The angle of depression of a ship from the top a tower of height 50 m is `30^(@)`. Find the horizontal distance between the ship and the tower.

Answer

To solve the problem, we need to find the horizontal distance between a ship and a tower given the height of the tower and the angle of depression from the top of the tower to the ship.

### Step-by-Step Solution:

1. **Understand the Problem:**
- We have a tower of height \( AC = 50 \) m.
- The angle of depression from the top of the tower (point A) to the ship (point B) is \( 30^\circ \).
- We need to find the horizontal distance \( BC \) between the base of the tower (point C) and the ship (point B).

2. **Identify the Angles:**
- The angle of depression from point A to point B is \( 30^\circ \).
- Since line AC is vertical and line BC is horizontal, the angle of elevation from point C to point A is also \( 30^\circ \) (alternate interior angles).

3. **Use Trigonometric Ratios:**
- In triangle ABC, we can use the tangent function, which relates the angle to the opposite side and the adjacent side:
\[ \tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AC}{BC} \]
- Here, \( AC \) is the height of the tower (50 m), and \( BC \) is the distance we want to find.

4. **Substitute Known Values:**
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
- Substituting the values into the equation gives:
\[ \frac{1}{\sqrt{3}} = \frac{50}{BC} \]

5. **Solve for BC:**
- Rearranging the equation to find \( BC \):
\[ BC = 50 \cdot \sqrt{3} \]

6. **Calculate the Value:**
- The horizontal distance \( BC \) is therefore:
\[ BC = 50\sqrt{3} \text{ meters} \]

### Final Answer:
The horizontal distance between the ship and the tower is \( 50\sqrt{3} \) meters.