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.
 A: I'm not seeing a straightforward way to calculate $B$.  However, this is multiple choice, and you can eliminate the impossible values, leaving the correct one.  Ask yourself: Given what you know, what is the largest $B$ could possibly be?
UPDATE: This works is $\angle ADC=45$, not $\angle ADB=45$.  Apparently there is some confusion about this.
A: 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 $\sin(a\pm b)$.

A: 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 $t = 30^\circ, \angle B = 105^\circ$
A: Way late, but here's a geometric solution.

Let $D$ be the midpoint of $\overline {BC}$. Let $M$ be the midpoint of       $\overline{AC}$ and let $E$ be the foot of the altitude through $A$.
Assume $\angle ADC=45$. Then $DE=AE$ since $\triangle AED$ is right. Since  $\triangle     AEC$ is right with $\angle ACE=30$, deduce $AE=\frac12 AC$. Since $M$ is a midpoint, we know $\frac12       AC=MC$. But $\triangle EMC$ is isosceles with $MC=ME$.
This shows that $DE=ME$, so $\triangle DEM$ is isosceles. Then $\angle MDE$ is half of the exterior angle $\angle MEC$, which equals $30$ since$ \triangle EMC$ is  isosceles. So $\angle MDE=15$.
Finally, $D$ and $M$ are midpoints, so $\triangle ABC\sim\triangle    MDC$.
Conclude $\angle ABC=\angle MDC=15$.
(This question is solved by the same triangle as this question, hence the same diagram and similar line of reasoning.)
