Question

在△ABC中,角A,B,C的对边分别为a,b,c,且sin^2B+sin^2C+1/2cos2A=1/2+sinBsinC.

(1)求角A；（2）若AC向量*AB向量=4求△ABC的面积及周长的最小值

Answer 1

sin^2B+sin^2C+1/2cos2A=1/2+sinBsinC.
（1）sin^2B+sin^2C-sin^2A+1／2=1/2+sinBsinC.
sin^2B+sin^2C-sin^2A=sinBsinC.①
余弦定理：a^2 = b^2+ c^2 - 2·b·c·cosA②
正弦定理：a/sinA=b/sinB=c/sinC=2R③
联立两式可得：sin^2B+sin^2C-sin^2A=2sinBsinCcosA④
2cosA=1
所以
A=30º
（2）AC向量*AB向量=|AC||AB|cosA=4
|AC||AB|=8
ABC的面积=|AC||AB|sinA／2=4√3