To solve for the angle $\angle XZY$ in this right triangle, we can use trigonometric ratios. Specifically, we have the lengths of the adjacent side $XY = 6 \, \text{cm}$ and the hypotenuse $XZ = 15 \, \text{cm}$. Since we are dealing with the angle adjacent to the given sides, we can use the cosine function:

$\cos(\angle XZY) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XY}{XZ} = \frac{6}{15}$

Now, we can calculate the angle $\theta = \angle XZY$ by taking the inverse cosine (arccos) of the ratio:

$\theta = \cos^{-1}\left(\frac{6}{15}\right)$

Let me compute the value.The size of angle $\angle XZY$ is approximately $66.4^\circ$, rounded to one decimal place.

Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. How would you find the size of angle $\angle XYZ$?
2. What trigonometric ratio would you use if you were given the opposite side instead of the adjacent?
3. How can you verify if a triangle is a right triangle using the Pythagorean theorem?
4. What is the sine of angle $\angle XZY$?
5. How would you find the area of this triangle using the given dimensions?

Tip: In a right triangle, knowing any two sides allows you to calculate any angle using basic trigonometric ratios.