题目内容
如图所示,在圆锥PO中,已知

【答案】分析:过点C作CD⊥AB,垂足为D,过点D作DF⊥PA,垂足为F,连接CF,可证∠CFD为二面角B-PA-C的平面角,求出DF=
,CF=
,即可求得二面角B-PA-C的余弦值.
解答:
解:过点C作CD⊥AB,垂足为D,过点D作DF⊥PA,垂足为F,连接CF,则
∵PO⊥圆O,CD?圆O,∴PO⊥CD
∵AB∩PO=O,∴CD⊥平面PAB
∵DF⊥PA,∴CF⊥PA
∴∠CFD为二面角B-PA-C的平面角
∵⊙O的直径AB=2,点C在弧AB上,且∠COB=60°
∴CD=
,AD=
∵
,∴PA=
由AD×PO=PA×DF,可得DF=
∴CF=
∴cos∠CFD=
=
故答案为:
点评:本题考查二面角的平面角,解题的关键是正确作出二面角的平面角,利用等面积计算DF的长,属于中档题.


解答:

∵PO⊥圆O,CD?圆O,∴PO⊥CD
∵AB∩PO=O,∴CD⊥平面PAB
∵DF⊥PA,∴CF⊥PA
∴∠CFD为二面角B-PA-C的平面角
∵⊙O的直径AB=2,点C在弧AB上,且∠COB=60°
∴CD=


∵


由AD×PO=PA×DF,可得DF=

∴CF=

∴cos∠CFD=


故答案为:

点评:本题考查二面角的平面角,解题的关键是正确作出二面角的平面角,利用等面积计算DF的长,属于中档题.

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