SOLUTION
The gradient angle can be determined using the diagram below
Now to find the gradient angle, we will use SOHCAHTOA because the triangle is a right-angled triangle
[tex]\begin{gathered} \text{From SOHCAHTOA} \\ \text{SOH sin}\theta=\frac{opposite}{\text{hypotenuse}} \\ \\ CAH\text{ }\cos \theta\frac{\text{adjacent}}{\text{hypotenuse}} \\ \\ \text{TOA }\tan \theta=\frac{opposite}{\text{adjacent}} \\ \\ \text{But we will make use of } \\ \text{TOA }\tan \theta=\frac{\text{opposite}}{\text{adjacent}} \\ \\ \tan \theta=\frac{73}{\text{1}02} \\ \\ \tan \theta=0.7157 \\ \theta=\tan ^{-1}0.7157 \\ \theta=35.59^o \end{gathered}[/tex]
