在三角形ABC中若(SINA)(SINA)=(SINB)(SINB)+(SINB)(SINC)+(SINC)(SINC),则角A为多少
问题描述：
在三角形ABC中若(SINA)(SINA)=(SINB)(SINB)+(SINB)(SINC)+(SINC)(SINC),则角A为多少
答
(sinA)(sinA)=(sinB)(sinB)+(sinB)(sinC)+(sinC)(sinC)
由正弦定理:
a/2R=sinA
b/2R=sinB
c/2R=sinC
则有:
(a^2/4R^2)=(b^2/4R^2)+(bc/4R^2)+(c^2/4R^2)
a^2=b^2+bc+c^2
b^2+c^2-a^2=-bc
则：cosA=(b^2+c^2-a^2)/2bc
=-1/2
则：A=120度