2009-II-47

Let $N$ be the mid-point of $HG$. Then the angle between $MX$ and the plane $BCHG$ is $\mathrm{\angle }XMN$. By applying the Pythagoras Theorem to the $\mathrm{\Delta }MNG$, we have

$\begin{array}{rcl}M{N}^2& =& M{G}^2+G{N}^2\\ MN& =& \sqrt{{15}^2+8^2}\\ & =& 17\text{ cm}\end{array}$

Note that $XN=6\text{ cm}$. Consider $\mathrm{\Delta }XMN$, we have

$\begin{array}{rcl}\tan \mathrm{\angle }XMN& =& {\displaystyle \frac{XN}{MN}}\\ \tan \theta & =& {\displaystyle \frac{6}{17}}\end{array}$