设BC中点为D.
CC1//AA1
故AB与CC1所成角=AB与AA1所成角
设AA1=AB=BC=CA=2a
BD=DC=a
AD^2+DC^2=AC^2
AD=sqrt(3)a
AD^2+A1D^2=AA1^2
A1D=a
A1D^2+DB^2=A1B^2
A1B^2=2a
A1b=sqrt(2)a
在三角形AA1B中
AA1=AB=2a
A1B=sqrt(2)a
由余弦定理得
A1B^2=AA1^2+AB^2-2AA1*AB*cos(夹角)
2=8-8cos(夹角)
所求夹角cos值为3/4