[tex]\tan\Theta=\dfrac{|BC|}{|CA|}[/tex]
Use Pythagorean theorem:
[tex]|AB|^2=|BC|^2+|CA|^2\\\\9^2=6^2+|CA|^2\\\\81=36+|CA|^2\ \ \ \ |-36\\\\|CA|^2=45\to|CA|=\sqrt{45}\\\\|CA|=\sqrt{9\cdot5}\\\\|CA|=\sqrt9\cdot\sqrt5\\\\|CA|=3\sqrt5[/tex]
[tex]\tan\Theta=\dfrac{6}{3\sqrt5}=\dfrac{2}{\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}=\dfrac{2\sqrt5}{5}[/tex]
