In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be equal to
$(A)15^\circ\hspace{1cm}(B)75^\circ\hspace{1cm}(C)115^\circ\hspace{1cm}(D)135^\circ$

Let AD be the median.Angle $ADC=45^\circ,ACD=30^\circ,CAD=105^\circ$.and $BD=CD$Then i am stuck,how to find angle $B?$Please help me.Thanks.

If you have access to Geogebra you can easily find the answer yourself:

Now we you only need to prove it. :)

Edit: It seems that the inclination of the median is not clear, but making the two choices on the same figure brings an immediate solution:
In triangle $AED$ we have $\frac{EC}{DC} = \frac{\sin 45^\circ}{\sin 105^\circ}=\sqrt{3}-1$. In triangle $ADC$ we have $\frac{DC}{AC} = \frac{\sin 15^\circ}{\sin 135^\circ}=\frac{\sqrt{3}-1}{2}$
This means that $\frac{2DC}{AC} = \frac{EC}{DC}$ so $DC \cdot BC = AC \cdot EC$. Thus $A,B,D,E$ are on the same circle and $\angle ABC = 105^\circ$ and $\angle EBC = \angle EAD = 15^\circ$.
So the right position of $A$ is in fact the position of $E$ in the second figure, and $\angle B = 15^\circ$.
I used freely the well known facts: $\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\sin 105^\circ = \frac{\sqrt{6}+\sqrt{2}}{4}$. These can be deduced immediately using formulas for $\sin(a\pm b)$.

