第二题,步骤
2个回答
2014-03-10
展开全部
由b^2=ac知道a,b.c成等比数列,
则根据正弦定理,a/SinA =b/SinB =c/SinC
知,SinA SinB SinC也成等比数列.
Cos(A-C)+CosB=Cos(A-C)-Cos(A+C),展开得2SinASinC=3/2
则: 2·Sin^2 B = 3/2 ;
SinB=√3/2
→B=60°或120°
又根据原题的条件知,CosB>0,
∴B只能为60°
则根据正弦定理,a/SinA =b/SinB =c/SinC
知,SinA SinB SinC也成等比数列.
Cos(A-C)+CosB=Cos(A-C)-Cos(A+C),展开得2SinASinC=3/2
则: 2·Sin^2 B = 3/2 ;
SinB=√3/2
→B=60°或120°
又根据原题的条件知,CosB>0,
∴B只能为60°
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