The acute angle between the lines $y=\left(2-\sqrt{3}\right)\left(x-6\right)$  and $y=\left(2+\sqrt{3}\right)\left(x+8\right)$ is

If $\theta$ is acute angle between two lines  $y=m_1\left(x-a\right)$ and  $y=\left(2+\sqrt{3}\right)\left(x+8\right)$ then  $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
Hence, 
$\begin{array}{c}\tan \theta =\left|\frac{\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)}{1+\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\right|\\ =\left|\frac{-2\sqrt{3}}{1+4-3}\right|\\ =\sqrt{3}\end{array}$

It gives 

$\theta =60°$

Therefore, the acute angle between the given lines is  $\theta =60°$