[tex]\begin{gathered} \text{ In the triangle, } \\ \text{opposite to angle 30 = x} \\ \text{adjacent to angle 30 = }10m \end{gathered}[/tex][tex]\begin{gathered} \text{ Using SOH CAH TOA} \\ \text{ Tan 30 = }\frac{\text{opposite }}{\text{adjacent}} \end{gathered}[/tex][tex]\begin{gathered} \text{ tan 30 = }\frac{x}{10} \\ 0.5773\text{ = }\frac{x}{10} \\ x\text{ = 10 x 0.5773} \\ x=\text{ 5.773} \\ x=\text{ 5.77m (to 2 decimal places)} \end{gathered}[/tex]