Express $\frac{30+18\sqrt{3}}{2+\sqrt{3}}$ in the form $r + s\sqrt{3}$, where $r$ and $s$ are integers. The diagram shows a triangle $XYZ$. $XY$ is $(4 + 2\sqrt{3})$ cm and $YZ$ is $(a + b\sqrt{3})$ cm, where $a$ and $b$ are integers.
The included angle $XYZ$ is $150^\circ$.
Given that the area of the triangle is $(15 + 9\sqrt{3})$ cm$^2$, find the value of $a$ and of $b$.

1. Multiply the numerator and denominator by the conjugate of the denominator

$\frac{30+18\sqrt{3}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

2. Expand the numerator and denominator

$\frac{(30)(2) - 30\sqrt{3} + 18\sqrt{3}(2) - 18(\sqrt{3})^2}{2^2 - (\sqrt{3})^2}$
$= \frac{60 - 30\sqrt{3} + 36\sqrt{3} - 18(3)}{4 - 3}$

3. Simplify the expression to the required form

$= \frac{60 + 6\sqrt{3} - 54}{1}$
$= 6 + 6\sqrt{3}$

1. Set up the equation for the area of the triangle

$\text{Area} = \frac{1}{2}(XY)(YZ)\sin(\angle XYZ)$
$15 + 9\sqrt{3} = \frac{1}{2}(4 + 2\sqrt{3})(a + b\sqrt{3})\sin 150^\circ$

2. Substitute the exact value of $\sin 150^\circ$ and simplify

Since $\sin 150^\circ = \frac{1}{2}$:
$15 + 9\sqrt{3} = \frac{1}{2}(4 + 2\sqrt{3})(a + b\sqrt{3})\left(\frac{1}{2}\right)$
$15 + 9\sqrt{3} = \frac{1}{4}(4 + 2\sqrt{3})(a + b\sqrt{3})$
$60 + 36\sqrt{3} = (4 + 2\sqrt{3})(a + b\sqrt{3})$

3. Rearrange to make $(a + b\sqrt{3})$ the subject

$a + b\sqrt{3} = \frac{60 + 36\sqrt{3}}{4 + 2\sqrt{3}}$
Factorising the numerator and denominator by 2:
$a + b\sqrt{3} = \frac{2(30 + 18\sqrt{3})}{2(2 + \sqrt{3})}$
$a + b\sqrt{3} = \frac{30 + 18\sqrt{3}}{2 + \sqrt{3}}$

4. Rationalise the denominator (or use the result from part (i))

Rationalising the denominator:
$a + b\sqrt{3} = \frac{30 + 18\sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$
$a + b\sqrt{3} = \frac{60 - 30\sqrt{3} + 36\sqrt{3} - 54}{2^2 - (\sqrt{3})^2}$
$a + b\sqrt{3} = \frac{6 + 6\sqrt{3}}{4 - 3}$
$a + b\sqrt{3} = 6 + 6\sqrt{3}$

5. Compare coefficients to find $a$ and $b$

Comparing the rational and irrational parts:
$a = 6$
$b = 6$