为了竖一块广告牌,要制造三角形支架.三角形支架如图,要求∠ACB=6

为了竖一块广告牌,要制造三角形支架.三角形支架如图,要求∠ACB=60°,BC长度大于1米,且AC比AB长0.5米.为了广告牌稳固,要求AC的长度越短越好,求AC最短为多少米?且当AC最短时,BC长度为多少米?

答案

解:设BC=a(a>1),AB=c,AC=b,b-c=.

    c2=a2+b2-2abcos60°,

    将c=b-代入得(b-)2=a2+b2-ab,

    化简得b(a-1)=a2-.

    ∵a>1,∴a-1>0.

    b===(a-1)++2≥+2.

    当且仅当a-1=时,取“=”,即a=1+时,b有最小值2+.

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