To find angle $B$ in the given right triangle $\triangle ABC$, where $\angle A$ is the right angle, we can use trigonometric ratios. Given:

• $AB = 25$ (adjacent side to angle $B$)
• $AC = 30$ (hypotenuse)

We can use the cosine function, defined as: $\cos(B) = \frac{\text{adjacent}}{\text{hypotenuse}}$

Thus, $\cos(B) = \frac{AB}{AC} = \frac{25}{30} = \frac{5}{6}$

Now, to find angle $B$, we take the inverse cosine (arccos): $B = \arccos\left(\frac{5}{6}\right)$

Using a calculator to find the arccos: $B \approx \arccos\left(\frac{5}{6}\right) \approx 33.56^\circ$

So, angle $B$ is approximately $33.56^\circ$.

Would you like further details or have any questions about this solution?

Here are some related questions you might find interesting:

1. How can we find angle $C$ in the triangle?
2. What is the sine of angle $B$?
3. How can we use the Pythagorean theorem to verify the side lengths?
4. What is the tangent of angle $B$?
5. How do we calculate the area of $\triangle ABC$?

Tip: For right triangles, using trigonometric ratios (sine, cosine, tangent) is a straightforward way to find angles when side lengths are known.