Solution of triangles

Let's identify the information of the problem using the following figure:

Let's observe that we know one side and two angles. Let's apply the sine theorem: $$\dfrac{a}{\sin(A)}=\dfrac{b}{\sin(B)}=\dfrac{c}{\sin(C)}$$

We see, that it would be necessary to know the angle $A$, but this is not a problem, since $$A=180^\circ-B-C=180^\circ-45^\circ-105^\circ=30^\circ$$ Thus: $$b=\dfrac{a\cdot\sin(B)}{\sin(A)}=\dfrac{a\cdot\sin(45^\circ)}{\sin(30^\circ)}=\dfrac{6\dfrac{\sqrt{2}}{2}}{\dfrac{1}{2}}=6\sqrt{2} \ \mbox{m}$$

Now we already know 2 sides and 2 angles. We can then apply the sine or the cosine theorem. We are going to apply again the sine one. So we have: $$c=\dfrac{a\cdot\sin(C)}{\sin(A)}=\dfrac{6\cdot\sin(105^\circ)}{\sin(30^\circ)}=\dfrac{6\dfrac{1}{4}(\sqrt{6}+\sqrt{2})}{\dfrac{1}{2}}=3(\sqrt{6}+\sqrt{2}) \ \mbox{m}$$