急 sin^A=sin^B=sinC(sinB+sinC)

题目

急 sin^A=sin^B=sinC(sinB+sinC)
三角形ABC中,若sin^A=sin^B=sinC(sinB+sinC),则角A=
呃……是sin^2A=sin^2B+sinC(sinB+sinC)

答案

由正弦定理 (a/sinA)=(b/sinB)=(c/sinC)
a^2 = b^2 + c^2 + bc
由余弦定理
cosA = (b^2+c^2-a^2)/2bc = -1/2
A = 120度