∵sinC=2√3sinB，由正弦定理，得c=2√3b又∵a^2-c^2=√3bc由余弦定理，cosA=（b^2+c^2-a^2）/2bc =（b^2-√3bc）/2bc =（bc/2√3-√3bc）/2bc =-5√3╱12∴A =arccos﹙-5√3╱12﹚
