# How do you evaluate tan^-1(sqrt3/3) without a calculator?

Sep 29, 2016

${\tan}^{- 1} \left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$

#### Explanation:

Consider a triangle with sides in ratio $1 : \sqrt{3} : 2$

This is a right angled triangle since:

${1}^{2} + {\sqrt{3}}^{2} = 1 + 3 = 4 = {2}^{2}$

In fact, it is one half of an equilateral triangle, hence has angles $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$...

Now $\tan \left(\theta\right) = \text{opposite"/"adjacent}$

Hence we find:

$\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Since $\frac{\pi}{6}$ radians is in Q1, this is the value of ${\tan}^{- 1} \left(\frac{\sqrt{3}}{3}\right)$