In the figure, the polygons shown are regular. Find $"x"$. For reference:

my progress:
I marked the angles I could find but couldn't finish
$a_{i9} = \frac{180(n-2)}{n}=\frac{180(9-2}{9}=\angle140^\circ\\a_{i6}= \frac{180(6-2)}{6}=\angle120^\circ\\\angle ADB = \frac{360^\circ-2(140^\circ)}{2} = \angle 40^\circ\\\angle BJI = \frac{\angle 180^\circ-\angle 120^\circ}{2}=30^\circ\\a_{i5}= \frac{180(5-2)}{5}=\angle108^\circ\\\\
S_{ai5} = 180(n-2) = 180(5-2) =\angle 540^\circ \\
\angle HDE = \frac{540 - 3(140)}{2}=\angle 60^\circ$


Hint: $\triangle BAN$ is isosceles (because $ABCN$ is a rhombus), and therefore so is $\triangle NAL$.
