Trigonometry (11th Edition) Clone

$60^o$
The value of $\theta$ must be in the interval $(0^o, 180^o)$. $\theta=\cot^{-1}{\left(\frac{\sqrt3}{3}\right)}$ means that $\cot{\theta} = \frac{\sqrt3}{3}$. Since cotangent is the reciprocal of tangent, then $\cot{\theta} = \frac{\sqrt3}{3}$ means $\tan{\theta}=\frac{3}{\sqrt3}=\sqrt3$. Note that $\tan{60^o} = \sqrt3$. Thus, $\cot^{-1}{\left(\frac{\sqrt3}{3}\right)}=60^o$