连接c'F,因为E,F分别是aa',bb'中点
所以c'F平行且等于d'E
即求∠a'Fc'的余弦
△a'Fc'中有余弦定理:
cos∠a'Fc'=[(a'F)^2+(Fc')^2-(a'c')^2]/[2(a'F)(Fc')]
=[(5/4)a^2+(5/4)a^2-2*(a^2)]/[2*(5/4)a^2]…………【设立方体棱长为a】
={[(5/2)-2]a^2}/[(5/2)a^2]
=1/5
连接c'F,因为E,F分别是aa',bb'中点
所以c'F平行且等于d'E
即求∠a'Fc'的余弦
△a'Fc'中有余弦定理:
cos∠a'Fc'=[(a'F)^2+(Fc')^2-(a'c')^2]/[2(a'F)(Fc')]
=[(5/4)a^2+(5/4)a^2-2*(a^2)]/[2*(5/4)a^2]…………【设立方体棱长为a】
={[(5/2)-2]a^2}/[(5/2)a^2]
=1/5