find the raduis

Let $\angle {ABC}=α$ . Then $\angle {BAC}=π-2α$ .

Now, $\tan α=\dfrac{R}{2}\implies R=2\tan α$ .

$\tan (π-2α)=\dfrac{R}{3}=-\tan 2α=-\dfrac{2\tan α}{1-\tan^2 α}$ .

So, $1-\tan^2 α=-3\implies \tan α=2\implies R=2\tan α=\boxed 4$ .

Let $\angle A = \theta$ and draw the perpendiculars to the two tangent points. Then we get:

$\begin{aligned} \tan \theta & = \frac R3 \\ \tan \frac \theta 2 & = \frac 2R & \small \blue{\text{Since }\tan x = \frac {2\tan \frac x2}{1-\tan^2 \frac x2}} \\ \implies \frac R3 & = \frac {\frac 4R}{1-\frac 4{R^2}} = \frac {4R}{R^2-4} \\ \frac 13 & = \frac 4{R^2-4} \\ R^2 - 4 & = 12 \\ R & = \boxed 4 \end{aligned}$