From the top of a lighthouse 120 m above the sea, the angle of depression of a boat is `15^(@).` what is the distance of the boat from the lighthouse?

Options

A. A. 120(2+√3)m
B. B. 120 m
C. 444 m
D. None of these

Answer

To solve the problem step by step, we will use the concept of angles of depression and trigonometry.

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a lighthouse that is 120 meters tall, and we need to find the distance from the base of the lighthouse to a boat in the sea, given that the angle of depression from the top of the lighthouse to the boat is \(15^\circ\).

2. **Draw the Diagram**:
- Let point A be the top of the lighthouse, point B be the base of the lighthouse, and point C be the position of the boat.
- The height of the lighthouse (AB) is 120 m.
- The angle of depression from A to C is \(15^\circ\).
- Since BC is horizontal, the angle of elevation from C to A is also \(15^\circ\).

3. **Set Up the Right Triangle**:
In triangle ABC:
- AB (height of the lighthouse) = 120 m (perpendicular)
- BC (distance from the lighthouse to the boat) = X (adjacent side)
- Angle CAB = \(15^\circ\)

4. **Use the Tangent Function**:
The tangent of an angle in a right triangle is defined as the ratio of the opposite side to the adjacent side. Therefore:
\[ \tan(15^\circ) = \frac{AB}{BC} = \frac{120}{X} \]

5. **Rearrange the Equation**:
From the tangent equation, we can rearrange it to find X:
\[ X = \frac{120}{\tan(15^\circ)} \]

6. **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:
\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

7. **Substitute Back into the Equation**:
Now substitute \(\tan(15^\circ)\) back into the equation for X:
\[ X = \frac{120}{\frac{\sqrt{3} - 1}{\sqrt{3} + 1}} = 120 \cdot \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

8. **Rationalize the Denominator**:
To simplify, multiply the numerator and denominator by \(\sqrt{3} + 1\):
\[ X = 120 \cdot \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = 120 \cdot \frac{3 + 2\sqrt{3} + 1}{3 - 1} = 120 \cdot \frac{4 + 2\sqrt{3}}{2} \]
Simplifying gives:
\[ X = 120 \cdot (2 + \sqrt{3}) \]

9. **Final Answer**:
Therefore, the distance of the boat from the lighthouse is:
\[ X = 120(2 + \sqrt{3}) \text{ meters} \]