<s>[INST] 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.
 [/INST] here is way to compute the angle $\angle BAD$ using the rule of sines where $D$ is mid point of $BC.$
let me use $t = \angle BAD.$  
suing the rule of sin on $\Delta ABD, \Delta ADC$  we have $$\frac{BD}{\sin t} = \frac{AD}{\sin(t + 45^\circ)}, \frac{CD}{\sin(15^\circ)} = \frac{AD}{\sin 30^\circ} $$ 
from the two equations, we get $$\frac{AD}{BD} = \frac{sin(t+45^\circ)}{\sin t}=\frac{\sin 30^\circ}{\sin 15^\circ}=2\cos 15^\circ \to \tan t =\frac 1{2\sqrt 2 \cos 15^\circ - 1}  = \frac 1 {\sqrt 3}. $$
therefore $t = 30^\circ, \angle B = 105^\circ$
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