因为A1D1//B1C1,故角BD1A1 = BD1和B1C1所成的角.
连接:A1B,在三角形A1D1B中,A1D1 = a,D1B =(根号3)a ,A1B = (根号2)a.
由余弦定理得:cos角BD1A1 = [1+3 -2]/[2*1*(根号3)] = 1/(根号3) = (根号3)/3.
即为所求.
因为A1D1//B1C1,故角BD1A1 = BD1和B1C1所成的角.
连接:A1B,在三角形A1D1B中,A1D1 = a,D1B =(根号3)a ,A1B = (根号2)a.
由余弦定理得:cos角BD1A1 = [1+3 -2]/[2*1*(根号3)] = 1/(根号3) = (根号3)/3.
即为所求.