立体几何，求详细过程

Question

如题所述

Answer 1

BD,BC,AD的中点为E,F,G.
连接EG,EF，
∴EG∥=½AB,EF∥=½DC
∴∠GEF或其补角为所求.
∴∠ABC=∠GEF
cos∠abc=6²ï¼‹6²-8²/2x6x6=1/9
∴∠ABC=arccos1/9或π-arccos1/9
所以为：arccos1/9或π-arccos1/9。

Answer 2

取BD中点G，延长CG使CG=GC'=根号20=AG
cosACG=(64+20-20)/(2*8*根号20)=4/根号20
AC' * AC'= AC * AC + CC' *CC' -2 AC*CC' *cosACG=64+80 - 2*8*2*根号20 *4/根号20=16
AC'=4
BC'//DC 角ABC' 即为所求
cosABC' =( AB*AB + BC'*BC'-AC'*AC')/(2 AB*BC')
=(36+36-16)/(2*6*6)=7/9