Question!!

Raymond

2012-07-08 4:10 am

正四面體與底的交角是多少,請詳細列出步驟

和

解釋為什麽它們的交角會大於正四面體與底的夾角(60度)

回答 (2)

用戶27207

2012-07-09 4:21 am

✔ 最佳答案

Can I use vector?

2012-07-08 20:21:06 補充：

圖片參考：http://imgcld.yimg.com/8/n/HA00016323/o/701207070055113873412250.jpg

http://i1099.photobucket.com/albums/g395/jasoncube/4e11242b.png In regular tetrahedron VABC, V' is projection of V on ABC, M is mid - ptof BC.= Required angle= ∠VMA=cos⁻¹[MV'/MV]= cos⁻¹(⅓MA/MV) <-(Since V' is centroid and MA is medianof ABC)= cos⁻¹(⅓MV/MV)= cos⁻¹⅓≈ 70.5° (Which is > 60°) I HOPE THIS CAN HELP YOU! ^_^

2012-07-09 17:37:57 補充：
(No need vector!~)

參考: My Maths World

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Raymond

2012-07-08 6:52 am

ok...........

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