A man walks 10 m towards a lamp post and notices that the angle of elevation of the top of the post increases from `30^(@)` to `45^(@)` the height of the lamp post is

Options

A. A. 10 m
B. B. `( 5 sqrt3 + 5) m `
C. C. ` ( 5 sqrt3 - 5) m`
D. D. ` (10 sqrt3 + 10 ) m `

Answer

To solve the problem step by step, we will use trigonometric relationships in right triangles formed by the man, the lamp post, and the angles of elevation.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A man walks 10 meters towards a lamp post.
- The angle of elevation from his initial position to the top of the lamp post is \(30^\circ\).
- After walking 10 meters closer, the angle of elevation becomes \(45^\circ\).
- We need to find the height of the lamp post, denoted as \(h\).

2. **Setting Up the Diagram**:
- Let \(A\) be the position of the man when the angle of elevation is \(30^\circ\).
- Let \(B\) be the position of the man after walking 10 meters towards the lamp post.
- Let \(C\) be the top of the lamp post.
- Let \(D\) be the base of the lamp post.
- The distance \(AD\) (from the man’s initial position to the lamp post) is \(x\).
- The distance \(BD\) (from the man’s new position to the lamp post) is \(x - 10\).

3. **Using Trigonometry for Triangle ABC**:
- In triangle \(ABC\) (with angle \(30^\circ\)):
\[ \tan(30^\circ) = \frac{h}{x} \]
- We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x} \]
- Rearranging gives us:
\[ h = \frac{x}{\sqrt{3}} \quad \text{(Equation 1)} \]

4. **Using Trigonometry for Triangle ABD**:
- In triangle \(ABD\) (with angle \(45^\circ\)):
\[ \tan(45^\circ) = \frac{h}{x - 10} \]
- We know that \(\tan(45^\circ) = 1\), so:
\[ 1 = \frac{h}{x - 10} \]
- Rearranging gives us:
\[ h = x - 10 \quad \text{(Equation 2)} \]

5. **Equating the Two Expressions for Height**:
- From Equation 1 and Equation 2, we have:
\[ \frac{x}{\sqrt{3}} = x - 10 \]
- Multiplying through by \(\sqrt{3}\) to eliminate the fraction:
\[ x = \sqrt{3}(x - 10) \]
- Expanding gives:
\[ x = \sqrt{3}x - 10\sqrt{3} \]
- Rearranging to isolate \(x\):
\[ x - \sqrt{3}x = -10\sqrt{3} \]
\[ x(1 - \sqrt{3}) = -10\sqrt{3} \]
- Solving for \(x\):
\[ x = \frac{-10\sqrt{3}}{1 - \sqrt{3}} \]

6. **Finding Height \(h\)**:
- Substitute \(x\) back into Equation 2:
\[ h = x - 10 \]
- After calculating \(x\), substitute to find \(h\):
\[ h = \frac{-10\sqrt{3}}{1 - \sqrt{3}} - 10 \]

7. **Rationalizing and Simplifying**:
- Rationalize the expression for \(h\) and simplify to find the final height in meters.

8. **Final Result**:
- After calculations, we find:
\[ h = 5\sqrt{3} + 5 \text{ meters} \]