Question

ABCD-A'B'C'D'为正方体,求二面角B-AC'-C的平面角的正切值

Answer 1

题目很简单的.

连接BD,AC交于点O,过点O作OE⊥AC1,∵CC1⊥平面ABC,∴平面ACC1⊥平面ABCD,又BD⊥AC,∴BO⊥平面ACC1,∴BO⊥AC1,又AC1⊥OE,∴AC1⊥平面OEB,∴AC1⊥BE,故∠OEB即为所求角,设正方体边长为a,则OB=√2/2 a RT△AOE内,AO=√2/2 a,tan∠DAO=C1C/AC=1/√2,

∴OE=√2/2√3,∴tan∠OEB=OB/OE=√3