GCSE level trigonometry question In the UK we do GCSE exams and this is one of the GCSE questions I can't solve at this time of night lol. First part is easy but I can't see a way around part B. Any help would be much appreciated. 
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$\angle DBC = 45^\circ$ and $\angle BCD = \arctan \dfrac{18} 6 $.  Therefore $\angle BDC = 180^\circ - 45^\circ - \arctan\dfrac{18} 6$.
So then you have the three angles of $\triangle BDC$ and you have the length of one of the sides, so you can use the law of sines.
$$\sin x = \sin \angle BCD = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{18}{\sqrt{18^2+6^2}} = \frac{3}{\sqrt{3^2 + 1^2}}. $$
$$\sin \angle DBC = \frac{\sqrt 2} 2$$
\begin{align}
\sin \angle BDC & = \sin\left( 180^\circ - 45^\circ - \arcsin\frac 3 {\sqrt{10}} \right) \\[10pt]
& = \sin\left( 45^\circ + \arcsin\frac 3 {\sqrt{10}} \right) \\[10pt]
& = \sin 45^\circ \cos\arcsin\frac 3 {\sqrt{10}} + \cos 45^\circ \sin\arcsin\frac 3 {\sqrt{10}} \\[10pt]
& = \frac{\sqrt 2}2 \cdot \frac 1 {\sqrt{10}} + \frac{\sqrt 2} 2\cdot\frac 3 {\sqrt{10}} = \frac 1 {2\sqrt 5} + \frac 3 {2\sqrt 5} \\[10pt]
& = \frac 2 {\sqrt 5}.
\end{align}
The law of sines then tells us that
$$
\frac{6\text{ cm}}{2/\sqrt 5} = \frac{BD}{3/\sqrt{10}}.
$$
Dividing both sides by $\sqrt 5$, we get
$$
\frac{6\text{ cm}} 2 = \frac{BD\cdot\sqrt 2} 3.
$$
