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15问答网 > 直三棱柱ABC-A1B1C1中，若角BAC=90°，AB=AC=AA1，则异面直线BA1与BC1所成角为多少度

直三棱柱ABC-A1B1C1中，若角BAC=90°，AB=AC=AA1，则异面直线BA1与BC1所成角为多少度

2025-04-07 02:17:47

推荐回答（2个）

回答1：

再画一个一样的三棱锥重叠起来（如延长AA1到A2，同样B2C2）！很容易找到平行线与异面直线相交（AC1平行A1C2）！
在三角形BA1C2中，BA1=A1C2=根号2，BC2=根号6
用余弦定理得：cosA=-1/2；
角A=120度
异面直线间的夹角为180-120=60度

回答2：

60度

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