d
\(\begin{array}{l}
{F_A} = \frac{{\mu mg}}{{\sin \theta  - \mu \cos \theta }}\\
similarly,\\
{F_B} = \frac{{\mu mg}}{{\sin \theta  + \mu \cos \theta }}\\
\,\therefore \,\frac{{{F_A}}}{{{F_B}}} = \frac{{\frac{{\mu mg}}{{\sin \theta  - \mu \cos \theta }}}}{{\frac{{\mu mg}}{{\sin \theta  + \mu \cos \theta }}}}\,\,\,\,\,\\
\, = \frac{{\sin {{60}^ \circ } - \frac{\begin{array}{l}
\mu mg\\
\sqrt 3 
\end{array}}{5}\cos \,{{60}^ \circ }}}{{\sin {{30}^ \circ } + \frac{\begin{array}{l}
\mu mg\\
\sqrt 3 
\end{array}}{5}\cos \,{{30}^ \circ }}}\,\,
\end{array}\)

\(\begin{array}{l}
\,\,\left[ {\mu  = \frac{{\sqrt 3 }}{5}\,given} \right]\,\\
 = \frac{{\sin \,{{30}^{ \circ \,}} + \frac{{\sqrt 3 }}{5} - \cos \,{{30}^ \circ }}}{{\sin \,{{60}^ \circ } - \frac{{\sqrt 3 }}{5}\cos \,{{60}^ \circ }}}\\
 = \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{5} \times \frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{5} \times \frac{1}{2}}}
\end{array}\)

\(\begin{array}{l}
 = \frac{{\frac{1}{2}\left( {1 + \frac{3}{5}} \right)}}{{\frac{{\sqrt 3 }}{5}\left( {1 - \frac{1}{5}} \right)}} = \frac{{\frac{1}{2} \times \frac{8}{5}}}{{\frac{{\sqrt 3  \times 4}}{{10}}}}\\
 = \frac{{\frac{8}{{10}}}}{{\frac{{\sqrt 3  \times 4}}{{10}}}} = \frac{8}{{\sqrt 3  \times 4}} = \frac{2}{{\sqrt 3 }}
\end{array}\)