Question

如图,在三棱柱ABC-A1B1C1中,侧面AA1C1C⊥底面ABC,AA1=A1C=AC=2,AB= BC,且AB⊥BC？

Answer 1

作OD∥BC交AB于D,连A1D,作OE⊥A1D于E,连AE,作CF⊥AE于F.
AB⊥BC,
∴OD⊥BC,
A1O⊥平面ABC,
∴A1D⊥BC,AB⊥平面A1OD,
∴平面A1AB⊥平面A1OD,
∴OE⊥平面A1AB,
∴平面AOE⊥平面A1AB,
∴CF⊥平面A1AB,
∴∠CA1F是A1C与平面A1AB所成的角.
AC=2,AB=BC=√2,O为AC中点,
∴OE=BC/2=√2/2,A1A=A1C=2,
∴A1O=√3,
∴A1D=√(A1O^2+OD^2)=√(7/2),
∴OE=A1O*OD/A1D=√(3/7),
∴CF=2OE=2√(3/7),
∴sinCA1F=CF/A1C=√21/7,为所求.,2,如图,在三棱柱ABC-A1B1C1中,侧面AA1C1C⊥底面ABC,AA1=A1C=AC=2,AB= BC,且AB⊥BC,

如图,只需回答第二问

建系还没教,不能用,