Given the information on the problem, we can draw the following right triangle:
using the tangent function, we have the following:
[tex]\begin{gathered} \tan (58)=\frac{\text{opposite side}}{adjacent\text{ side}}=\frac{10}{s} \\ \Rightarrow\tan (58)=\frac{10}{s} \end{gathered}[/tex]solving for 's', we get:
[tex]\begin{gathered} \tan (58)=\frac{10}{s} \\ \Rightarrow s\cdot\tan (58)=10 \\ \Rightarrow s=\frac{10}{\tan (58)}=6.2 \end{gathered}[/tex]therefore, the length of the shadow is 6.2 ft