1. **Problem statement:** Find the value of $\theta$ given: a) $\cot \theta = \frac{\sqrt{3}}{3}$ b) $\tan \theta = \frac{\sqrt{3}}{3}$ --- 2. **Recall the definitions and key values:** - $\cot \theta = \frac{1}{\tan \theta}$ - Common exact values for $\tan \theta$: - $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ - $\tan 45^\circ = 1$ - $\tan 60^\circ = \sqrt{3}$ - Angles in radians: - $30^\circ = \frac{\pi}{6}$ - $45^\circ = \frac{\pi}{4}$ - $60^\circ = \frac{\pi}{3}$ --- 3. **Part a) Solve for $\theta$ when $\cot \theta = \frac{\sqrt{3}}{3}$:** - Since $\cot \theta = \frac{1}{\tan \theta}$, then: $$\tan \theta = \frac{1}{\cot \theta} = \frac{1}{\frac{\sqrt{3}}{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$$ - From the common values, $\tan 60^\circ = \sqrt{3}$. - Therefore, $\theta = 60^\circ$. - Check options: 4. 60° matches. --- 4. **Part b) Solve for $\theta$ when $\tan \theta = \frac{\sqrt{3}}{3}$:** - From the common values, $\tan 30^\circ = \frac{\sqrt{3}}{3}$. - In radians, $30^\circ = \frac{\pi}{6}$. - Also, tangent is positive in the first and third quadrants, so another solution is $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$, but this is not in the options. - Check options: 1. $\frac{\pi}{6}$ matches. --- **Final answers:** a) $\theta = 60^\circ$ b) $\theta = \frac{\pi}{6}$