From the top of a light house 60 m high with its base at sea level the angle of depression of aboat is `15^(@)`. The distance of the boat from the light house is

Options

A. `((sqrt(3)-1)/(sqrt(3)+1))60m`
B. `((sqrt(3)+1)/(sqrt(3)-1))60m`
C. `((sqrt(3)-1)/(sqrt(3)+1))^(2)m`
D. None

Answer

To solve the problem step by step, we will use trigonometric ratios to find the distance of the boat from the lighthouse.

### Step-by-Step Solution:

1. **Understand the Problem:**
We have a lighthouse that is 60 meters high. The angle of depression from the top of the lighthouse to the boat is 15 degrees. We need to find the distance of the boat from the base of the lighthouse.

2. **Draw a Diagram:**
- Draw a vertical line representing the lighthouse (60 m).
- From the top of the lighthouse, draw a horizontal line to represent the line of sight to the boat.
- Draw a line from the top of the lighthouse to the boat, forming a right triangle with the height of the lighthouse and the distance from the base of the lighthouse to the boat.

3. **Identify the Right Triangle:**
- The height of the lighthouse (perpendicular) = 60 m.
- The distance from the base of the lighthouse to the boat (base) = d (unknown).
- The angle of depression from the top of the lighthouse to the boat = 15 degrees.

4. **Use the Angle of Depression:**
The angle of depression from the top of the lighthouse corresponds to the angle of elevation from the boat to the top of the lighthouse. Therefore, the angle at the boat is also 15 degrees.

5. **Apply the Tangent Function:**
We can use the tangent function, which relates the opposite side (height of the lighthouse) to the adjacent side (distance from the lighthouse to the boat):
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
Here, \(\theta = 15^\circ\), opposite = 60 m, and adjacent = d.
\[ \tan(15^\circ) = \frac{60}{d} \]

6. **Rearrange the Equation:**
Rearranging the equation gives us:
\[ d = \frac{60}{\tan(15^\circ)} \]

7. **Calculate \(\tan(15^\circ)\):**
We can use the tangent subtraction formula:
\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ) \tan(30^\circ)} \]
Where:
- \(\tan(45^\circ) = 1\)
- \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)

Plugging in these values:
\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

8. **Substitute Back to Find d:**
Now substituting \(\tan(15^\circ)\) back into the equation for d:
\[ d = \frac{60}{\frac{\sqrt{3} - 1}{\sqrt{3} + 1}} = 60 \cdot \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

9. **Final Calculation:**
This gives us the distance of the boat from the lighthouse:
\[ d = 60 \cdot \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Final Answer:
The distance of the boat from the lighthouse is \( 60 \cdot \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \) meters.