I am not sure at the moment if Chris and I have interpreted the question the same.

(This diagram is drawn to scale)

Here is my interpretation.

$$\\

Let length BC = x

Now I will use the cosine rule:

$$\boxed{cosA=\frac{b^2+c^2-a^2}{2bc}}\\\\

so\\\\

cos\theta=\frac{8^2+16^2-x^2}{2\times8\times16}\quad and \quad cos\theta=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\

Hence\\\\

\frac{8^2+16^2-x^2}{2\times8\times16}=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\

\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\

\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\

\frac{320-x^2}{8}=\frac{480-x^2}{ 3\sqrt{21}}\\\\

3\sqrt{21}(320-x^2)=8(480-x^2)\\\\

3\sqrt{21}\times320-3\sqrt{21}\;x^2\;=\;3840-8x^2\\\\

960\sqrt{21}-3840\;=\;(-8+3\sqrt{21})\;x^2\\\\$$

$$\\x^2=\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}\\\\

x=+\sqrt{\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}}\\\\$$

$${\sqrt{{\frac{\left({\mathtt{960}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}{\mathtt{\,-\,}}{\mathtt{3\,840}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}\right)}}}} = {\mathtt{9.864\: \!242\: \!223\: \!858\: \!689}}$$

so

The forth side is approx 9.86metres long

Since our answers are the same I guess we interpreted the question the same after all