $\angle ADE=\angle DEF=75^{\circ }.$ Therefore, our answer is $\angle DAC+\angle ADE=80^{\circ }+75^{\circ }=\boxed{155^{\circ }}.$