EXPLANATIONS:
Given;
We are given the following expression;
[tex]arctan(\frac{1}{\sqrt{3}})[/tex]Required;
We are required to find the angle measure of this in both radians, and degrees.
Step-by-step solution;
For the angle whose tangent is given as 1 over square root of 3, on the unit circle, we would have
[tex]\begin{gathered} tan\theta=\frac{1}{\sqrt{3}} \\ Rationalize: \\ \\ \frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ =\frac{\sqrt{3}}{\sqrt{3}\times\sqrt{3}} \\ \\ =\frac{\sqrt{3}}{3} \end{gathered}[/tex]On the unit circle, the general solution for this value as shown would be;
[tex]tan^{-1}(\frac{\sqrt{3}}{3})=\frac{\pi}{6}[/tex]To convert this to degree measure, we will use the following equation;
[tex]\frac{r}{\pi}=\frac{d}{180}[/tex]We now substitute for the value of r;
[tex]\begin{gathered} \frac{\frac{\pi}{6}}{\pi}=\frac{d}{180} \\ \\ \frac{\pi}{6}\div\frac{\pi}{1}=\frac{d}{180} \\ \\ \frac{\pi}{6}\times\frac{1}{\pi}=\frac{d}{180} \\ \\ \frac{1}{6}=\frac{d}{180} \end{gathered}[/tex]We now cross multiply;
[tex]\begin{gathered} \frac{180}{6}=d \\ \\ 30=d \end{gathered}[/tex]Therefore;
ANSWER:
[tex]\begin{gathered} radians=\frac{\pi}{6} \\ \\ degrees=30\degree \end{gathered}[/tex]