$D$ is on side $BC$ of $\triangle{ABC}$ such that $BD=2$, $CD=1$, $\angle{ADB}=60^{\circ}$, $\angle{ACB}=30^{\circ}$, find $\angle{ABC}$.

Solve:

\(\begin{multline} \shoveleft \text{Let } E \text{ on }AD \text{ such that }CD=DE \\ \shoveleft \implies BE=CE=\sqrt{3}, BE\perp DE\\ \shoveleft \implies \angle{CBE}=\angle{BCE}=30^{\circ}=\angle{CED}\\ \shoveleft \implies \angle{ACE}=45^{\circ}-30^{\circ}=15^{\circ}\\ \shoveleft \implies \angle{ACE}=15^{\circ}\implies CE=AE=BE\\ \shoveleft \implies \angle{ABE}=\angle{BAE}=45^{\circ}\\ \shoveleft \implies \angle{ABC}=\bbox[5px, border: 1px solid black]{75^{\circ}} \end{multline}\)