如图所示，一个直棱柱的底面为三角形 $ABC$，其中角 $BAC = 60^\circ$ 且 $AB = (2 + \sqrt{3})\text{ cm}$。该棱柱的高为 $\left(\frac{20}{\sqrt{3}}\right)\text{ cm}$，体积为 $(25 + 10\sqrt{3})\text{ cm}^3$。 不使用计算器，将 $AC$ 的长度表示为 $(a + b\sqrt{3})$ 的形式，其中 $a$ 和 $b$ 为常数。

1. Find the area of the triangular base $ABC$ using the volume and height of the prism.

$\text{Volume} = \text{Base Area} \times \text{Height}$
$25 + 10\sqrt{3} = \text{Area}(ABC) \times \left(\frac{20}{\sqrt{3}}\right)$
$\text{Area}(ABC) = \frac{(25 + 10\sqrt{3})\sqrt{3}}{20}$
$\text{Area}(ABC) = \frac{25\sqrt{3} + 30}{20} = \frac{5\sqrt{3} + 6}{4}$

2. Express the area of triangle $ABC$ using the sine formula for area.

$\text{Area}(ABC) = \frac{1}{2} \times AB \times AC \times \sin(BAC)$
$\text{Area}(ABC) = \frac{1}{2}(2 + \sqrt{3})(AC)\sin(60^\circ)$
$\text{Area}(ABC) = \frac{1}{2}(2 + \sqrt{3})(AC)\left(\frac{\sqrt{3}}{2}\right)$
$\text{Area}(ABC) = \frac{\sqrt{3}(2 + \sqrt{3})}{4} \times AC$
$\text{Area}(ABC) = \frac{2\sqrt{3} + 3}{4} \times AC$

3. Equate the two expressions for the area to solve for $AC$.

$\frac{2\sqrt{3} + 3}{4} \times AC = \frac{5\sqrt{3} + 6}{4}$
$(2\sqrt{3} + 3) \times AC = 5\sqrt{3} + 6$
$AC = \frac{5\sqrt{3} + 6}{2\sqrt{3} + 3}$

4. Rationalise the denominator to express $AC$ in the required form.

$AC = \frac{6 + 5\sqrt{3}}{3 + 2\sqrt{3}} \times \frac{3 - 2\sqrt{3}}{3 - 2\sqrt{3}}$
$AC = \frac{18 - 12\sqrt{3} + 15\sqrt{3} - 10(3)}{3^2 - (2\sqrt{3})^2}$
$AC = \frac{18 + 3\sqrt{3} - 30}{9 - 12}$
$AC = \frac{-12 + 3\sqrt{3}}{-3}$

5. Simplify the fraction to get the final answer.

$AC = 4 - \sqrt{3}$