The sketch of the question can be shown on a line diagram below
[tex]\begin{gathered} where \\ \theta=gr\text{ adient angle} \\ \text{opposite = 6 ft} \\ \text{adjacent = 82 ft} \end{gathered}[/tex]To get the gradient angle, we make use of the trigonometrical ratio of tan
[tex]\begin{gathered} \tan \theta=\frac{\text{opposite}}{\text{adjacent}} \\ \tan \theta=\frac{6}{82} \\ \tan \theta=0.0732 \\ \theta=\tan ^{-1}(0.0732) \\ \theta=4.187^0 \\ \end{gathered}[/tex]Therefore, the gradient angle of the road is 4.19 degrees.
