Let $\theta = \arcsin \frac{5}{13}$. Then $\sin \theta = \frac{5}{13}$.

We want to find $\cos \theta$. Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we can solve for $\cos \theta$:

$\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}$.

Since $\cos \theta$ must be positive (since $\theta$ lies in the first quadrant), we have $\cos \theta = \frac{12}{13}$.

Therefore, $\cos \left( \arcsin \frac{5}{13} \right) = \boxed{\frac{12}{13}}$.The answer is: \frac{12}{13}
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Determine the value of $\cos \left( \arcsin \frac{5}{13} \right)$.

### Answer:
