如图所示:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b3119313b07eca80f222b27c832397dda1448302?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
如上图所示,过C作BD垂线,交BD于E,交AB于F.因为C'E⊥BD,FE⊥BD,因此二面角A-BD-C'=∠C'EF.
在RT△BCD中,CE为高,则CE=C'E=√6/3,而△CED∽△FEB,解得EF=√6/6,CF=√6/2.则RT△CBF中,BF=√2/2.即F为AB中点.
如中图所示,在△AC'B中,AC'=√6/3,C'B=1,AB=√2,C'F为AB边中线,根据余弦定理解得C'F=√3/3.
如右图所示,在△C'FE中,C'F=√3/3,EF=√6/6,C'E=√6/3,根据余弦定理解得cos∠C'EF=3/4.