Here is one solution, which mostly uses basic geometry but uses trigonometry at the end.
Here is an extension of your diagram:
I dropped a perpendicular from point $A$ to side $BC$ and called the intersection $H$, as you did. Since we are free to choose our units, let's set length $AH$ to one.
Now let's find other segment lengths. Since $HD$ is the side of a $45-45-90$ triangle, its length is the same as the other side: one. Since $HC$ is the long side of a $30-60-90$ triangle and the other side is one, its length is $\sqrt 3$. Subtracting those two, we get $DC=\sqrt 3-1$. By assumption, $BD=DC$ so $BD=\sqrt 3-1$. Finally, we get $BH = BD-HD=(\sqrt 3 - 1)-1 = \sqrt 3-2$.
Oops! The calculated length of $BH$, which is $\sqrt 3-2$, is negative! So something is wrong with the diagram. We see that point $H$ must lie to the left of point $B$ and angle $ABC$ must be obtuse, not acute as the original diagram supposed. If we correct the diagram, we get:
The other lengths stay the same, but now $HB$ is $2-\sqrt 3$.
I added angle $Y$ which is angle $BAH$, in the diagram. Using triangle $BAH$ we see that
$$\tan Y=2-\sqrt 3$$
There are multiple ways to see that this means
You can get this by using the angle difference formula for $\tan(45°-30°)$. This can also be checked by expanding $\tan(15°+30°)$ or $\tan(2\cdot 15°)$. I'll leave that to you.
Finally, by looking at the $45-45-90$ triangle $ADH$ we see that $X+Y=45°$. Since $Y=15°$ we get our final answer:
One way to check this is to draw my second diagram in Geogebra, which confirms the values for $X$ and $Y$.