Solve the following triangle, knowing that the side $$a=6$$m and that two angles that are on this side are $$B=45^\circ$$ and $$C=105^\circ$$.
Development:
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: $$$c=\dfrac{a\cdot\sin(C)}{\sin(A)}=\dfrac{6\cdot\sin(105^\circ)}{\sin(30^\circ)}=\dfrac{6\dfrac{1}{4}(\sqrt{6}+\sqrt{2})}{\dfrac{1}{2}}=3(\sqrt{6}+\sqrt{2}) \ \mbox{m}$$$
Solution:
$$A=30^\circ$$
$$b=6\sqrt{2}$$ m
$$b=3(\sqrt{6}+\sqrt{2})$$ m