1. The problem states that $\cot \theta = \frac{\sqrt{3}}{2}$. We need to find $\theta$ or related trigonometric values. 2. Recall that $\cot \theta = \frac{\cos \theta}{\sin \theta}$. So, we have: $$\frac{\cos \theta}{\sin \theta} = \frac{\sqrt{3}}{2}$$ 3. This implies: $$\cos \theta = \frac{\sqrt{3}}{2} \sin \theta$$ 4. Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, substitute $\cos \theta$: $$\sin^2 \theta + \left(\frac{\sqrt{3}}{2} \sin \theta\right)^2 = 1$$ 5. Simplify: $$\sin^2 \theta + \frac{3}{4} \sin^2 \theta = 1$$ $$\left(1 + \frac{3}{4}\right) \sin^2 \theta = 1$$ $$\frac{7}{4} \sin^2 \theta = 1$$ 6. Solve for $\sin^2 \theta$: $$\sin^2 \theta = \frac{4}{7}$$ 7. Take the square root: $$\sin \theta = \pm \frac{2}{\sqrt{7}}$$ 8. Find $\cos \theta$ using step 3: $$\cos \theta = \frac{\sqrt{3}}{2} \times \pm \frac{2}{\sqrt{7}} = \pm \frac{\sqrt{3}}{\sqrt{7}}$$ 9. Therefore, the possible values for $\sin \theta$ and $\cos \theta$ are: $$\sin \theta = \pm \frac{2}{\sqrt{7}}, \quad \cos \theta = \pm \frac{\sqrt{3}}{\sqrt{7}}$$ 10. The angle $\theta$ can be found by: $$\theta = \cot^{-1} \left(\frac{\sqrt{3}}{2}\right)$$ Final answer: $$\sin \theta = \pm \frac{2}{\sqrt{7}}, \quad \cos \theta = \pm \frac{\sqrt{3}}{\sqrt{7}}, \quad \theta = \cot^{-1} \left(\frac{\sqrt{3}}{2}\right)$$