一道数学空间向量题.

哪来的B'啊 .
如果题目是,
底面是正三角形ABC的三棱柱A'-ABC,角A'AB=45度,角A'AC=60度.
求二面角B-AA'-C的余弦.
解法如下,
1）三角形A'AB内,由B做直线AA'的垂线,垂足为D.
2）由B做平面AA'C的垂线,垂足为E.
3)连接DE,DE的延长线交直线AC于F.
因,BE垂直于平面AA'C,
所以,BE垂直于AA'.
又,BD垂直于AA'.
AA'垂直于三角形BDE所在的平面.
故,AA'垂直于DF.
4)因此,在三角形BDF中,角BDF的余弦即为所求.
记AB = 2^(1/2)L
BD = AB/SIN（角A'AB）= L.
AD = BD = L.
DF = AD*TAN(角A'AC) = 3^(1/2)L.
AF = AD/COS(角A'AC) = 2L
BF^2 = AB^2 + AF^2 - 2AB*AF*COS(角BAC) = 2L^2 + 4L^2 - 4*2^(1/2)L^2
= [6 - 4(2)^(1/2)]L^2
二面角B-AA'-C的余弦 = COS(角BDF) = [BD^2 + DF^2 - BF^2]/[2*BD*DF]
= {L^2 + 3L^2 - [6 - 4(2)^(1/2)]L^2}/{2*L*3^(1/2)L}
= [1 + 3 - 6 + 4(2)^(1/2)]/[2(3)^(1/2)]
= [2(2)^(1/2) - 1]3^(-1/2)
= [2(6)^(1/2) - 3^(1/2)]/3