From the top of a lighthouse 70 m high with its base at 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. A. `70 ( 2- sqrt3) m`
B. B. ` (70 ( 2 + sqrt3) m `
C. C. `70 ( 3- sqrt3) m`
D. D. `70 ( 3 + sqrt3) m`

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 that is 70 m tall. The angle of depression from the top of the lighthouse to a boat is 15 degrees. We need to find the horizontal distance from the base of the lighthouse to the boat.

2. **Draw a Diagram**:
- Draw a vertical line representing the lighthouse (70 m).
- Mark the top of the lighthouse as point A and the base as point B.
- Mark the position of the boat as point C.
- The angle of depression from point A to point C is 15 degrees. Since the angle of depression is equal to the angle of elevation from point C to point A, we can denote this angle as ∠ACB = 15 degrees.

3. **Identify the Right Triangle**:
- In triangle ABC, AB is the height of the lighthouse (70 m), and BC is the distance from the foot of the lighthouse to the boat (which we need to find, let's call it x).
- The angle ∠ACB = 15 degrees.

4. **Use the Tangent Function**:
- From trigonometry, we know that:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
- Here, the opposite side is AB (70 m) and the adjacent side is BC (x).
- Therefore, we can write:
\[ \tan(15^\circ) = \frac{70}{x} \]

5. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ x = \frac{70}{\tan(15^\circ)} \]

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

7. **Substituting Back**:
- Substitute \(\tan(15^\circ)\) back into the equation for x:
\[ x = \frac{70}{\frac{\sqrt{3} - 1}{\sqrt{3} + 1}} = 70 \cdot \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

8. **Rationalizing the Denominator**:
- To simplify further, multiply the numerator and denominator by \(\sqrt{3} + 1\):
\[ x = 70 \cdot \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2} = 70 \cdot \frac{3 + 2\sqrt{3} + 1}{3 - 1} = 70 \cdot \frac{4 + 2\sqrt{3}}{2} = 70 \cdot (2 + \sqrt{3}) \]

9. **Final Calculation**:
- Thus, the distance from the foot of the lighthouse to the boat is:
\[ x = 70(2 + \sqrt{3}) \text{ meters} \]