作业帮已知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/](/uploads/image/z/1961794-10-4.jpg?t=%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E8%BE%B9%2C%E8%8B%A5a%2Cb%2Cc%E7%9A%84%E5%80%92%E6%95%B0%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82%E8%AF%81%E8%A7%92B%E4%B8%BA%E9%94%90%E8%A7%92.2%2Fb%3D1%2Fa%2B1%2Fc2%2FsinB%3D1%2FsinA%2B1%2FsinC2%2FsinB%3D%28sinA%2BsinC%29%2FsinAsinC%E6%89%80%E4%BB%A5sinAsinC%3D-%281%2F2%29%5Bcos%28A%2BC%29-cos%28A-C%29%5D%EF%BC%9E0cos%28A%2BC%29-cos%28A-C%29%EF%BC%9C0-cosB-cos%28A-C%29%EF%BC%9C0cosB%EF%BC%9Ecos%28C-A%29%E2%88%B51%2F)
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
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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