2010-II-48

連結 $NC$。設 $2x$ 為該立方體的邊長，則 $MB=x$。利用畢氏定理於 $\mathrm{\Delta }MBC$，可得

$\begin{array}{rcl}M{C}^2& =& M{B}^2+B{C}^2\\ MC& =& \sqrt{x^2+(2x{)}^2}\\ & =& \sqrt{5}x\end{array}$

留意 $MC=NC$，則 $NC=\sqrt{5}x$。利用畢氏定理於 $\mathrm{\Delta }BCH$，可得

$\begin{array}{rcl}H{B}^2& =& B{C}^2+H{C}^2\\ HB& =& \sqrt{(2x{)}^2+(2x{)}^2}\\ & =& \sqrt{8}x\end{array}$

留意 $HB=NM$，則 $NM=\sqrt{8}x$。利用餘弦公式於 $\mathrm{\Delta }MNC$，可得

$\begin{array}{rcl}\cos \theta & =& {\displaystyle \frac{M{N}^2+M{C}^2–N{C}^2}{2(MN)(MC)}}\\ & =& {\displaystyle \frac{(\sqrt{8}x{)}^2+(\sqrt{5}x{)}^2–(\sqrt{5}x{)}^2}{2(\sqrt{8}x)(\sqrt{5}x)}}\\ & =& {\displaystyle \frac{8}{2\sqrt{40}}}\\ & =& {\displaystyle \frac{2}{\sqrt{10}}}\\ & =& {\displaystyle \frac{2}{\sqrt{10}}}\times {\displaystyle \frac{\sqrt{10}}{\sqrt{10}}}\\ & =& {\displaystyle \frac{\sqrt{10}}{5}}\end{array}$