解答:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b90e7bec54e736d10c92e28f98504fc2d46269b7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:以BP为边作等边三角形BPD,连接AD,
则BD=BP=DP=
,∠DBP=∠BDP=60°,
∵△ABC是等边三角形,
∴AB=BC,∠ABC=60°,
∵∠ABD+∠ABP=∠CBP+∠ABP=60°,
∴∠ABD=∠CBP,
在△ABD与△CBP中,
,
∴△ABD≌△CBP(SAS),----------------------------(3分)
∴∠BPC=∠BDA,AD=PC=1,
在△ADP中,∵PA=2,PD=
,AD=1,
∴AP
2=DP
2+AD
2,
∴△APD是直角三角形,---------------------------(4分)
∴∠ADP=90°,
∴∠ADB=∠ADP+∠BDP=150°,
∴∠BPC=150°.-----------------------------------(5分)