在三角形ABC中,A=2B,sinB=3分之根号3.若B=2,求三角形ABC的面积
根据正弦定理sinA/sinB=a/b故a/b=2cosB=2√6/3
得到a=4√6/3
再根据余弦定理cosB=（a^2+c^2-b^2）/(2ac)=√6/3
求出（3c-6）（3c-10）=0
c=2（舍去）或10/3
故S=1/2acsinB=20√2/9
绛旓細鏍规嵁姝ｅ鸡瀹氱悊sinA/sinB=a/b 鏁卆/b=2cosB=2鈭6/3 寰楀埌a=4鈭6/3 鍐嶆牴鎹綑寮﹀畾鐞哻osB=锛坅^2+c^2-b^2锛/(2ac)=鈭6/3 姹傚嚭锛3c-6锛夛紙3c-10锛=0 c=2锛堣垗鍘伙級鎴10/3 鏁匰=1/2acsinB=20鈭2/9
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