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是锐角

问题描述:

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是锐角

证明: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...