Find the x

Since $\triangle ABC$ is isosceles, $\angle ABC=140^\circ$ . It follows that $\angle ADC=150^\circ$ .

Let $AB=BC=1$ . By cosine law, we have,

$(AC)^2=1^2+1^2-2(1)(1) \cos~140=2-2 \cos~140$ $\implies$ $AC=\sqrt{2-2 \cos~140}$

By sine law, we have

$\dfrac{DC}{\sin~20}=\dfrac{\sqrt{2-2 \cos~140}}{\sin~150}$ $\implies$ $DC=\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}$

By cosine law, we have

$(DB)^2=\left(\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}\right)^2+1^2-2\left(\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}\right) \cos~30$

$DB=\sqrt{\left(\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}\right)^2+1-2\left(\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}\right) \cos~30}$

By sine law,

$\dfrac{DB}{\sin~30}=\dfrac{1}{\sin~x}$ $\implies$ $\sin~x=\dfrac{\sin~30}{DB}$ $\implies$ $\sin~x=\dfrac{\sin~30}{\sqrt{\left(\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}\right)^2+1-2\left(\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}\right) \cos~30}}$

$x=\sin^{-1}\left(\dfrac{\sin~30}{\sqrt{\left(\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}\right)^2+1-2\left(\dfrac{\sin~20}{\sin~150}\sqrt{2-2 \cos~140}\right) \cos~30}}\right)=$ $\boxed{50^\circ}$