如图,AD‖BC,∴∠CBC1=异面直线AD与BC1所成角,
∠AKC1=120°.∴∠CKC1=60°.⊿CKC1为正三角形,
CC1=KC.BC=BC1=√2KC.
cos∠CBC1=(BC²+BC1²-CC1²)/(2×BC×BC1)
=(2+2-1)CC1²/(2×√2×√2)CC1²=3/4.
异面直线AD与BC1所成角的余弦值是3/4.
如图,AD‖BC,∴∠CBC1=异面直线AD与BC1所成角,
∠AKC1=120°.∴∠CKC1=60°.⊿CKC1为正三角形,
CC1=KC.BC=BC1=√2KC.
cos∠CBC1=(BC²+BC1²-CC1²)/(2×BC×BC1)
=(2+2-1)CC1²/(2×√2×√2)CC1²=3/4.
异面直线AD与BC1所成角的余弦值是3/4.