Using Pythagoras theorem,

\(\displaystyle{8}^{{2}}={b}^{{2}}+{\left({4}\sqrt{{3}}\right)}^{{2}}\)

\(\displaystyle{b}^{{2}}={8}^{{2}}–{\left({4}\sqrt{{3}}\right)}^{{2}}=\)

\(\displaystyle={64}–{48}={16}\)

\(\displaystyle{b}^{{2}}={16}\)

\(\displaystyle{b}={4}\)

\(\displaystyle{8}^{{2}}={b}^{{2}}+{\left({4}\sqrt{{3}}\right)}^{{2}}\)

\(\displaystyle{b}^{{2}}={8}^{{2}}–{\left({4}\sqrt{{3}}\right)}^{{2}}=\)

\(\displaystyle={64}–{48}={16}\)

\(\displaystyle{b}^{{2}}={16}\)

\(\displaystyle{b}={4}\)