已知a,b,c是三角形ABC的三边,若a,b,c的倒数成等差数列,求证角B为锐角.2/b=1/a+1/c2/sinB=1/sinA+1/sinC2/sinB=(sinA+sinC)/sinAsinC所以sinAsinC=-(1/2)[cos(A+C)-cos(A-C)]>0cos(A+C)-cos(A-C)<0-cosB-cos(A-C)<0cosB>cos(C-A)∵1/
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已知a,b,c是三角形ABC的三边,若a,b,c的倒数成等差数列,求证角B为锐角.2/b=1/a+1/c2/sinB=1/sinA+1/sinC2/sinB=(sinA+sinC)/sinAsinC所以sinAsinC=-(1/2)[cos(A+C)-cos(A-C)]>0cos(A+C)-cos(A-C)<0-cosB-cos(A-C)<0cosB>cos(C-A)∵1/
已知a,b,c是三角形ABC的三边,若a,b,c的倒数成等差数列,求证角B为锐角.
2/b=1/a+1/c
2/sinB=1/sinA+1/sinC
2/sinB=(sinA+sinC)/sinAsinC
所以sinAsinC=-(1/2)[cos(A+C)-cos(A-C)]>0
cos(A+C)-cos(A-C)<0
-cosB-cos(A-C)<0
cosB>cos(C-A)
∵1/c>1/a
a>c
∠A>∠C
C-A<0
cos(C-A)>0
∴cosB>0
B为锐角
已知a,b,c是三角形ABC的三边,若a,b,c的倒数成等差数列,求证角B为锐角.2/b=1/a+1/c2/sinB=1/sinA+1/sinC2/sinB=(sinA+sinC)/sinAsinC所以sinAsinC=-(1/2)[cos(A+C)-cos(A-C)]>0cos(A+C)-cos(A-C)<0-cosB-cos(A-C)<0cosB>cos(C-A)∵1/
a,b,c的倒数成等差数列 => a>b>c 或a大角对大边,角B不是最大角,所以只能是锐角(内角和180)
2/b=1/a+1/c
2/sinB=1/sinA+1/sinC
2/sinB=(sinA+sinC)/sinAsinC
所以sinAsinC=-(1/2)[cos(A+C)-cos(A-C)]>0
cos(A+C)-cos(A-C)<0
-cosB-cos(A-C)<0
cosB>cos(C-A)
∵1/c>1/a
a>c
...
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2/b=1/a+1/c
2/sinB=1/sinA+1/sinC
2/sinB=(sinA+sinC)/sinAsinC
所以sinAsinC=-(1/2)[cos(A+C)-cos(A-C)]>0
cos(A+C)-cos(A-C)<0
-cosB-cos(A-C)<0
cosB>cos(C-A)
∵1/c>1/a
a>c
∠A>∠C
C-A<0
cos(C-A)>0
∴cosB>0
B为锐角
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