Find \( \cot (\theta) \) if \( \sin (\theta)=\frac{5}{13} \) and \( \theta \) terminates in QI. Answer exactly, \( \cot (\theta)=\square \)

To find \( \cot (\theta) \), we start with the given value of \( \sin (\theta) = \frac{5}{13} \). Since \( \theta \) is in the first quadrant (QI), we know that both sine and cosine are positive. 1. We can find \( \cos (\theta) \) using the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1. \] Plugging in \( \sin(\theta) \): \[ \left(\frac{5}{13}\right)^2 + \cos^2(\theta) = 1, \] \[ \frac{25}{169} + \cos^2(\theta) = 1. \] Subtracting \( \frac{25}{169} \) from both sides: \[ \cos^2(\theta) = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169}. \] Now taking the square root (and since \( \theta \) is in QI, we take the positive root): \[ \cos(\theta) = \frac{12}{13}. \] 2. Now we can find \( \cot(\theta) \) since \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \): \[ \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5}. \] Thus, \( \cot(\theta) = \frac{12}{5} \).