Question

在Rt△ABC中，∠A=90°，有一個銳角為60°，BC=6．若點P在直線AC上（不與點A，C重合），且∠ABP=30°，則CP的長為　　　　　　．                                                                          

Answer 1

　6或2或4　．                                                                       

【考點】解直角三角形．                                                                         

【專題】壓軸題；分類討論．                                                                  

【分析】根據(jù)題意畫出圖形，分4種情況進行討論，利用直角三角形的性質解答．                    

【解答】解：如圖1：                                                                        

                                                            

當∠C=60°時，∠ABC=30°，與∠ABP=30°矛盾；                                            

如圖2：                                                                                             

                                                          

當∠C=60°時，∠ABC=30°，                                                                   

∵∠ABP=30°，                                                                                  

∴∠CBP=60°，                                                                                  

∴△PBC是等邊三角形，                                                                         

∴CP=BC=6；                                                                                    

如圖3：                                                                                             

                                                                            

當∠ABC=60°時，∠C=30°，                                                                   

∵∠ABP=30°，                                                                                  

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

∴PC=PB，                                                                                        

∵BC=6，                                                                                           

∴AB=3，                                                                                           

∴PC=PB===2；                                                              

如圖4：                                                                                             

                                                                           

當∠ABC=60°時，∠C=30°，                                                                   

∵∠ABP=30°，                                                                                  

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

∴PC=BC÷cos30°=4．                                                                         

故答案為：6或2或4．                                                                  

【點評】本題考查了解直角三角形，熟悉特殊角的三角函數(shù)值是解題的關鍵．