Triangle is a right triangle because has a 90° angle
then we can use trigonometric ratios to find sides
if I use sine

whre alpha is the reference angle, O the opposite side to the angle and h the hypotenuse of the triangle
using 60° as reference angle
![\sin (60)=\frac{x}{4\sqrt[]{3}}](https://img.qammunity.org/2023/formulas/mathematics/college/i8n82dyi86a77kd9qlo984adcclogi6hwu.png)
then solve for x
![\begin{gathered} x=4\sqrt[]{3}\sin (60) \\ \\ x=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/s9ajoio85lthh8qv92avuhufob4gghxs45.png)
the value of side x is 6 units
If I use cosine

where alpha is the reference angle, a the adjacent side to the references angle and H the hypotenyse of the triangle
using 60° as reference angle
![\cos (60)=\frac{y}{4\sqrt[]{3}}](https://img.qammunity.org/2023/formulas/mathematics/college/ifquuej1tt3uch87ml0loz1hl1jtrsyrkp.png)
then solve for y
![\begin{gathered} y=4\sqrt[]{3}\cos (60) \\ \\ y=2\sqrt[]{3} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vxsll04i8mz27cs244dnevz7wefskl71m9.png)
the value of side x is 2root3 units