∵(sinB+sinC)/(cosB+cosC) =[2sin(B+C)/2cos(B-C)/昌悉蚂2]/[2cos(B+C)/2cos(B-C)/2] =tan(B+C)/2 =tan(π/2-A/2) =cotA/2 =cosA/2/sinA/2 sinA=2sinA/2cosA/2∴cosA/2/sinA/2=2sinA/2cosA/陆芦2∴ 2sin²A/2=1∴ 1-2sin²A/耐埋2=0∴ cosA=0∴ A=90°选A