Question

在Rt△ABC中，∠A=90°，有一个锐角为60°，BC=6．若点P在直线AC上（不与点A，C重合），且∠ABP=30°，则CP的长为　　　　　　．                                                                          

Answer 1

　6或2或4　．                                                                       

【考点】解直角三角形．                                                                         

【专题】压轴题；分类讨论．                                                                  

【分析】根据题意画出图形，分4种情况进行讨论，利用直角三角形的性质解答．                    

【解答】解：如图1：                                                                        

                                                            

当∠C=60°时，∠ABC=30°，与∠ABP=30°矛盾；                                            

如图2：                                                                                             

                                                          

当∠C=60°时，∠ABC=30°，                                                                   

∵∠ABP=30°，                                                                                  

∴∠CBP=60°，                                                                                  

∴△PBC是等边三角形，                                                                         

∴CP=BC=6；                                                                                    

如图3：                                                                                             

                                                                            

当∠ABC=60°时，∠C=30°，                                                                   

∵∠ABP=30°，                                                                                  

∴∠PBC=60°﹣30°=30°，                                                                        

∴PC=PB，                                                                                        

∵BC=6，                                                                                           

∴AB=3，                                                                                           

∴PC=PB===2；                                                              

如图4：                                                                                             

                                                                           

当∠ABC=60°时，∠C=30°，                                                                   

∵∠ABP=30°，                                                                                  

∴∠PBC=60°+30°=90°，                                                                         

∴PC=BC÷cos30°=4．                                                                         

故答案为：6或2或4．                                                                  

【点评】本题考查了解直角三角形，熟悉特殊角的三角函数值是解题的关键．