We want to calculate tan 49°20'
First, we convert 20' into degrees.
[tex]\begin{gathered} \frac{60^{\prime}}{20^{\prime}}=\frac{1\degree}{x} \\ x=\frac{20^{\prime}}{60^{\prime}}\cdot1\degree=0.333\degree \\ \therefore20^{\prime}=0.333\degree \end{gathered}[/tex]Therefore, we have: tan 49°20' = tan 49.333° = 1.164
