如图,在直三棱柱ABC--A1B1C1中,AB=1,AC=AA!=根号3,BC=2.求二面角A-A1C-B的余弦值

显示全部楼层 · 2013-1-23 14:38:46

连结A1C、AC、，交于M。连结A1B、BM，根据勾股定理，A1C=√（AA1^2+AC^2)=√6,A1B=√（A1A^2+AB^2)=2=BC,∴△BCA1和△AA1C都是等腰△，∵四边形ACC1A1是正方形，∴M是A1C的中点，（平行四边形线角线互相平分）∴BM⊥A1C，AM⊥A1C，∴〈BMA是二面角A-A1C-B的平面角，BM=√（BC^2-CM^2/)=√（4-6/4）=√10/2，AM=A1C/2=√6/2，AB=1，∵AB^2+AM^2=5/2=BM^2,∴△BAM是RT△，∴〈BAM=90°，cos<AMB=AM/BM=(√6/2）/（√10/2）=√15/5，∴二面角A-A...

千问 · 2013-1-23 14:38:46

解：（向量法）∵AB=1,AC=AA1=√3,BC=2,AA1⊥面A1B1C1∴BC2 =AB2 +AC2,BC⊥AC以A为坐标原点，建立直角坐标系设A（0,0,0）D为A1C中点，连接AD，BD∵AC=AA1且角A1AC=90°∴AD⊥A1C∵A1B=2=BC∴BD⊥A1C∴∠ADB即为A-...