To find leg $ a $ use formula:
$$ \tan \left( \beta \right) = \dfrac{ b }{ a } $$After substituting $ \beta = 60^o $ and $ b = 4 $ we have:
$$ \tan( 60^o ) = \dfrac{ 4 }{ a } $$ $$ \sqrt{ 3 } = \dfrac{ 4 }{ a } $$ $$ a = \dfrac{ 4 }{ \sqrt{ 3 } } $$ $$ a = \frac{ 4 \sqrt{ 3}}{ 3 } $$