31. In $\triangle A B C, D C=2 B D, \angle A B C=45^{\circ}$ and $\angle A D C=60^{\circ}$. Find $\angle A C B$ in degrees.
31. Answer: 75
Construct a point $M$ on $A D$ so that $C M$ is perpendicular to $A D$. Join $B$ and $M$. Since $\angle A D C=60^{\circ}, \angle M C D=30^{\circ}$. As $\sin 30^{\circ}=\frac{1}{2}$, so $2 M D=D C$. This means that $B D=M D$ and $\triangle M D B$ is isosceles. It follows that $\angle M B D=30^{\circ}$ and $\angle A B M=15^{\circ}$.
We further observe that $\triangle M B C$ is also isosceles and thus $M B=M C$.
Now $\angle B A M=\angle B M D-\angle A B M=15^{\circ}$, giving us yet another isosceles triangle $\triangle B A M$. We now have $M C=M B=M A$, so $\triangle A M C$ is also isosceles. This allows us to calculate $\angle A C M=45^{\circ}$ and finally $\angle A C B=30^{\circ}+45^{\circ}=75^{\circ}$.