NEED HELP ASAP!!!!!

By the angle sum/difference formula for cos,

cos(a - b)  =  cos a cos b + sin a sin b

And so...

${\cos(\frac{\pi}{6}-x)}\ =\ \cos(\frac{\pi}{6})\cos(x)+\sin(\frac{\pi}{6})\sin(x)\\~\\ \phantom{\cos(\frac{\pi}{6}-x)}\ =\ \sin(\frac{\pi}{6})\sin(x)+\cos(\frac{\pi}{6})\cos(x)$

Now we can see that

$A\ =\ \sin(\frac{\pi}{6})\ =\ \frac12 \\~\\ B\ =\ \cos(\frac{\pi}{6})\ =\ \frac{\sqrt3}{2}$

And so...

$\dfrac{B}{A}\ =\ \dfrac{(\frac{\sqrt3}{2})}{(\frac{1}{2})}\ =\ \sqrt3$    _