<s>[INST] What's the measure of the radius of the circle inscribed in the triangle ATB? For reference: Calculate the radius of the circle inscribed
in the triangle ATB

My progress:  I put the trigonometric solution...would a geometric solution be possible?
 [/INST] 
First note that $\triangle ATB$ is right triangle with $\angle ATB = 90^\circ$.
Drop a perp from $P$ to $BQ$. Then, $AB = PD = 2 \sqrt2$ (as $PQ = 3, QD = 1$).
Next drop a perp from $B$ to $PQ$. Then, $ \displaystyle \triangle QBH \sim \triangle QPD \implies QH = \frac{2}{3}, BH = \frac{4 \sqrt2}{3}$
$ \displaystyle TH = TQ - HQ = \frac43$
$ \displaystyle BT = \sqrt{TH^2 + BH^2} = \frac{4}{\sqrt3}$
Now for a right triangle, inradius is $  \displaystyle \frac{a + b - c}{2} ~, ~ $ where $c$ is hypotenuse and $a$ and $b$ are perpendicular sides.
So inradius of $ \displaystyle \triangle BTH = \frac{2 (\sqrt2 + 1 - \sqrt3)}{3}  $
As $\triangle BTH \sim \triangle BAT, ~ $ inradius of $\triangle BAT$,
$ \displaystyle r = \sqrt{\frac32} \cdot \frac{2 (\sqrt2 + 1 - \sqrt3)}{3} = \frac{2 + \sqrt2 - \sqrt6}{\sqrt3}$
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