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解题思路:(1)根据四棱锥P-ABCD的底面是边长为1的正方形,侧棱PC⊥底面ABCD,知高为PC="2." 应用体积计算公式即得;
(2)连结AC,根据ABCD是正方形,得到BD⊥AC ,由PC⊥底面ABCD 得到BD⊥PC,推出BD⊥平面PAC;由于不论点E在何位置,都有AE
![](https://img.yulucn.com/upload/c/33/c33f6c6aed2571a9926ac3c862bf0f2a_thumb.jpg)
平面PAC,故得BD⊥AE;
(3)设
![](https://img.yulucn.com/upload/0/34/034ad28c520104f047252c639797a9a6_thumb.jpg)
相交于
![](https://img.yulucn.com/upload/7/96/796b1db60617ca5290535ff9ead8b8a7_thumb.jpg)
,连
![](https://img.yulucn.com/upload/f/dd/fdd1baddfdd0eb7bed57be07e8c21f7b_thumb.jpg)
,可知
![](https://img.yulucn.com/upload/e/76/e7652f16fdb9e8e620e1dd46c682aecf_thumb.jpg)
是二面角P-BD-C的的一个平面角,计算其正切即得二面角P-BD-C的正切值.
试题解析:(1)该四棱锥P-ABCD的底面是边长为1的正方形,
侧棱PC⊥底面ABCD,且PC="2."
∴
![](https://img.yulucn.com/upload/e/f1/ef173760978cac86da85d0f9acdd4828_thumb.jpg)
4分
(2)连结AC,∵ABCD是正方形
∴BD⊥AC ∵PC⊥底面ABCD 且
![](https://img.yulucn.com/upload/b/70/b70df6de126ef61d47b512f6b6527d0b_thumb.jpg)
平面
![](https://img.yulucn.com/upload/a/56/a5643b667b5d0b3648f3fd7eecf02910_thumb.jpg)
∴BD⊥PC
又∵
![](https://img.yulucn.com/upload/a/11/a11da89c0a8225f6d35d1e220632f56d_thumb.jpg)
∴BD⊥平面PAC
∵不论点E在何位置,都有AE
![](https://img.yulucn.com/upload/c/33/c33f6c6aed2571a9926ac3c862bf0f2a_thumb.jpg)
平面PAC
∴BD⊥AE 8分
(3)设
![](https://img.yulucn.com/upload/0/34/034ad28c520104f047252c639797a9a6_thumb.jpg)
相交于
![](https://img.yulucn.com/upload/7/96/796b1db60617ca5290535ff9ead8b8a7_thumb.jpg)
,连
![](https://img.yulucn.com/upload/f/dd/fdd1baddfdd0eb7bed57be07e8c21f7b_thumb.jpg)
,由四棱锥P-ABCD的底面是边长为1的正方形,PC⊥底面ABCD知,
![](https://img.yulucn.com/upload/e/76/e7652f16fdb9e8e620e1dd46c682aecf_thumb.jpg)
是二面角P-BD-C的的一个平面角,
![](https://img.yulucn.com/upload/8/99/899610d1f1c90c6d8fb6f2ce300b284e_thumb.jpg)
,即二面角P-BD-C的正切值为
![](https://img.yulucn.com/upload/5/39/539aacffdc66606d5d2cfa1ea0783a58_thumb.jpg)
.
(1)
;(2)见解析;(3)
.
1年前
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