在三角形ABC中,cos方2分之A=b+c/2c,(a,b分别为角ABC的对边),则三角形ABC的形状为?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 10:56:53
![在三角形ABC中,cos方2分之A=b+c/2c,(a,b分别为角ABC的对边),则三角形ABC的形状为?](/uploads/image/z/1693288-64-8.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2Ccos%E6%96%B92%E5%88%86%E4%B9%8BA%3Db%2Bc%2F2c%2C%28a%2Cb%E5%88%86%E5%88%AB%E4%B8%BA%E8%A7%92ABC%E7%9A%84%E5%AF%B9%E8%BE%B9%EF%BC%89%2C%E5%88%99%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%BD%A2%E7%8A%B6%E4%B8%BA%3F)
在三角形ABC中,cos方2分之A=b+c/2c,(a,b分别为角ABC的对边),则三角形ABC的形状为?
在三角形ABC中,cos方2分之A=b+c/2c,(a,b分别为角ABC的对边),则三角形ABC的形状为?
在三角形ABC中,cos方2分之A=b+c/2c,(a,b分别为角ABC的对边),则三角形ABC的形状为?
a/sinA=b/sinB=c/sinC
所以cos²A/2=(1+cosA)/2=(sinB+sinC)/2sinC
sinC+sinCcosA=sinB+sinC
sinCcosA=sin(180-A-C)=sin(A+C)=sinAcosC+cosAsinC
所以
sinAcosC=1
A是三角形内角则0
由正弦定理:
b/2R=sinB,
c/2R=sinC
所以(b+c)/2c
=[(2RsinB)+(2RsinC)]/[2(2RsinC)]
=(sinB+sinC)/2sinC
所以:
cos^2(A/2)=(sinB+sinC)/2sinC
(cosA+1)/2=(sinB+sinC)/2...
全部展开
由正弦定理:
b/2R=sinB,
c/2R=sinC
所以(b+c)/2c
=[(2RsinB)+(2RsinC)]/[2(2RsinC)]
=(sinB+sinC)/2sinC
所以:
cos^2(A/2)=(sinB+sinC)/2sinC
(cosA+1)/2=(sinB+sinC)/2sinC
(cosA+1)sinC=sinB+sinC
cosAsinC=sinB
=sin(π-A-C)
=sin(A+C)
=sinAcosC+cosAsinC
所以sinAcosC=0
因为A是三角形内角,所以sinA>0
故cosC=0
C=90°
所以三角形是直角三角形
收起