The given parts in the question are
[tex]\begin{gathered} \text{opposite}=6 \\ \text{adjacent}=5 \\ \text{angle}=k \end{gathered}[/tex]
To calculate the value of angle k, we will use the trigonometric ratio
SOH ,CAH, TOA
[tex]\begin{gathered} \sin \theta=\frac{opposite}{hypotenus} \\ \cos \theta=\frac{adjacent}{hypotenus} \end{gathered}[/tex]
[tex]\begin{gathered} \text{Tan}\theta=\frac{opposite}{adjacent} \\ \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} \text{Tan}\theta=\frac{opposite}{adjacent} \\ \tan \text{ k =}\frac{6}{5} \\ \tan \text{ k = 1.2} \end{gathered}[/tex]
To find the value of k, we will calculate the value of the arc tan of 1.2 using a calculator or a tangent table
[tex]\begin{gathered} \tan \text{ k = 1.2} \\ k=\tan ^{-1}1.2 \\ k=50.19^0 \\ k\approx50^0 \end{gathered}[/tex]
Hence,
The value of angle k = 50° (50 degrees)
