<s>[INST] Find the length of DQ and the angle QDB on the triangle Find the length of DQ and the angle QDB.

To find the length of DQ, I use cosine rule:
\begin{align}
DQ^2&=QB^2+BD^2-2\cdot QB\cdot BD \cos(\angle QBD)\\
&=9+32-24\sqrt{2}\cos(\angle QBD)\\
&=41-24\sqrt{2}\cos(\angle QBD).
\end{align}
I don't know the angle $\angle QBD$. Now I try to find it using sine rule.
Consider that $\angle TBD =\angle QBD$.
\begin{align}
\dfrac{TD}{\sin(\angle TBD)}=\dfrac{DB}{\sin(\angle DTB)}.
\end{align}
We can see that to find $\angle QBD=\angle TBD$, need the $\angle DTB$. But we don't know the value $\angle DTB$. I'm get stuck here. I can't compute the length DQ and the $\angle QDB$.
Anyone know how to find it?
 [/INST] 
$\triangle BTD$ is an isosceles triangle. So if $TH$ is the perp from $T$ to $DB$,
$TH^2 = 5^2 - (2\sqrt2)^2 \implies TH = \sqrt{17}$
Drop another perp from $Q$ to $DB$. As $QG || TH$,
$BG = \frac{3}{5} BH = \frac{3}{10} BD, DG = \frac{7}{10} BD$
$QG = \frac{3}{5} TH$
$DQ^2 = DG^2 + QG^2 = \frac{1}{100} \left(49 BD^2 + 36 TH^2 \right) = \cfrac{109}{5}$
$\tan \angle QDB = \frac{QG}{DG} = \cfrac{3 \sqrt{17}}{14 \sqrt2}$
</s>