三角形ABC,对边分别为abc,cos(A-C)+cosB=3/2,b的平方等于ac,求角Bcos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2sinAsinC=3/4根据正弦定理,a/sinA=b/sinB=c/sinC=2Rb^2=sin^B*4R^2 a=sinA*2R c=sinC*2R所
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/19 09:22:19
![三角形ABC,对边分别为abc,cos(A-C)+cosB=3/2,b的平方等于ac,求角Bcos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2sinAsinC=3/4根据正弦定理,a/sinA=b/sinB=c/sinC=2Rb^2=sin^B*4R^2 a=sinA*2R c=sinC*2R所](/uploads/image/z/651477-21-7.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2ABC%2C%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAabc%2Ccos%28A-C%29%2BcosB%3D3%2F2%2Cb%E7%9A%84%E5%B9%B3%E6%96%B9%E7%AD%89%E4%BA%8Eac%2C%E6%B1%82%E8%A7%92Bcos%28A-C%29%2BcosB%3Dcos%28A-C%29-cos%28A%2BC%29%3DcosAcosC%2BsinAsinC-cosAcosC%2BsinAsinC%3D2sinAsinC%3D3%2F2sinAsinC%3D3%2F4%E6%A0%B9%E6%8D%AE%E6%AD%A3%E5%BC%A6%E5%AE%9A%E7%90%86%2Ca%2FsinA%3Db%2FsinB%3Dc%2FsinC%3D2Rb%5E2%3Dsin%5EB%2A4R%5E2+a%3DsinA%2A2R+c%3DsinC%2A2R%E6%89%80)
三角形ABC,对边分别为abc,cos(A-C)+cosB=3/2,b的平方等于ac,求角Bcos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2sinAsinC=3/4根据正弦定理,a/sinA=b/sinB=c/sinC=2Rb^2=sin^B*4R^2 a=sinA*2R c=sinC*2R所
三角形ABC,对边分别为abc,cos(A-C)+cosB=3/2,b的平方等于ac,求角B
cos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC
=2sinAsinC=3/2
sinAsinC=3/4
根据正弦定理,a/sinA=b/sinB=c/sinC=2R
b^2=sin^B*4R^2 a=sinA*2R c=sinC*2R
所以,sin^B=sinA*sinC=3/4
因为B
三角形ABC,对边分别为abc,cos(A-C)+cosB=3/2,b的平方等于ac,求角Bcos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2sinAsinC=3/4根据正弦定理,a/sinA=b/sinB=c/sinC=2Rb^2=sin^B*4R^2 a=sinA*2R c=sinC*2R所
条件给了cos(A-C)+cosB=3/2,后面算出B的两种可能120或60,
假设B=120,cosB=-1/2,将其代入原式
cos(A-C)+cosB=cos(A-C)-1/2=3/2
即:cos(A-C)=2
由于cos的取值范围为-1到1
所以不可能等于2
所以B=120不成立
-1
正弦sinx和余弦cosx 他们的取值范围都是:-1《sinx《1 -1《cosx《1
既然它们的取值最大都才能取到1,那你这个cos(A-C)=2当然不成立了啊
cos的取值范围在负1到正1之间,2则不能取