Question

在三角形ABC中,若sinA^2+sinAsinB=sinC^2-sinB^2,求角C

Answer 1

sinA^2+sinAsinB=sinC^2-sinB^2,
sinA^2+sinB^2-sinC^2=-sinAsinB,
由正弦定理,sinA=a/2R,.,代入,约去2R,得
a^2+b^2-c^2=-ab,
由余弦定理,
cosC=(a^2+b^2-c^2)/(2ab)=-1/2,
C=2π/3.