An interesting angle

Draw in (red) lines $AF, DF, CG$ at angles and between points as shown in graphic. Triangle $DGF$ is equilateral, and all other integer angles as shown are easily found through angle chasing.

Because triangle $AFC$ is isosceles, $AF=CF$ . From this, we know that triangles $AFE$ and $CFG$ are congruent.

Therefore $GF=EF$ , and since $DF=GF$ , we know $DF=EF$ , and conclude that triangle $DFE$ is isosceles.

Therefore, $\angle DEF=50$ , and from this we have $\angle AED=20$

$\angle CEB = 180-70-20-60=30$

$\angle BDC = 180 - 60-10-70=40$

Let $CB=1$ . Sine law on $\triangle BCD$ .

$\dfrac{CD}{\sin 60}=\dfrac{1}{\sin 40}$ $\implies$ $CD\approx 1.347$

Sine law on $\triangle BCE$ .

$\dfrac{CE}{\sin 80}=\dfrac{1}{\sin 30}$ $\implies$ $CE \approx 1.97$

Cosine law on $\triangle CDE$ .

$DE^2=1.97^2+1.347^2-2(1.97)(1.347)(\cos 10)$ $\implies$ $DE \approx 0.685$

Sine law on $\triangle CDE$ .

$\dfrac{\sin x}{1.347}=\dfrac{\sin 10}{0.685}$ $\implies$ $x=20$