Angle of depression from the top of a light house of two boats are `45^(@) and 30^(@)` due east which are 60 m apart. The height of the light house is

Options

A. `60sqrt(3)`
B. `30(sqrt(3)+1)`
C. `30(sqrt(3)-1)`
D. None

Answer

To solve the problem of finding the height of the lighthouse given the angles of depression to two boats and their distance apart, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a lighthouse (AB) and two boats (C and D) that are 60 meters apart. The angles of depression from the top of the lighthouse to the boats are 45° and 30°, respectively.

2. **Drawing the Diagram**:
- Draw a vertical line representing the lighthouse (AB).
- Mark point A as the top of the lighthouse and B as the bottom (ground level).
- Draw horizontal lines from A to points C and D, representing the angles of depression to the boats.
- Label the distance between the boats (C and D) as 60 m.

3. **Identifying the Angles**:
- The angle of depression to boat C is 45°, which means the angle of elevation from C to A is also 45°.
- The angle of depression to boat D is 30°, which means the angle of elevation from D to A is also 30°.

4. **Setting Up the Triangles**:
- For triangle ABC (with angle 45°):
- Let the height of the lighthouse be \( h \).
- The distance from B to C (the base) can be denoted as \( x \).
- Using the tangent function:
\[ \tan(45°) = \frac{h}{x} \implies h = x \]

- For triangle ABD (with angle 30°):
- The distance from B to D (the base) is \( 60 + x \).
- Using the tangent function:
\[ \tan(30°) = \frac{h}{60 + x} \implies h = (60 + x) \cdot \tan(30°) = (60 + x) \cdot \frac{1}{\sqrt{3}} \]

5. **Setting Up the Equation**:
- From the two equations derived:
\[ h = x \quad \text{(1)} \]
\[ h = \frac{60 + x}{\sqrt{3}} \quad \text{(2)} \]

6. **Substituting Equation (1) into Equation (2)**:
- Replace \( h \) in equation (2) with \( x \):
\[ x = \frac{60 + x}{\sqrt{3}} \]

7. **Solving for x**:
- Multiply both sides by \( \sqrt{3} \):
\[ x \sqrt{3} = 60 + x \]
- Rearranging gives:
\[ x \sqrt{3} - x = 60 \]
\[ x(\sqrt{3} - 1) = 60 \]
\[ x = \frac{60}{\sqrt{3} - 1} \]

8. **Finding h**:
- Substitute \( x \) back into equation (1):
\[ h = x = \frac{60}{\sqrt{3} - 1} \]

9. **Rationalizing the Denominator**:
- Multiply numerator and denominator by \( \sqrt{3} + 1 \):
\[ h = \frac{60(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{60(\sqrt{3} + 1)}{3 - 1} = \frac{60(\sqrt{3} + 1)}{2} = 30(\sqrt{3} + 1) \]

### Final Answer:
The height of the lighthouse is \( h = 30(\sqrt{3} + 1) \) meters.