Question

在正方体ABCD-A1B1C1D1中,E是棱CC1的中点,求截面A1BD和截面EBD所成二面角的度数

Answer 1

设正方体棱长为1,连结AC交BD于O,连结A1O、EO、A1C1,

则A1O⊥BD、EB=ED,则EO⊥BD,∴∠A1OE是平面A1BD与平面EBD所成二面角的平面角,

Rt△A1AO中,AO=√2/2,A1O = √(A1A^2+AO^2) = √(1+(√2/2)^2) = √6/2,

Rt△ECO中,EO = √(EC^2+CO^2) = √((1/2)^2+((√2/2)^2)) = √3/2,

Rt△A1C1E中,A1E = √(A1C1^2+EC1^2) = √(√2)^2+(1/2)^2) = 3/2,

△A1OE中,A1O^2+EO^2 = (√6/2)^2+(√3/2)^2 = 9/4,A1C1^2 = (3/2)^2 = 9/4,

∴∠A1OE=90°,解毕.