Show that $\sin 105^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$. The diagram below shows triangle $ABC$, such that $AB = 4\ \text{cm}$, angle $ABC = 105^\circ$ and the area of triangle $ABC$ is $10 + 2\sqrt{3}\ \text{cm}^2$. Find $BC$ in the form $\sqrt{2}\bigl(a\sqrt{3} + b\bigr)$, where $a$ and $b$ are integers.

1. Express 105 degrees as a sum of special angles

$\sin 105^\circ = \sin(60^\circ + 45^\circ)$

2. Apply the sine addition formula

$\sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ$

3. Substitute exact values and simplify to the required form

$= \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)$
$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$
$= \frac{\sqrt{6} + \sqrt{2}}{4}$

1. Set up the equation for the area of the triangle using the formula $\frac{1}{2}ab\sin C$.

$\text{Area} = \frac{1}{2}(AB)(BC)\sin(ABC)$
$10 + 2\sqrt{3} = \frac{1}{2}(4)(BC)\sin 105^\circ$

2. Substitute the exact value of $\sin 105^\circ$ from part (a) and isolate $BC$.

$10 + 2\sqrt{3} = 2(BC)\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)$
$10 + 2\sqrt{3} = BC\left(\frac{\sqrt{6} + \sqrt{2}}{2}\right)$
$BC = \frac{2(10 + 2\sqrt{3})}{\sqrt{6} + \sqrt{2}}$

3. Rationalise the denominator by multiplying the numerator and denominator by the conjugate.

$BC = \frac{2(10 + 2\sqrt{3})(\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})}$
$BC = \frac{2(10\sqrt{6} - 10\sqrt{2} + 2\sqrt{18} - 2\sqrt{6})}{6 - 2}$
$BC = \frac{2(8\sqrt{6} - 10\sqrt{2} + 6\sqrt{2})}{4}$
$BC = \frac{2(8\sqrt{6} - 4\sqrt{2})}{4}$

4. Simplify the expression to the required form.

$BC = \frac{16\sqrt{6} - 8\sqrt{2}}{4}$
$BC = 4\sqrt{6} - 2\sqrt{2}$
$BC = \sqrt{2}(4\sqrt{3} - 2)$