From the top of a lighthouse 60 meters high with its base at the sea level, the angle of depression of a boat is `15^(@)`. The distance of the boat from the foot of the lighthouse is

Options

A. `(sqrt(3)-1)/(sqrt(3)+1)`. 60 meters
B. `(sqrt(3)+1)/(sqrt(3)-1)`. 60 meters.
C. `(sqrt(3)+1)/(sqrt(3)-1)` meters
D. None of these

Answer

To solve the problem step by step, we will analyze the situation using trigonometry.

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

### Step 2: Draw a Diagram
- Draw a vertical line representing the lighthouse (AB) with height 60 meters.
- Point A is the top of the lighthouse, and point B is the base.
- Draw a horizontal line from point A to the right, representing the line of sight to the boat.
- Let point C be the position of the boat.
- The angle of depression from A to C is 15 degrees.

### Step 3: Identify the Right Triangle
- The angle of depression from A to C is equal to the angle of elevation from C to A, which is also 15 degrees.
- We can form a right triangle ABC, where:
- AB = 60 meters (height of the lighthouse)
- BC = x meters (distance from the base of the lighthouse to the boat)
- Angle ACB = 15 degrees.

### Step 4: Use the Tangent Function
Using the tangent function, we have:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
In our case:
\[ \tan(15^\circ) = \frac{AB}{BC} = \frac{60}{x} \]
Rearranging gives:
\[ x = \frac{60}{\tan(15^\circ)} \]

### Step 5: Calculate \(\tan(15^\circ)\)
To find \(\tan(15^\circ)\), 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:
- \(\tan(45^\circ) = 1\)
- \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)

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

### Step 6: Substitute \(\tan(15^\circ)\) into the Equation
Now substituting \(\tan(15^\circ)\) back into the equation for x:
\[ x = \frac{60}{\frac{\sqrt{3} - 1}{\sqrt{3} + 1}} = 60 \cdot \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Step 7: Simplify the Expression
To simplify:
\[ x = 60 \cdot \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = 60 \cdot \frac{(\sqrt{3} + 1)^2}{2} \]
Calculating \((\sqrt{3} + 1)^2\):
\[ (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \]
Thus:
\[ x = 30(4 + 2\sqrt{3}) = 120 + 60\sqrt{3} \]

### Final Answer
The distance of the boat from the foot of the lighthouse is:
\[ x = 120 + 60\sqrt{3} \text{ meters} \]