${\left(\frac{\sqrt{3}}{4}\right)}^2+{\cos }^2\alpha =1$
$\frac{3}{16}+{\cos }^2\alpha =1$
${\cos }^2\alpha =\frac{13}{16}|\sqrt{}$
$\cos \alpha =\frac{\sqrt{13}}{4}$     lub     $\cos \alpha =-\frac{\sqrt{13}}{4}$

Kąt $\alpha$ jest rozwarty, czyli $\cos \alpha =-\frac{\sqrt{13}}{4}$.