Find the unknown length

This solution is same as Luis' s & Rakibul's solution , but with a little latex :-) $\Rightarrow$

In right $\Delta$ $ABD$ ,

$\tan {2 \theta}$ $=$ $\dfrac {4}{3}$ ........... $(1)$

Similarly, in $\Delta$ $ABC$ ,

$\tan {\theta}$ $=$ $\dfrac {BC}{3}$ .......... $(2)$

Now , again from $(1)$ , we have :

$\dfrac {2 \tan {\theta}}{1- \tan^{2} \theta}$ $=$ $\dfrac {4}{3}$

Solving the above quadratic equation in $\tan {\theta}$ , we get :

$\tan{\theta}$ $=$ ${-2 , \dfrac {1}{2}}$

Since $\tan {\theta} > 0$ ,

Therefore , $\tan{\theta}$ $=$ $\dfrac {1}{2}$ ...... $(3)$

Combining $(2)$ & $(3)$ , we get :

$\dfrac {BC}{3} = \dfrac {1}{2}$

$\Rightarrow$ $\color{#20A900}{\boxed {BC \ = \ \dfrac {3}{2}}}$

First of all, we must considerate both angles equal. So, the great angle is 2 times the angle 0. In this way, 4/3 is the tg of 2(0), and x/3 tg of (0). We considerate this, and transform tg(2(0))=4/3 in a expresion with tg(0) with the trigonometric equation of doble angle. Then of that, we make a sistem by sustitution cause tg(0)=x/3. The solution is 1.5

Here Tan@ = x/3 (x=BC); Also Tan2@= 4/3; 2tan@/(1-tan@^2)=4/3; (2X/3)/(1-x^2/9)=4/3; Solving equation we will get, X=-6 or x=3/2; Hence length cannot be negetive ao answer is 3/2=1.5;

ie. See formula for tan (2A) for the 3rd line.