发布网友 发布时间:2024-05-03 19:03
共1个回答
热心网友 时间:2024-08-17 10:54
做上底面法向量 A1H , MG
解法1:
已知 AA1 = 1 可推出
AE = 1 * cos60° = 1/2
AH = AE / cos60° =√3/3
MG = A1H = √(1² - AH²) =√6/3
又 HG = A1M = √3/2 可推出
AG = AH + HG = √3/3 + √3/2 = 5√3 / 6
所以
AM = √(MG ² + AG²) =√( (√6/3) ² + (5√3 / 6)² ) =√11/2
解法1:余弦定理
同上 可知 sin∠AA1H =√3/3
cos(∠AA1C1) = cos (90° +∠AA1H) = cos∠AA1H cos90° - sin∠AA1H sin90° =-√3/3
AM = √(AA1² + AM²-2*AA1*AM*cos∠AA1C1)=√(1² +(√3/2)² - 2 * 1* (√3/2) * (-√3/3) )
=√11/2