Find the missing angle

Let $BA=AE=ED=DC=1$ . Since $\triangle EDC$ is isosceles, $\angle DEC=ECD=40$ . Also, $\triangle EAB$ is isosceles so $\angle ABE=BEA=25$ . Now $\angle BEC=120-25-40=55$ .

By law of cosines on $\triangle EDC$ , we have

$w^2=1^2+1^2-2(1)(1)(\cos~100)$ $\implies$ $w=1.5321$

By law of cosines in $\triangle EAB$ , we have

$y^2=1^2+1^2-2(1)(1)(\cos~130)$ $\implies$ $y=1.8126$

By law of cosines in $\triangle BEC$ , we have

$z^2=1.8126^2+1.5321^2-2(1.8126)(1.5321)(\cos~55)$ $\implies$ $z=1.56432$

By law of sines in $\triangle BEC$ , we have

$\dfrac{\sin~\theta}{1.8426}=\dfrac{\sin~55}{1.5643}$

$\theta=71.65$

Finally, $x=\angle ECD+\theta=40+71.65=111.65 \approx$ $\boxed{112}$