Find $x$

By trigonometry.

$\angle AEB=180-60-80=40$

$\angle ADB=180-80-50=50$

$\triangle ADB$ is isosceles.

Let $AB=AD=1$ .

By sine law on $\triangle ADB$ , we have

$\dfrac{DB}{\sin 80}=\dfrac{1}{\sin 50}$

$DB=\dfrac{\sin 80}{\sin 50}$

By sine law on $\triangle AEB$ , we have

$\dfrac{EB}{\sin 60}=\dfrac{1}{\sin 40}$

$EB=\dfrac{\sin 60}{\sin 40}$

By cosine law on $\triangle DEB$ , we have

$DE^2=\left(\dfrac{\sin 60}{\sin 40}\right)^2+\left(\dfrac{\sin 80}{\sin 50}\right)^2-2\left(\dfrac{\sin 60}{\sin 40}\right)\left(\dfrac{\sin 80}{\sin 50}\right)(\cos 30)$

$DE \approx 0.68404$

By sine law on $\triangle ADE$ , we have

$\dfrac{\sin x}{1}=\dfrac{\sin 20}{0.68404}$

$\boxed{x=30^\circ}$