How to find the angle of a non right angled triangle in a cube? I have to find $\angle MHN$ ($\angle H$ in $\Delta HMN$). It is inside a cube that has side lengths of $12$ cm. $M$ is the midpoint of the diagonal $BD$ and $N$ is the midpoint of edge $GF$. Here's the diagram:

I'm completely lost on how I would find $\angle MHN$  because the triangle is skewed if I try to imagine it in 3D (the top is further back than the bottom of the triangle).
I have however calculated the length of line $MH$, in exact form it is $3\sqrt{24}$ cm, which is also $14.696938...$ cm. I could also probably find the lengths of line $MN$ and $HN$ but I would need to know at least one other angle to find $\angle MHN$ using the sine or cosine rule.
I don't know how to figure out the size of another angle.
I'd appreciate the help.
 A: Use Pythagorean theorem to find all sides of $\Delta HMN$ as follows
$$\text{in right}\ \Delta MLN, \ \ \ \ \ \  MN^2=ML^2+LN^2=12^2+6^2=180$$
$$\text{in right}\ \Delta MLH, \ \ \ \ \ \ MH^2=ML^2+LH^2=12^2+(6\sqrt2)^2=216$$
$$\text{in right}\ \Delta HGN, \ \ \ \ \ \ HN^2=HG^2+GN^2=12^2+6^2=180$$
Method-1: Let $\angle MHN=\angle HMN=\theta\ $ in isosceles $\Delta HMN$, then $\angle MNH=\pi-2\theta$. Using Sine rule in $\Delta HMN$ as follows
$$\frac{\sin\angle MHN}{MN}=\frac{\sin\angle MNH}{MH}\iff \frac{\sin\theta}{\sqrt{180}}=\frac{\sin(\pi-2\theta)}{\sqrt{216}}$$
$$\sin\theta\left(\cos\theta-\frac{3}{\sqrt{30}}\right)=0\quad \quad (\sin\theta\ne 0)$$
$$\therefore \ \theta=\cos^{-1}\left(\frac{3}{\sqrt{30}}\right)\approx \color{blue}{56.79^\circ}$$
Method-2: Since all three sides of $\Delta MHN$ are known hence use Cosine rule as follows
$$\cos \angle MHN=\frac{MH^2+HN^2-MN^2}{2(MH)(HN)}=\frac{216+180-180}{2(\sqrt{216})(\sqrt{180})}$$
$$\angle MHN=\cos^{-1}\left(\frac{3}{\sqrt{30}}\right)\approx \color{blue}{56.79^\circ}$$
A: Suppose $H$ is at the origin $(0,0,0)$ and $B$ is at the point $(12,12,12)$.  Then we can define vectors $\vec{HM}=(6,6,12)\implies|\vec{HM}|=6\sqrt{6}$ and $\vec {HN}=(12,6,0)\implies|\vec{HN}|=6\sqrt{5}$.  Let $\theta$ represent the angle between $\vec{HM}$ and $\vec{HN}$.  As a result, we have
\begin{align}
\cos\theta&=\frac{(6,6,12)\cdot(12,6,0)}{(6\sqrt6)(6\sqrt5)}=\frac{108}{36\sqrt{30}}=\frac{3}{\sqrt{30}}\\
\theta&=\cos^{-1}\left(\frac{3}{\sqrt{30}}\right)\doteq56.789^\circ.
\end{align}
A: Calculate $MN$, $MH$, and $HN$. Then using generalized Pythagoras's theorem:
$$\cos\angle MHN=\frac{MH^2+HN^2-MN^2}{2\ MH\ HN}$$
