To find leg $ a $ use formula:
$$ \tan \left( \alpha \right) = \dfrac{ a }{ b } $$After substituting $ \alpha = 30^o $ and $ b = 5 $ we have:
$$ \tan( 30^o ) = \dfrac{ a }{ 5 } $$ $$ \frac{\sqrt{ 3 }}{ 3 } = \dfrac{ a }{ 5 } $$$$ a = \frac{\sqrt{ 3 }}{ 3 } \cdot 5 $$$$ a = \frac{ 5 \sqrt{ 3}}{ 3 } $$