Five. (Full marks 14 points) As shown in the figure, $P$ is a point on $BC$ of $\triangle ABC$, and $PC=2PB$. It is known that $\angle ABC=45^{\circ}$, $\angle APC=60^{\circ}$. Find $\angle ACB$.

保留源文本的换行和格式，直接输出翻译结果如下：

Five. (Full marks 14 points) As shown in the figure, $P$ is a point on $BC$ of $\triangle ABC$, and $PC=2PB$. It is known that $\angle ABC=45^{\circ}$, $\angle APC=60^{\circ}$. Find $\angle ACB$.
Five, make the symmetric point $C_{1}$ of point $C$ with respect to line $A P$, and connect $B C_{1}$, $P C_{1}$, and $A C_{1}$. Then $C_{1} P = C P = \angle B P$, $\angle C_{1} P B = 180^{\circ} - \angle A P C - \angle A P C_{1} = 60^{\circ}$. Therefore, $A C^{\prime} P B$ is similar to a right triangle with a hypotenuse of 2 and a leg of 1, so $\angle C_{1} B P^{\prime} = 90^{\circ}$.

Given $\angle A B P = 45^{\circ}$, $A B$ is the angle bisector of $\angle C_{1} B C$. $\times A P$ is the angle bisector of $\angle C_{1} P C$, so point $A$ is equidistant from lines $C P$, $P C_{1}$, and $B C_{1}$. Thus, $A C_{1}$ is the angle bisector of $\angle P C_{1} D$ (where point $D$ is a point on the extension of line segment $B C_{1}$ beyond $C_{1}$).
That is, $\angle A C_{1} P = \frac{1}{2} (180^{\circ} - 30^{\circ}) = 75^{\circ}$.
Therefore, $\angle A C B = \angle A C_{1} P = 75^{\circ}$.