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

Options

A. `(sqrt3-1)/(sqrt3+1)` . 60 metres
B. `(sqrt3+1)/(sqrt3-1)` . 60 metres
C. `(sqrt3+1)/(sqrt3-1)` metres
D. none of these

Answer

To solve the problem step by step, we will use trigonometric concepts related to angles of depression and right triangles.

### Step-by-Step Solution:

1. **Understand the Problem**: We have a lighthouse (AB) that is 60 m high, and we need to find the distance (BC) of a boat (C) from the foot (B) of the lighthouse when the angle of depression from the top of the lighthouse to the boat is 15 degrees.

2. **Draw the Diagram**:
- Draw a vertical line representing the lighthouse (AB) with a height of 60 m.
- Mark point B as the base of the lighthouse at sea level.
- Mark point C as the position of the boat.
- Draw a horizontal line from point A (top of the lighthouse) to point C.
- The angle of depression from A to C is 15 degrees.

3. **Identify the Angles**:
- The angle of depression from A to C is 15 degrees.
- By alternate interior angles, the angle CAB (the angle between the line from A to C and the vertical line AB) is also 15 degrees.

4. **Set Up the Right Triangle**:
- In triangle ABC, we have:
- Perpendicular (height of the lighthouse) = AB = 60 m
- Base (distance from the foot of the lighthouse to the boat) = BC = x (which we need to find)
- Angle CAB = 15 degrees

5. **Use the Tangent Function**:
- From the definition of tangent in a right triangle:
\[ \tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} \]
- Here, we can write:
\[ \tan(15^\circ) = \frac{AB}{BC} = \frac{60}{x} \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ x = \frac{60}{\tan(15^\circ)} \]

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

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

9. **Final Calculation**:
- Simplifying gives:
\[ x = 60 \cdot \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = 60 \cdot \frac{(\sqrt{3} + 1)^2}{2} \]
- Calculate \((\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}\).
- Thus, \(x = 30(4 + 2\sqrt{3})\).

10. **Conclusion**:
- The distance of the boat from the foot of the lighthouse is \(x = 30(4 + 2\sqrt{3})\) meters.