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651 B
651 B
- 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}