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?
 A: Use the cosine rule with $\triangle TDB$ to get
$$\begin{equation}\begin{aligned}
TD^2 & = TB^2 + BD^2 - 2(TB)(DB)\cos(\measuredangle QBD) \\
25 & = 25 + 32 - 2(5)(4\sqrt{2})\cos(\measuredangle QBD) \\
-32 & = -40\sqrt{2}\cos(\measuredangle QBD) \\
\cos(\measuredangle QBD) & = \frac{2\sqrt{2}}{5}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Alternatively, as suggested by alex's question comment, since $\triangle TDB$ is isosceles (with $TD = TB = 5$), the perpendicular from $T$ bisects $DB$, say at point $E$, so $\triangle TEB$ is right-angled at $E$. Since $EB$ is half of $DB$, i.e., it's $2\sqrt{2}$, this can be used instead to more directly get that
$$\cos(\measuredangle QBD) = \frac{EB}{TB} = \frac{2\sqrt{2}}{5} \tag{2}\label{eq2A}$$
In either case, substituting this into your equation for $DQ^2$ gives
$$\begin{equation}\begin{aligned}
DQ^2 & = 41 - 24\sqrt{2}\left(\frac{2\sqrt{2}}{5}\right) \\
& = 41 - \frac{48(2)}{5} \\
& = \frac{205 - 96}{5} \\
& = \frac{109}{5}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Finally, with $DQ = \sqrt{\frac{109}{5}}$, the cosine rule can be used again in $\triangle QDB$ to determine $\cos(\measuredangle QDB)$ and, from that, $\measuredangle QDB$. I'll leave this for you to do.
A: 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
$$\cos\angle QDB = \frac{DQ^2+DB^2-QB^2}{2\times DQ \times DB}$$
$$\cos\angle QDB = \frac{\frac{109}{5}+32-9}{2\times \sqrt\frac{109}{5}\times 4\sqrt2}$$
$$\cos\angle QDB =\frac{14\sqrt{1090}}{545}$$
A: 
$\triangle BTD$ 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 $QG || TH$,
$BG = \frac{3}{5} BH = \frac{3}{10} BD, DG = \frac{7}{10} BD$
$QG = \frac{3}{5} TH$
$DQ^2 = DG^2 + QG^2 = \frac{1}{100} \left(49 BD^2 + 36 TH^2 \right) = \cfrac{109}{5}$
$\tan \angle QDB = \frac{QG}{DG} = \cfrac{3 \sqrt{17}}{14 \sqrt2}$
A: 
Let $M$ be $(0,0)$. Clearly, we can see that the triangle $DTB$ is isoceles. $MB$ is along the positive $x$ axis and $MT$ along positive $y$ axis.
$M$ is the foot of perpendicular from $T$ to $DB$. So, it is the mid point of side $DB$.
Hence the co-ordinates of $B$ and $D$ are $(2√2,0)$ and $(-2√2,0)$.
Also, application of Pythagoras theorem in triangle $DTM$ helps us see that $TM$ is $√17$.
As $Q$ divides the side $TB$ in ratio $2:3$, by section formula, $Q$ is $$(\frac{4√2}{5},\frac{3√17}{5})$$.
By distance formula, $$QD=\sqrt{\frac{109}{5}}$$.
Once $DQ$ is known, using sine rule in triangle $DQB$, the angle $QDB$ can be known.
