Question

三角形ABC的三边为abc若2b=a+c求角B 的范围

Answer 1

=(a+c)/2

b^2=(a+c)^2/4

cosB=(a^2+c^2-b^2)/2ac

=(a^2+c^2-a^2/4-ac/2-c^2/4)/2ac

=(3/4a^2+3/4c^2-ac/2)/2ac

=(3/8)*a/c+(3/8)*c/a-1/4

(3/8)*a/c+(3/8)*c/a>=2根号[(3/8)*a/c*(3/8)*c/a]=3/4

所以cosB>=3/4-1/4=1/2=cos60度

B