Three sides of a quadrilateral are $$a = 4\sqrt{3}, b = 9$$ and $$c =\sqrt{3}$$. The sides $$𝑎$$ and $$𝑏$$ enclose an angle of $$30^{\circ}$$, and the sides $$𝑏$$ and $$𝑐$$ enclose an angle of $$90^{\circ}$$. If the acute angle between the diagonals is $$x^{\circ}$$, what is the value of $$𝑥$$?

Let the quadrilateral be $$ABCD$$ in that order with the given consecutive sides

$$AB = a = 4\sqrt{3},\;\;BC = b = 9,\;\;CD = c = \sqrt{3}$$

and with the interior angles

$$\angle ABC = 30^{\circ},\;\;\angle BCD = 90^{\circ}.$$

We place the figure on the Cartesian plane by the following convenient choices.

Step 1: Assign vectors for the sides
Take vertex $$B$$ at the origin and draw $$\overrightarrow{BA}$$ along the positive $$x$$-axis.

$$\overrightarrow{BA}= (4\sqrt{3},\,0).$$

The side $$BC$$ makes an angle of $$30^{\circ}$$ with $$BA$$, so

$$\overrightarrow{BC}= \bigl(9\cos 30^{\circ},\,9\sin 30^{\circ}\bigr) =\left(\frac{9\sqrt3}{2},\,\frac{9}{2}\right).$$

Step 2: Draw $$CD$$ perpendicular to $$BC$$
Because $$\angle BCD = 90^{\circ},\;\overrightarrow{CD}$$ is perpendicular to $$\overrightarrow{BC}$$. The clockwise unit vector perpendicular to $$BC$$ is $$\bigl(\sin 30^{\circ},-\cos 30^{\circ}\bigr)=\left(\frac12,\,-\frac{\sqrt3}{2}\right).$$ Multiplying by $$|CD|=c=\sqrt3$$ gives

$$\overrightarrow{CD}= \left(\frac{\sqrt3}{2},\,-\frac{3}{2}\right).$$

(The anticlockwise direction would give a mirror image of the same convex quadrilateral; the acute angle between the diagonals is the same.)

Step 3: Express the two diagonals as vectors

Diagonal $$BD$$: $$\overrightarrow{BD}= \overrightarrow{BC}+\overrightarrow{CD} =\left(\frac{9\sqrt3}{2}+\frac{\sqrt3}{2},\,\frac92-\frac32\right) =\left(5\sqrt3,\,3\right).$$

Diagonal $$AC$$: $$\overrightarrow{AC}= \overrightarrow{AB}+\overrightarrow{BC} -\overrightarrow{BA} =\overrightarrow{BC}-\overrightarrow{BA} =\left(\frac{9\sqrt3}{2}-4\sqrt3,\,\frac{9}{2}-0\right) =\left(\frac{\sqrt3}{2},\,\frac{9}{2}\right).$$

Step 4: Use the dot-product to find the angle $$x$$ between the diagonals

Dot product $$\overrightarrow{BD}\,\cdot\,\overrightarrow{AC} = (5\sqrt3)\left(\frac{\sqrt3}{2}\right)+3\left(\frac{9}{2}\right) =\frac{15}{2}+\frac{27}{2}=21.$$

Magnitudes $$|\overrightarrow{BD}|=\sqrt{(5\sqrt3)^2+3^2} =\sqrt{75+9}=\sqrt{84}=2\sqrt{21},$$ $$|\overrightarrow{AC}|=\sqrt{\left(\frac{\sqrt3}{2}\right)^2+\left(\frac92\right)^2} =\sqrt{\frac34+\frac{81}{4}} =\sqrt{\frac{84}{4}}=\sqrt{21}.$$

Hence $$\cos x =\frac{\overrightarrow{BD}\cdot\overrightarrow{AC}} {|\overrightarrow{BD}|\;|\overrightarrow{AC}|} =\frac{21}{\bigl(2\sqrt{21}\bigr)\bigl(\sqrt{21}\bigr)} =\frac{21}{42}=\frac12.$$

Therefore $$x = \arccos\!\left(\frac12\right)=60^{\circ}.$$

The acute angle between the diagonals is $$\boxed{60}$$.