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