The image shows a triangle ABC with a point D on BC. AD is perpendicular to AC. If AC = 5, AB = 13, and AD = 12, find the length of CD.

The image shows $\triangle CAB$ with $\angle CAB = 90^\circ$.
The length of side $AC = 5$.
The length of side $AB = 12$.
$AD$ is an altitude to $BC$, which means $\angle ADB = 90^\circ$ and $\angle ADC = 90^\circ$.
We need to find the angles $x = \angle DAB$ and $y = \angle CAD$.

In the right-angled triangle $\triangle CAB$:
We can use the tangent function, which is $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$.
For $\angle ABC$:

For $\angle ACB$:

In the right-angled triangle $\triangle ADB$, the sum of angles is $180^\circ$.
We have $\angle ADB = 90^\circ$, $\angle DAB = x$, and $\angle ABD = \angle ABC$.
Therefore, $x + \angle ABC + 90^\circ = 180^\circ$.

Substitute the value of $\angle ABC$:

Using the trigonometric identity $90^\circ - \arctan(k) = \arctan\left(\frac{1}{k}\right)$:

Calculating the numerical value:

In the right-angled triangle $\triangle ADC$, the sum of angles is $180^\circ$.
We have $\angle ADC = 90^\circ$, $\angle CAD = y$, and $\angle ACD = \angle ACB$.
Therefore, $y + \angle ACB + 90^\circ = 180^\circ$.

Substitute the value of $\angle ACB$:

Using the trigonometric identity $90^\circ - \arctan(k) = \arctan\left(\frac{1}{k}\right)$:

Calculating the numerical value:

The values of the angles are:
$x = \boxed{\mathbf{\arctan\left(\frac{12}{5}\right) \approx 67.38^\circ}}$
$y = \boxed{\mathbf{\arctan\left(\frac{5}{12}\right) \approx 22.62^\circ}}$