Obliczamy |BC|:

$\text{tg}\alpha =2$  

$\frac{\left|BC\right|}{\left|AC\right|}=2$  

$\frac{\left|BC\right|}{3}=2\mid \cdot 3$  

$\left|BC\right|=6$  

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Z tw. Pitagorasa obliczamy |AB|:

${\left|AC\right|}^2+{\left|BC\right|}^2={\left|AB\right|}^2$  

$3^2+6^2={\left|AB\right|}^2$  

$9+36={\left|AB\right|}^2$  

$45={\left|AB\right|}^2$  

$\left|AB\right|=\sqrt{45}=\sqrt{9\cdot 5}=3\sqrt{5}$