From the top of a light house 60 metre high, with its base at the sea leave, the angel of depression of a boat is `30^(@)` . The distance of the boat from the foot of the light - house is

Options

A. `60 sqrt(3)` metres
B. `(60)/( sqrt(3))` metres
C. 60 metres
D. `30 sqrt(2)` metres

Answer

To solve the problem step by step, we will use the concept of trigonometry, specifically the tangent function, to find the distance of the boat from the foot of the lighthouse.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a lighthouse that is 60 meters high.
- The angle of depression to the boat is 30 degrees.
- We need to find the horizontal distance from the foot of the lighthouse to the boat.

2. **Drawing the Diagram:**
- Imagine a right triangle formed by the lighthouse, the line of sight from the top of the lighthouse to the boat, and the horizontal distance from the lighthouse to the boat.
- Let:
- \( A \) be the top of the lighthouse,
- \( B \) be the foot of the lighthouse,
- \( C \) be the position of the boat.
- The height \( AB = 60 \) meters (perpendicular).
- The angle of depression \( \angle ACB = 30^\circ \).

3. **Identifying the Angles:**
- The angle of depression from point \( A \) to point \( C \) is \( 30^\circ \).
- Therefore, the angle \( \angle ABC \) (the angle of elevation from the boat to the top of the lighthouse) is also \( 30^\circ \) due to alternate interior angles.

4. **Using Trigonometric Ratios:**
- In triangle \( ABC \):
- The opposite side to angle \( 30^\circ \) is \( AB = 60 \) meters.
- The adjacent side is \( BC \) (the distance we want to find).
- We can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
- For \( \theta = 30^\circ \):
\[ \tan(30^\circ) = \frac{AB}{BC} = \frac{60}{BC} \]

5. **Finding the Value of \( BC \):**
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
- Thus, we can write:
\[ \frac{1}{\sqrt{3}} = \frac{60}{BC} \]
- Cross-multiplying gives:
\[ BC = 60 \cdot \sqrt{3} \]

6. **Final Calculation:**
- Therefore, the distance of the boat from the foot of the lighthouse is:
\[ BC = 60 \cdot \sqrt{3} \text{ meters} \]

### Final Answer:
The distance of the boat from the foot of the lighthouse is \( 60 \sqrt{3} \) meters.

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