The given triangle is right angled.
Using trigonometric properties for a right triangle,
[tex]\begin{gathered} \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}} \\ \end{gathered}[/tex][tex]\begin{gathered} \tan 15\degree=\frac{10}{l} \\ l=\frac{10}{\tan 15\degree} \\ l=37.32 \end{gathered}[/tex][tex]undefined[/tex]
