正方体ABCD-A1B1C1D1,(1)求二面角A1-BD1-D的余弦值.2)求二面角A1-BD1-A的余弦值没见过这么变态的题目,555.....
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![正方体ABCD-A1B1C1D1,(1)求二面角A1-BD1-D的余弦值.2)求二面角A1-BD1-A的余弦值没见过这么变态的题目,555.....](/uploads/image/z/5214746-2-6.jpg?t=%E6%AD%A3%E6%96%B9%E4%BD%93ABCD-A1B1C1D1%2C%EF%BC%881%EF%BC%89%E6%B1%82%E4%BA%8C%E9%9D%A2%E8%A7%92A1-BD1-D%E7%9A%84%E4%BD%99%E5%BC%A6%E5%80%BC.2%EF%BC%89%E6%B1%82%E4%BA%8C%E9%9D%A2%E8%A7%92A1-BD1-A%E7%9A%84%E4%BD%99%E5%BC%A6%E5%80%BC%E6%B2%A1%E8%A7%81%E8%BF%87%E8%BF%99%E4%B9%88%E5%8F%98%E6%80%81%E7%9A%84%E9%A2%98%E7%9B%AE%2C555.....)
正方体ABCD-A1B1C1D1,(1)求二面角A1-BD1-D的余弦值.2)求二面角A1-BD1-A的余弦值没见过这么变态的题目,555.....
正方体ABCD-A1B1C1D1,(1)求二面角A1-BD1-D的余弦值.2)求二面角A1-BD1-A的余弦值
没见过这么变态的题目,555.....
正方体ABCD-A1B1C1D1,(1)求二面角A1-BD1-D的余弦值.2)求二面角A1-BD1-A的余弦值没见过这么变态的题目,555.....
在平面A1BD1上作A1E⊥BD1,连结DE,
∵A1B=BD,(都是正方形对角线),
A1D1=DD1,
BD1=BD1,(公用边),
∴△A1BD≌△DBD1,(SSS),
∴DE⊥BD1,
∴〈A1ED是二面角A1-BD1-D的平面角,
A1E=DE,(同是BD1边上的高).
认A1D1⊥平面ABB1A1,A1B∈平面ABB1A1,
∴A1D1⊥A1B,即〈D1A1B=90°,
∴△A1BD1是RT△,
根据等面积原理,
BD1*A1E=A1D1*A1B,
设正方体棱长为1,则根据勾股定理,A1B=√2,BD1=√3,
∴A1E=1*√2/√3=√6/3,
DE=A1E=√6/3,
A1D=√2,
在△A1ED中,根据余弦定理,
cos<A1ED=(A1E^2+DE^2-A1D^2)/(2A1E*DE)
=(6/9+6/9-2)/(2*√6/3*√6/3)
=-1/2,
∴二面角A1-BD1-D的余弦值为-1/2.
2、同第一题方法,连结BC1,作A1E⊥BD1,连结C1E,
同前所述,<A1EC是二面角A1-BD1-C1的平面角,
cos<A1EC1=-1/2,
∵AB//=C1D1,
∴四边形ABC1D1是平行四边形,
∴BC1//AD1,
∴A、B、C1、D1四点共面,
∴二面角A1-BD1-A是二面角A1BD1-C1的补角,
∴二面角A1-BD1-A的余弦值为1/2.