A heavy box is to be dragged along a rough horizontal floor. To do...

${F}_{A}\cos 30 ° = \mu\left(mg + {F}_{A}\sin 30 °\right)$

${F}_{B}\cos 60 ° = \mu\left(mg - {F}_{B}\sin 60 °\right)$

${F}_{A}\left(\cos 30 ° - \mu sin30 °\right) = \mumg$

${F}_{B}\left(\cos 60 ° + \mu sin60 °\right) = \mumg$

Divide the equation,

we get,

$⇒ \dfrac{{F}_{A}}{{F}_{B}} = \dfrac{\mumg / \left(\cos 30 ° - \mu sin 30 °\right)}{\mumg / \left(\cos 60 ° + \mu sin 60 °\right)} = \dfrac{2}{\sqrt{3}}$