三角形ABC中,A、B、C所对的边为a、b、c,A=120度,求(b-c)/[acos(60+C)]

问题描述:

三角形ABC中,A、B、C所对的边为a、b、c,A=120度,求(b-c)/[acos(60+C)]

用正弦定理sinA/a=sinB/b=sinC/c,得出
(b-c)/[acos(60+C)]
=(sinB-sinC)/[sinAcos(60+C)]
=[2cos(B+C)/2*sin(B-C)/2]/[sin120cos(60+C)]
={2con30*sin[(B+C)/2-C]}/[sin120cos(60+C)]
=[2cos30*sin(30-C)]/[sin120cos(60+C)]
=[2cos30*cos(60+C]/[sin120cos(60+C)]
=2cos30/sin120
=2