错题集中不会的三角问题~~来解决一下
2sinA(cosB+cosC)=3(sinB+sinC),则: 4sinAcos((B+C)/2)cos((B-C)/2)=6sin((B+C)/2)cos((B-C)/2) 因|B-C|<π,故cos((B-C)/2)≠0 则2sinAcos((B+C)/2)=3sin((B+C)/2) 4(sin(A/2))^2cos(A/2)=3cos(A/2) (sin(A/2))^2=3/4 sin(A/2)=√3/2 A/2=60° A=120°
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