14.(6分)在$\Delta ABC$中,$\angle B=60\circ$,$AB=2$,$M$是$BC$的中点,$AM=2\sqrt{3}$,则$AC=$____;$\cos \angle MAC=$____. 分析:在$\Delta ABM$、$\Delta ABC$和$\Delta AMC$中用余弦定理即可解决此题. 解:在$\Delta ABM$中:$AM^{2}=BA^{2}+BM^{2}-2BA\cdot BM\cos 60\circ$,$\therefore (2\sqrt{3})^{2}=2^{2}+BM^{2}-2\times 2\cdot BM\cdot \dfrac{1}{2}$,$\therefore BM^{2}-2BM-8=0$,解得:$BM=4$或$-2$(舍去). $\because$点$M$是$BC$中点,$\therefore MC=4$,$BC=8$,在$\Delta ABC$中:$AC^{2}=2^{2}+8^{2}-2\times 2\times 8\cos 60\circ =52$,$\therefore AC=2\sqrt{13}$; 在$\Delta AMC$中:$\cos \angle MAC=\dfrac{(2\sqrt{3})^{2}+(2\sqrt{13})^{2}-{4}^{2}}{2\times 2\sqrt{3}{\times 2\sqrt{13}}}=\dfrac{2\sqrt{39}}{13}$. 故答案为:$2\sqrt{13}$;$\dfrac{2\sqrt{39}}{13}$. 点评:本题考查余弦定理应用,考查数学运算能力,属于中档题.
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