Step-by-step explanation:
Using Pythagoras Theorem,
[tex]BA=\sqrt{6^2-3^2}[/tex]
[tex]BA=\sqrt{36-9}[/tex]
[tex]BA=\sqrt{27}[/tex]
[tex]BA=3\sqrt{3}[/tex]
[tex]x=BA-3[/tex]
[tex]x=3\sqrt{3}-3[/tex]
∴[tex]x=3(\sqrt{3}-1)[/tex]
