Question

在三角形ABC中,a,b,c分别为角A,B,C所对的边,且2sinBsinC－cos(B－C)＝1/2. （1）求：角A

Answer 1

(1) 2sinBsinC－cos(B－C)＝1/22sinBsinC－cosBcosC-sinBsinC＝1/2sinBsinC-cosBcosC=1/2cos(B+C)=-1/2cosA=1/2,所以A=60°(2)由余弘定理得:b^2+c^2-bc=3==> (b+c)^2-3=3bc≤3[(b+c)/2]^2==> (b+c)^2≤12==> b+c≤2(...