一道立体几何题已知斜三棱柱ABC-A1B1C1中,面A1ACC1垂直面ABC,角ABC等于90度,BC=2,AC=2倍根号3,且AA1垂直A1C,AA1=A1C,求侧面A1ABB1与底面ABC所成的锐二面角的大小

Question

一道立体几何题
已知斜三棱柱ABC-A1B1C1中,面A1ACC1垂直面ABC,角ABC等于90度,BC=2,AC=2倍根号3,且AA1垂直A1C,AA1=A1C,求侧面A1ABB1与底面ABC所成的锐二面角的大小

Answer 1

60°
解题：过A1做A1H⊥AC,垂足于H,过点H做HG‖BC交AB于G,连接A1G
证明∠A1GH即为求侧面A1ABB1与底面ABC所成的角.
A1H⊥AC,A1ACC1垂直面ABC,AC为两个面的交线 → A1H⊥面ABC →
A1H⊥AB.①
AB⊥BC,HG‖BC → AB⊥HG.②
由①②可以判定,AB⊥面A1GH → ∠A1GH即为求侧面A1ABB1与底面ABC所
成的角.
设AH=a,HG=a/(√3),因为AA1垂直A1C,AA1=A1C → A1H=a,
tan∠A1GH=A1H/HG=√3
∠A1GH=60°
侧面A1ABB1与底面ABC所成的锐二面角为60°