Question

在三角形ABC中,a,b,c分别是角A,B,C的对边,已知b2+c2-a2=bc (1)求

Answer 1

(1)由余弦定理：cosA=(b²+c²-a²)/2bc=bc/2bc=½;所以：A=π/3;

(2)2sin²B/2+2sin²C/2=1-cosB+1-cosC=1; cosB+cosC=1

B+C=π-A=2π/3; C=2π/3-B;

cosB+cos(2π/3-B)=1; cosB-½cosB+(√3/2)sinB =1;

½cosB+(√3/2)sinB=1; sin(B+π/6)=1;

所以B+π/6=π/2; B =π/3; C=π/3

所以三角形ABC是正三角形