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

Options

A. 400 m
B. 421 m
C. 444 m
D. 460 m

Explanation

To solve the problem step by step, let's break it down clearly:

### Step 1: Understand the Problem
We have a lighthouse that is 120 meters tall, and the angle of depression to a boat from the top of the lighthouse is 15 degrees. We need to find the horizontal distance from the base of the lighthouse to the boat.

### Step 2: Draw the Diagram
1. Draw a vertical line representing the lighthouse (120m).
2. At the top of the lighthouse, draw a horizontal line to represent the line of sight to the boat.
3. The angle of depression (15 degrees) is formed between the horizontal line and the line of sight to the boat.

### Step 3: Identify the Right Triangle
From the top of the lighthouse:
- The height (opposite side) is 120 meters.
- The horizontal distance to the boat (adjacent side) is what we need to find, let's denote it as \( x \).

### Step 4: Use the Tangent Function
Using the tangent of the angle of depression:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
Here, \( \theta = 15^\circ \), the opposite side is 120m, and the adjacent side is \( x \):
\[ \tan(15^\circ) = \frac{120}{x} \]

### Step 5: Rearranging the Equation
Rearranging the equation to solve for \( x \):
\[ x = \frac{120}{\tan(15^\circ)} \]

### Step 6: Calculate \( \tan(15^\circ) \)
Using 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 \( \tan(45^\circ) = 1 \) and \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Step 7: Substitute Back into the Equation
Substituting \( \tan(15^\circ) \) back into the equation for \( x \):
\[ x = \frac{120}{\frac{\sqrt{3} - 1}{\sqrt{3} + 1}} = 120 \cdot \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Step 8: Rationalize the Denominator
To simplify further, multiply the numerator and denominator by the conjugate of the denominator:
\[ x = 120 \cdot \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = 120 \cdot \frac{(\sqrt{3} + 1)^2}{3 - 1} \]
\[ = 120 \cdot \frac{3 + 2\sqrt{3} + 1}{2} = 120 \cdot \frac{4 + 2\sqrt{3}}{2} = 60(4 + 2\sqrt{3}) = 240 + 120\sqrt{3} \]

### Final Answer
The distance of the boat from the lighthouse is:
\[ x = 240 + 120\sqrt{3} \text{ meters} \]