(Ⅰ)证明:在△BAD中,AB=2AD=2,∠BAD=60°,
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021013915-52487.gif) 由余弦定理得,BD=![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021013915-52923.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021013915-63888.gif) AD⊥BD ----------------------------(2分) 又GD⊥平面ABCD ∴GD⊥BD, GD AD=D, ∴BD⊥平面ADG……………………4分 (Ⅱ)解:以D为坐标原点,OA为x轴,OB为y轴,OG为z轴建立空间直角坐标系D—xyz 则有A(1,0,0),B(0, ,0),G(0,0,1),E(0, )
-------------------------------(6分) 设平面AEFG法向量为![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021013916-22366.gif) 则![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021013916-14504.gif) 取 --------------------------------(9分) 平面ABCD的一个法向量![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021013917-89088.gif) -------------------------(10分) 设面ABFG与面ABCD所成锐二面角为 , 则 ---------------------------------------(12 |