∵在三角形ABC中,A+B+C=π,A=π/3
∴B+C=π-A=π-π/3=2π/3
∵(sinA)^2+(sinB)^2=(sinC)^2,即∵(sin(π/3))^2+(sinB)^2=(sinC)^2,3/4+(sinB)^2=(sinC)^2
∴(sinC)^2＞(sinB)^2
又：三角形三个内角的正弦值都大于零,sinB＞0,sinC＞0
∴sinC＞sinB
又：B+C=2π/3
∴C＞π/3,B＜π/3
令C=2π/3-B,代入3/4+(sinB)^2=(sinC)^2得：
3/4+(sinB)^2 = (sin(2π/3-B))^2
3/4+(sinB)^2 = (sin2π/3cosB-cos2π/3sinB)^2
3/4+(sinB)^2 = 1/4(√3cosB+sinB)^2
3+4(sinB)^2 = (√3cosB+sinB)^2
3+4(sinB)^2 = 3(cosB)^2+(sinB)^2+2√3sinbcosB
3-3(cosB)^2+3(sinB)^2 - 2√3sinbcosB = 0
3(sinB)^2 +3(sinB)^2 - 2√3sinbcosB = 0
6(sinB)^2 - 2√3sinbcosB = 0
2sinB(3sinB-√3cosB)=0
sinB≠0
∴3sinB-√3cosB=0
∴3sinB =√3cosB
tanB=√3/3
B=π/6