Find sin θ = 5/13

Given $\sin \theta = \frac{5}{13}$.

$$\theta = \arcsin\left(\frac{5}{13}\right)$$

$$\theta \approx 22.61986^\circ$$

Rounding to one decimal place:

$$\theta \approx 22.6^\circ$$

The angle $\theta$ is approximately $\boxed{22.6^\circ}$.

Finding $\cos \theta$

The identity is $\sin^2 \theta + \cos^2 \theta = 1$.
Substitute $\sin \theta = \frac{5}{13}$:

$$\left(\frac{5}{13}\right)^2 + \cos^2 \theta = 1$$ $$\frac{25}{169} + \cos^2 \theta = 1$$

$$\cos^2 \theta = 1 - \frac{25}{169}$$ $$\cos^2 \theta = \frac{169 - 25}{169}$$ $$\cos^2 \theta = \frac{144}{169}$$

$$\cos \theta = \pm\sqrt{\frac{144}{169}}$$ $$\cos \theta = \pm\frac{12}{13}$$

Assuming $\theta$ is an acute angle (in the first quadrant), $\cos \theta$ is positive.

$$\cos \theta = \frac{12}{13}$$

The value of $\cos \theta$ is $\boxed{\frac{12}{13}}$.

Finding $\tan \theta$

Substitute the known values of $\sin \theta$ and $\cos \theta$:

$$\tan \theta = \frac{\frac{5}{13}}{\frac{12}{13}}$$

$$\tan \theta = \frac{5}{12}$$

The value of $\tan \theta$ is $\boxed{\frac{5}{12}}$.