Question

在△ABC中,边a,b,c的对角分别为A、B、C,且sin^2A+sin^2C-sinAsinC=sin^2B,则B=?

Answer 1

sin^2A+sin^2C-sinAsinC=sin^2B

由正玄定理原式转换为

a^2+c^2-ac=b^2

由余弦定理得

cosB=(a^2+c^2-b^2)/(2ac)=[a^2+c^2-(a^2+c^2-ac)]/(2ac)

=ac/(2ac)=1/2

B=60°