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 tower.

Answer

To find the horizontal distance between the ship and the tower, we can use the concept of trigonometry, specifically the tangent function, which relates the angle of depression to the opposite and adjacent sides of a right triangle.

### Step-by-Step Solution:

1. **Identify the Components**:
We have a tower of height \( AB = 50 \) m. The angle of depression from the top of the tower to the ship is \( 30^\circ \).

2. **Draw the Diagram**:
Draw a right triangle where:
- \( A \) is the top of the tower.
- \( B \) is the base of the tower (ground level).
- \( C \) is the position of the ship.
- The angle \( ACB \) is the angle of depression, which is \( 30^\circ \).
- The height \( AB \) is the opposite side (50 m), and the horizontal distance \( BC \) is the adjacent side.

3. **Use the Angle of Depression**:
The angle of depression from point \( A \) to point \( C \) is equal to the angle \( CAB \) (alternate interior angles). Therefore, \( CAB = 30^\circ \).

4. **Set Up the Tangent Function**:
Using the definition of tangent in the right triangle:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
Here, \( \theta = 30^\circ \), the opposite side is \( AB = 50 \) m, and the adjacent side is \( BC \):
\[ \tan(30^\circ) = \frac{AB}{BC} = \frac{50}{BC} \]

5. **Calculate \( \tan(30^\circ) \)**:
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

6. **Set Up the Equation**:
Substitute \( \tan(30^\circ) \) into the equation:
\[ \frac{1}{\sqrt{3}} = \frac{50}{BC} \]

7. **Cross-Multiply**:
Cross-multiplying gives:
\[ BC = 50 \cdot \sqrt{3} \]

8. **Final Calculation**:
Therefore, the horizontal distance \( BC \) between the ship and the tower is:
\[ BC = 50\sqrt{3} \text{ meters} \]

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

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