A quadrilateral has three sides of lengths $a=4 \sqrt{3}, b=9$ and $c=\sqrt{3}$. The angle between sides $a$ and $b$ is $30^{\circ}$, and the angle between sides $b$ and $c$ is $90^{\circ}$. What angle do the diagonals of the quadrilateral form?
I. solution. Let the intersection of the diagonals be $M$, and the intersection of the extensions of sides $a$ and $c$ be $E$ (Figure 1). Then

$$
\angle CEB = 180^\circ - (\angle EBC + \angle BCE) = 180^\circ - (30^\circ + 90^\circ) = 60^\circ
$$

Thus, the triangle $EBC$ is half of an equilateral triangle with height $b$. Therefore,

$$
EB = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{9}{\frac{\sqrt{3}}{2}} = 6\sqrt{3}, \quad \text{hence} \quad EA = EB - a = 2\sqrt{3}
$$

Since $CE$ is half of $EB$, its length is $3\sqrt{3}$. From this, $DE = CE - c = 2\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3f3794e933d997232083g-1.jpg?height=399&width=553&top_left_y=565&top_left_x=775)

In the triangle $ADE$, two sides are equal, and the angle between them is $60^\circ$, so it is equilateral. Thus, $d = 2\sqrt{3}$, $\angle EAD = \angle ADE = 60^\circ$, from which $\angle DAB = \angle CDA = 120^\circ$.

In the triangle $ABD$: $\angle ABD + \angle BDA = 180^\circ - \angle DAB = 60^\circ$. The triangles $ABD$ and $DAC$ are similar because

$$
\frac{AB}{DA} = \frac{4\sqrt{3}}{2\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3}} = \frac{DA}{CD}
$$

and the angle between these two sides is $\angle DAB = \angle CDA = 120^\circ$. Therefore, $\angle ABD = \angle DAC$ also holds.

From all this:

$$
\begin{aligned}
\angle AMD & = 180^\circ - (\angle DAM + \angle MDA) = 180^\circ - (\angle DAC + \angle BDA) = \\
& = 180^\circ - (\angle ABD + \angle BDA) = 180^\circ - 60^\circ = 120^\circ.
\end{aligned}
$$

Thus, the diagonals of the quadrilateral enclose an angle of $180^\circ - 120^\circ = 60^\circ$.

II. solution. From the previous solution, we use that the fourth side of the quadrilateral $ABCD$ is $2\sqrt{3}$, and the remaining two angles are $120^\circ$.

Place this quadrilateral in a "regular triangle grid" where the sides of the regular triangles are $2\sqrt{3}$ (Figure 2). It is easy to see that their height is 3.

![](https://cdn.mathpix.com/cropped/2024_05_02_3f3794e933d997232083g-1.jpg?height=363&width=523&top_left_y=1789&top_left_x=796)

Figure 2

Draw the diagonals on the diagram, then translate the shorter diagonal by $\sqrt{3}$ as shown on the diagram, then extend it to get point $F$. The triangle $DBF$ is equilateral because all three sides are the longest side of a triangle whose two shorter sides are $2\sqrt{3}$ and $4\sqrt{3}$, and the angle between them is $120^\circ$. Therefore, $\angle BDF = 60^\circ$. Let the intersection of the diagonals be $M$, then $\angle BMA$ and $\angle BDF$ are corresponding angles, so $\angle BMA = 60^\circ$, and the angle enclosed by the diagonals of the quadrilateral is $60^\circ$.

III. solution. Use the notations from Figure 1. Let $\angle BMA = \varphi$. According to the first solution, $BE = 6\sqrt{3}$, $CE = 3\sqrt{3}$, and the triangle $ADE$ is an equilateral triangle with side $2\sqrt{3}$, so its height is 3.

Calculate the area of the quadrilateral in two ways.

$$
T = T_{BCE} - T_{ADE} = \frac{3\sqrt{3} \cdot 9}{2} - \frac{2\sqrt{3} \cdot 3}{2} = \frac{21\sqrt{3}}{2}
$$

On the other hand,

$$
T = \frac{BD \cdot AC}{2} \cdot \sin \varphi
$$

We need the lengths of $BD$ and $AC$.

By the Pythagorean theorem: $BD = \sqrt{BC^2 + CD^2} = \sqrt{84} = 2\sqrt{21}$. Let the foot of the perpendicular from $A$ to $BC$ be $F$. It is clear that the triangle $BFA$ is half of an equilateral triangle, so $BF = 6$ and

$AF = 2\sqrt{3}$, and $FC = BC - BF = 3$. By the Pythagorean theorem for the triangle $AFC$, we get

$$
AC = \sqrt{AF^2 + FC^2} = \sqrt{21}
$$

Substituting the values