$x=2\cdot 6=12$

$y=2\cdot \frac{6\sqrt{3}}{2}=6\sqrt{3}$

$6^2+12^2=z^2$

$36+144=z^2$

$180=z^2\stackrel{z>0}{\to }z=\sqrt{180}=\sqrt{36\cdot 5}=6\sqrt{5}$

$\cos \alpha =\frac{12}{6\sqrt{5}}=\frac{2}{\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

Odp.: Szukany cosinus wynosi $\frac{2\sqrt{5}}{5}$.