According the question
Side AC of the triangle = 16
Side BC = 14
The opposite side of the triangle is AC = 14
The adjacent side of the triangle is BC = 16
Using SOH CAH TOA
[tex]\begin{gathered} \text{Tan }\theta\text{ = }\frac{opposite}{\text{adjacent}} \\ \text{Tan }\theta\text{ = }\frac{14}{16} \\ \text{Tan }\theta\text{ = 0.875} \\ \theta\text{ = arc tan of 0.875} \\ \theta=tan^{-1}(0.875) \\ \theta\text{ = 41. 18 degre}es \\ To\text{ the nearest tenth} \\ \theta\text{ = 41.2 degre}es \end{gathered}[/tex]