.在三角形ABC中,若sinA＝(sinB+sinC)/(cosB+cosC), 判断三角形ABC的形状. ：∵sinA＝(sinB+sinC)/(cosB+cosC) ∴sinA- (sinB+sinC)/(cosB+cosC) ＝0 ∴sinA- 2sin[(B+C)/2]cos[(B-C)/2]/ 2cos[(B+C)/2]cos[(B-C)/2]...