设BD=x,AP=2aAP中点E,因PC=AC,则CE⊥APCD⊥平面PAB===>DE⊥AP,DE=PE=AE,==>PD=AD=√2a又CE⊥AP,∠CED就是二面角C-PA-B大小CD^2=AC^2-AD^2=4-2a^2x^2+4-2a^2=x^2+2a^2=BC^2=AB^2a=1,===>AP=2,AD=CD=√2,DE=1,CE=√3co...