A man walks 10m 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 posts is :

Options

A. 10m
B. `(5sqrt(3)+5)m`
C. `(5sqrt(3)-5)m`
D. `(10sqrt(3)+10)m`

Explanation

To solve the problem step by step, we will use trigonometric principles related to the angles of elevation.

### Step 1: Understand the scenario
A man walks 10 meters towards a lamp post, and the angle of elevation changes from \(30^\circ\) to \(45^\circ\). We need to find the height of the lamp post.

### Step 2: Set up the diagram
Let:
- \(A\) be the top of the lamp post.
- \(B\) be the base of the lamp post.
- \(C\) be the initial position of the man (where the angle of elevation is \(30^\circ\)).
- \(D\) be the final position of the man (where the angle of elevation is \(45^\circ\)).
- Let the height of the lamp post \(AB = h\).
- The distance from \(B\) to \(C\) be \(x\).
- The distance from \(B\) to \(D\) be \(x - 10\) (since the man walks 10 meters towards the lamp post).

### Step 3: Use the first angle of elevation (30 degrees)
From point \(C\):
\[ \tan(30^\circ) = \frac{h}{x} \]
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x} \implies h = \frac{x}{\sqrt{3}} \quad \text{(1)} \]

### Step 4: Use the second angle of elevation (45 degrees)
From point \(D\):
\[ \tan(45^\circ) = \frac{h}{x - 10} \]
We know that \(\tan(45^\circ) = 1\), so:
\[ 1 = \frac{h}{x - 10} \implies h = x - 10 \quad \text{(2)} \]

### Step 5: Set the equations equal
From equations (1) and (2):
\[ \frac{x}{\sqrt{3}} = x - 10 \]

### Step 6: Solve for \(x\)
Rearranging gives:
\[ x - \frac{x}{\sqrt{3}} = 10 \]
Factoring out \(x\):
\[ x\left(1 - \frac{1}{\sqrt{3}}\right) = 10 \]
Calculating \(1 - \frac{1}{\sqrt{3}}\):
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]
Thus:
\[ x \cdot \frac{\sqrt{3} - 1}{\sqrt{3}} = 10 \implies x = \frac{10\sqrt{3}}{\sqrt{3} - 1} \]

### Step 7: Rationalize the denominator
Multiply numerator and denominator by \(\sqrt{3} + 1\):
\[ x = \frac{10\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{10\sqrt{3}(\sqrt{3} + 1)}{2} \]
Thus:
\[ x = 5\sqrt{3}(\sqrt{3} + 1) = 15 + 5\sqrt{3} \]

### Step 8: Substitute \(x\) back to find \(h\)
Using equation (2):
\[ h = x - 10 = (15 + 5\sqrt{3}) - 10 = 5 + 5\sqrt{3} \]

### Final Answer:
The height of the lamp post is:
\[ h = 5 + 5\sqrt{3} \text{ meters} \]