(1)取BC的中点F,连EF,DF则AB ∥ EF,AB与DE所成角即为EF与DE所成角
∵AD=BD= 2
2 ,∠ADB=90°,∴AB=4∴EF=2
又∵DE=DF=2,∴异面直线AB与DE所成角为60°
(2)如图,以C为顶点的侧面展开图,依题意即求DD 1的长
∵∠ACD=∠BCD=45°,AC=BC=AB,∴∠ACB=60°
∴∠DCD 1=150°,CD=CD 1= 2
2
∴ D
D 21 = (2
2 ) 2 + (2
2 ) 2 -2
2 •2
2 cos150°=16+8
3
(3)∵ 2R=
3• (2
2 ) 2 =2
6 ,∴ R=
6 , V=
4
3 π R 3 =8
6 π ∵ AB=4,R=
6 ,∴ cosθ=
(
6 ) 2 + (
6 ) 2 - 4 2
2•
6 •
6 =-
1
3
∴ θ=π-arccos
1
3 ,∴ A,B两点的球面距离为(π-arccos
1
3 )•
6
1年前
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