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解析:主要使用余弦定理来解答.
∵E为B1C1中点,且 EG和B1C1成45°角,
∴ 点G在: ① BB1的三分之一处,且BG1=1/3BB1,BG1=1cm,
② CC1的三分之一处,且CG2=1/3CC1,BG2=1cm,
① 在△EFG1中,有 EF=√ [ (EC1)²+(FC1)² ] = 2√2,
EG1=√ [ (EB1)²+(B1G1)² ] = 2√2,
FG1=√ [ (FB1)²+(B1G1)² ] = √ [(FC1)²+(C1B1)²+(B1G1)²] = 2√6,
由余弦定理有 cos∠FEG1= [(EF)²+(FG1)²-(EG1)²] / 2(EF)*(FG1)
= (√3)/2,
∴ ∠FEG1 = 150°.
② 在△EFG2中,有 EF=√ [ (EC1)²+(FC1)² ] = 2√2,
EG2=√ [ (EC1)²+(C1G2)² ] = 2√2,
FG2=√ [ (FC1)²+(C1G2)² ] = 2√2,
由余弦定理有 cos∠FEG2 = [(EF)²+(FG2)²-(EG2)²] / 2(EF)*(FG2)
=1/2 ,
∴ ∠FEG2 = 60°.
希望可以帮到你、
1年前
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