Ans: B
Let $N$ be the mid-point of $HG$. Then the angle between $MX$ and the plane $BCHG$ is $\angle XMN$. By applying the Pythagoras Theorem to the $\Delta MNG$, we have
$\begin{array}{rcl}
MN^2 & = & MG^2 + GN^2 \\
MN & = & \sqrt{15^2 + 8^2} \\
& = & 17 \text{ cm}
\end{array}$
Note that $XN = 6\text{ cm}$. Consider $\Delta XMN$, we have
$\begin{array}{rcl}
\tan \angle XMN & = & \dfrac{XN}{MN} \\
\tan \theta & = & \dfrac{6}{17}
\end{array}$
