(1)设sinA/a=sinB/b=sinC/c=k(正弦定理),那么有2a²k=(2b+c)bk+(2c+b)ck =>(b²+c²-a²)/2bc=-1/2=cosA(余弦定理),所以A=120°
(2)B+C=60°,所以sinB+sinC=sinB+sin(60°-B)=sin(60°+B)
(1)设sinA/a=sinB/b=sinC/c=k(正弦定理),那么有2a²k=(2b+c)bk+(2c+b)ck =>(b²+c²-a²)/2bc=-1/2=cosA(余弦定理),所以A=120°
(2)B+C=60°,所以sinB+sinC=sinB+sin(60°-B)=sin(60°+B)