题干

如图,四棱锥P﹣ABCD的底面是AB=2,BC= 2 的矩形,△PAB是等边三角形,侧面PAB⊥底面ABCD
(Ⅰ)证明:BC⊥面PAB    
(Ⅱ)求侧棱PC与底面ABCD所成的角.
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答案(点此获取答案解析)

证明:(Ⅰ)∵侧面PAB垂直于底面ABCD,

且侧面PAB与底面ABCD的交线是AB,

在矩形ABCD中,BC⊥AB,

∴BC⊥侧面PAB.

解:(Ⅱ)在侧面PAB内,过点P做PE⊥AB.垂足为E,连接EC,

∵侧面PAB与底面ABCD的交线是AB,PE⊥AB.

∴PE⊥底面ABCD.于是

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