For reference:
my progress:
I marked the angles I could find but couldn't finish
$a_{i9} = \frac{180(n-2)}{n}=\frac{180(9-2}{9}=\angle140^\circ\\a_{i6}= \frac{180(6-2)}{6}=\angle120^\circ\\\angle ADB = \frac{360^\circ-2(140^\circ)}{2} = \angle 40^\circ\\\angle BJI = \frac{\angle 180^\circ-\angle 120^\circ}{2}=30^\circ\\a_{i5}= \frac{180(5-2)}{5}=\angle108^\circ\\\\
S_{ai5} = 180(n-2) = 180(5-2) =\angle 540^\circ \\
\angle HDE = \frac{540 – 3(140)}{2}=\angle 60^\circ$
Best Answer
Lengthen $BA$ to the left and find the angles those it makes with $AM$ and $AL$. Then using them find $\angle MAL$ and $\angle LAN$. Then you can find $\angle ANL$ and eventually $x$.