In triangle ABC, AC = 18, CD = 14, angle C = 90 degrees. Find x and y.

Given $AC = 18$ and assuming $\triangle ADC$ is an isosceles right-angled triangle, then $DC = AC$.

$$DC = 18$$

The length of the base $BC$ is the sum of $BD$ and $DC$.

$$BC = BD + DC$$ $$BC = 14 + 18$$ $$BC = 32$$

$\triangle ABC$ is a right-angled triangle at $C$. We use the tangent function, which is the ratio of the opposite side to the adjacent side.

$$\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{BC}$$ $$\tan x = \frac{18}{32}$$ $$\tan x = \frac{9}{16}$$

To find $x$, take the inverse tangent:

$$x = \arctan\left(\frac{9}{16}\right)$$ $$x \approx 29.4^\circ$$

$\triangle ADC$ is a right-angled triangle at $C$. We use the tangent function.

$$\tan y = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{DC}$$ $$\tan y = \frac{18}{18}$$ $$\tan y = 1$$

To find $y$, take the inverse tangent:

$$y = \arctan(1)$$ $$y = 45^\circ$$

The values of the angles are:
$x \approx \boxed{\mathbf{29.4^\circ}}$
$y = \boxed{\mathbf{45^\circ}}$