In $\triangle ABC$, $BD$ is a median, $\angle DAB=15$ and $\angle ABD=30$. Find $\angle ACB$. As title implies, the goal is to find the measure of the missing angle in the following figure. While I have solved this, which I'll show below as an answer, I'm not quite sure if my answer is accurate, so I'm posting this here to see if my answer is correct and if there are any alternative approaches. Please post your own approaches to this!



WLOG, let $AD=DC=1$
Using sine rule yields
$$
\begin{aligned}
\frac{A B}{\sin 135^{\circ}}&=\frac{1}{\sin 30^{\circ}}
\Rightarrow A B =\sqrt{2}
\end{aligned}
$$
Using cosine rule gives
$$
\begin{aligned}
BC^2 &=2^2+\sqrt{2}^2-2(2) \sqrt{2} \cos 15^{\circ} \\
&=6-4 \sqrt{2} \cos 15^{\circ} \\
&=6-2(\sqrt{3}+1) \\
B C &=\sqrt{4-2 \sqrt{3}}=\sqrt{3}-1
\end{aligned}
$$
By sine rule again, we have
$$
\begin{aligned}
\frac{\sin C}{A B} &=\frac{\sin 15^{\circ}}{BC} \\
\sin C &=\frac{\sqrt{2} \sin 15^{\circ}}{\sqrt{3}-1}=\frac{\frac{\sqrt{3}-1}{2}}{\sqrt{3}-1}=\frac{1}{2} \\
C &=30^{\circ}
\end{aligned}
$$
