<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] Using Stewart Theorem
$$DQ^2=\frac{TD^2\times BQ+DB^2\times TQ}{BT}-BQ\times TQ$$
$$DQ^2=\frac{25 \times 3 +32\times 2}{5}-3\times 2$$
$$DQ^2=\frac {109}{5}$$
$$DQ=\sqrt{\frac {109}{5}}$$
Using Cosine Theorem
$$\cos\angle QDB = \frac{DQ^2+DB^2-QB^2}{2\times DQ \times DB}$$
$$\cos\angle QDB = \frac{\frac{109}{5}+32-9}{2\times \sqrt\frac{109}{5}\times 4\sqrt2}$$
$$\cos\angle QDB =\frac{14\sqrt{1090}}{545}$$
</s>