The shadow of a tower standing on a level plane is found to be 40 m longer when the sun 's altitude is `45^(@)` , than when it is `60^(@)` . The height of the tower is

Options

A. `30(3+sqrt(3))` metre
B. `40(3+sqrt(3))` metre
C. `20(3+sqrt(3))` metre
D. `10(3+sqrt(3))` metre

Answer

To solve the problem, we need to find the height of the tower using the information about the shadows at different angles of elevation.

### Step-by-Step Solution:

1. **Define Variables**:
Let the height of the tower be \( h \) meters. Let the length of the shadow when the sun's altitude is \( 60^\circ \) be \( x \) meters. When the altitude is \( 45^\circ \), the shadow is \( x + 40 \) meters.

2. **Use Trigonometric Ratios**:
- For the angle of elevation \( 60^\circ \):
\[ \tan(60^\circ) = \frac{h}{x} \]
We know that \( \tan(60^\circ) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{h}{x} \implies h = \sqrt{3} x \quad \text{(Equation 1)} \]

- For the angle of elevation \( 45^\circ \):
\[ \tan(45^\circ) = \frac{h}{x + 40} \]
Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{h}{x + 40} \implies h = x + 40 \quad \text{(Equation 2)} \]

3. **Equate the Two Equations**:
From Equation 1 and Equation 2, we have:
\[ \sqrt{3} x = x + 40 \]

4. **Rearrange the Equation**:
\[ \sqrt{3} x - x = 40 \]
\[ (\sqrt{3} - 1)x = 40 \]
\[ x = \frac{40}{\sqrt{3} - 1} \]

5. **Rationalize the Denominator**:
To simplify \( x \):
\[ x = \frac{40(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{40(\sqrt{3} + 1)}{3 - 1} = \frac{40(\sqrt{3} + 1)}{2} = 20(\sqrt{3} + 1) \]

6. **Find the Height of the Tower**:
Substitute \( x \) back into Equation 1 to find \( h \):
\[ h = \sqrt{3} x = \sqrt{3} \cdot 20(\sqrt{3} + 1) = 20(3 + \sqrt{3}) = 60 + 20\sqrt{3} \]

### Final Answer:
The height of the tower is \( h = 60 + 20\sqrt{3} \) meters.