![](http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/03/437de2e581f6305e882bf889337c50d1.png)

- 延长$AC$至$D$,使得$AD=AB$
$\triangle ABP \cong \triangle ADP$

也就是$\triangle APB$根据$AP$翻折后得到$\triangle APD$

$\because \angle APB=150^{\circ}$
$\therefore \angle BPD=60^{\circ}$
并且$BP=PD$
$\therefore \triangle PBD$是等边三角形
$\because \angle PBC=30^{\circ}$
$\therefore \angle CBD=30^{\circ}$

$\triangle PBC \cong \triangle BCD$,$SAS$
$\therefore PC=CD$
$\therefore \angle CPD=\angle CDP$
而$\angle PDC=\angle ABP=6^{\circ}$

$\therefore \angle ACP=2*6^{\circ}=12^{\circ}$