If the angle of depression of an object from a 75 m high tower is `30^@`, then the distance of the object from the tower is

Options

A. `25sqrt3` m
B. `50sqrt3` m
C. `75sqrt3` m
D. 150 m

Answer

To solve the problem, we need to find the distance of the object from the base of the tower using the given height of the tower and the angle of depression.

### Step-by-Step Solution:

1. **Understand the Problem**: We have a tower that is 75 meters high, and we need to find the horizontal distance (let's call it \( x \)) from the base of the tower to the object when the angle of depression from the top of the tower to the object is \( 30^\circ \).

2. **Draw a Diagram**:
- Draw a vertical line representing the tower (75 m high).
- Mark the top of the tower as point A and the bottom as point B.
- Mark the position of the object as point O.
- The angle of depression from point A to point O is \( 30^\circ \).

3. **Identify the Right Triangle**:
- In triangle ABO, where:
- AB is the height of the tower (75 m),
- AO is the horizontal distance from the tower to the object (x),
- The angle of depression \( \angle AOB \) is \( 30^\circ \).

4. **Use Trigonometric Ratios**:
- In right triangle ABO, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
- Here, \( \theta = 30^\circ \), the opposite side is AB (75 m), and the adjacent side is AO (x).
- Therefore, we have:
\[ \tan(30^\circ) = \frac{AB}{AO} = \frac{75}{x} \]

5. **Substitute the Value of \( \tan(30^\circ) \)**:
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
- So, we can write:
\[ \frac{1}{\sqrt{3}} = \frac{75}{x} \]

6. **Cross-Multiply to Solve for \( x \)**:
- Cross-multiplying gives:
\[ x = 75 \cdot \sqrt{3} \]

7. **Conclusion**:
- The distance of the object from the tower is:
\[ x = 75\sqrt{3} \text{ meters} \]

### Final Answer:
The distance of the object from the tower is \( 75\sqrt{3} \) meters.

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