The height of a tower is 50 meter. When the Sun's altitude increases from `30^(@)` to `45^(@)`, the length of the shadow of the tower is decreased by x meter. Find the approximate value of x in meter.

Answer

To solve the problem, we need to find the decrease in the length of the shadow of a tower as the sun's altitude changes from 30 degrees to 45 degrees. The height of the tower is given as 50 meters.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a tower of height \( AB = 50 \) meters.
- When the sun's altitude is \( 30^\circ \), the shadow length is \( BC \).
- When the sun's altitude increases to \( 45^\circ \), the shadow length is \( BD \).
- We need to find the decrease in shadow length, which is \( x = BC - BD \).

2. **Finding the Length of the Shadow at \( 45^\circ \)**:
- In the triangle \( ABD \):
- Here, \( AB \) is the height of the tower, and \( BD \) is the length of the shadow.
- We use the tangent function:
\[ \tan(45^\circ) = \frac{AB}{BD} \]
- Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{50}{BD} \implies BD = 50 \text{ meters} \]

3. **Finding the Length of the Shadow at \( 30^\circ \)**:
- In the triangle \( ABC \):
- Again, we use the tangent function:
\[ \tan(30^\circ) = \frac{AB}{BC} \]
- Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{50}{BC} \implies BC = 50\sqrt{3} \text{ meters} \]
- Approximating \( \sqrt{3} \approx 1.732 \):
\[ BC \approx 50 \times 1.732 \approx 86.6 \text{ meters} \]

4. **Calculating the Decrease in Shadow Length**:
- Now, we can find \( x \):
\[ x = BC - BD = 86.6 - 50 = 36.6 \text{ meters} \]

5. **Final Answer**:
- The approximate value of \( x \) is \( 36.6 \) meters.