∵tanA+B/2+tanC/2=4 ①
sinBsinC=cosA/2^2 ②
由① => cotC/2+tanC/2=4,令tanC/2=x,则x+1/x=4 => x=2±√3
∴C=2×arctan（2±√3）=30°or150°
由② => sinBsinC=(cosB+C/2)^2 =>
1-2sinBsinC=1-2(cosB+C/2)^2=cosB+C=cosBcosC-sinBsinC =>
cosBcosC+sinBsinC=1 即cos(B-C)=1
∴B=C
由①② => B=C=30°
∴A=120°
计算得 b=c=2