Possibilities for the side and angle of the triangle We have the triangle ABC with $c=18$, $\alpha=\frac{\pi}{6}$ and the median through $C$ has the length $5$. I want to determine all the possibilities for $a$ and $\gamma$.
I have done the following: 
From the median we have the following: 
$$m_c=\sqrt{\frac{2a^2+2b^2-c^2}{4}}\Rightarrow m_c^2=\frac{2a^2+2b^2-c^2}{4} \Rightarrow 4m_c^2=2a^2+2b^2-c^2 \\ \Rightarrow 4\cdot 25=2a^2+2b^2-18^2 \Rightarrow 2a^2+2b^2=100+324 \Rightarrow 2a^2+2b^2=424 \\ \Rightarrow a^2+b^2=212 \Rightarrow b^2=212-a^2$$ 
From the cosine law we have that $$a^2=b^2+c^2-2\cdot b\cdot c\cdot \cos \alpha \Rightarrow a^2=212-a^2+c^2-2\sqrt{212-a^2}\cdot c\cdot \frac{\sqrt{3}}{2} \\ \Rightarrow 2a^2=212+324-18\sqrt{3}\sqrt{212-a^2}\Rightarrow 2a^2=536-18\sqrt{3}\sqrt{212-a^2} \\ \Rightarrow a^2=268-9\sqrt{3}\sqrt{212-a^2} \ \ \
 \ (1)$$  
From the sine law we have that $$\frac{\sin\alpha}{a}=\frac{\sin \gamma}{c} \Rightarrow c\sin \alpha=a\sin \gamma \Rightarrow 18\cdot \frac{1}{2}=a\sin \gamma \Rightarrow a\sin \gamma=9 \ \ \ \ (2)$$ 
So from the equation we solve for $a$ and from the equation we solve for $\gamma$, right? 
Are they then all the posiibilities for $a$ and $\gamma$ ? 
 A: I think it is easier to find $C$ by straightedge and compass, then study $\gamma$.
Let $M$ be the midpoint of $AB$: $C$ must lie on a circle centered at $M$ with radius $5$, and on a fixed line from $A$. It follows that there are two solutions, namely $C_1$ and $C_2$ in the diagram below.

By the secant-tangent theorem we have $AC_1\cdot AC_2=4\cdot 14=56$ and by depicting the midpoint of $C_1 C_2$ (that is the foot of an altitude in the isosceles triangle $MC_1 C_2$) we also have
$$ \frac{AC_1+AC_2}{2}=9\cos 30^{\circ} = \frac{9}{2}\sqrt{3}$$
hence $AC_1$ and $AC_2$ are given by $\frac{9\sqrt{3}\pm\sqrt{19}}{2}$. In both cases $BC$ can be computed from the cosine theorem, then $\sin\gamma$ can be derived from the sine theorem.
A: Let $CM$ is a median of our triangle.
Thus, $$\frac{5}{\sin30^{\circ}}=\frac{9}{\sin\measuredangle ACM}$$
Hence, we have to cases.


*

*$\measuredangle ACM=\arcsin\frac{9}{10}$.


Thus, $$AC=\sqrt{9^2+5^2-2\cdot9\cdot5\cdot\cos\left(150^{\circ}-\arcsin\frac{9}{10}\right)}=$$
$$=\sqrt{106-90\left(-\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{19}}{10}+\frac{1}{2}\cdot\frac{9}{10}\right)}=$$
$$=\sqrt{65.5-4.5\sqrt{57}}=\frac{1}{2}\sqrt{262-18\sqrt{57}}=$$
$$=\frac{1}{2}\sqrt{243-18\sqrt{57}+19}=\frac{1}{2}(9\sqrt{3}-\sqrt{19})$$ and
$$BC=\sqrt{18^2+65.5-4.5\sqrt{57}-2\cdot18\cdot\frac{1}{2}(9\sqrt{3}-\sqrt{19})\cdot\frac{\sqrt3}{2}}=$$
$$=\frac{1}{2}\sqrt{586+18\sqrt{57}}.$$


*$\measuredangle ACM=180^{\circ}-\arcsin\frac{9}{10}$.
This case is the same.

