The shadow of a pole standing on a level ground is found to be 45 m longer when the sun's altitude is `30^@` then when it was `60^@` Determine the height of the pole.

Explanation

To determine the height of the pole, we can use the concept of trigonometry, specifically the tangent function, which relates the height of the pole to the length of the shadow based on the angle of elevation of the sun.

### Step-by-Step Solution:

1. **Define Variables:**
Let the height of the pole be \( h \) meters.
Let the length of the shadow when the sun's altitude is \( 60^\circ \) be \( x \) meters.
Therefore, when the sun's altitude is \( 30^\circ \), the length of the shadow will be \( x + 45 \) meters.

2. **Use Tangent Function:**
From trigonometry, we know that:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
For \( 60^\circ \):
\[ \tan(60^\circ) = \frac{h}{x} \]
For \( 30^\circ \):
\[ \tan(30^\circ) = \frac{h}{x + 45} \]

3. **Substitute Values of Tangent:**
We know:
\[ \tan(60^\circ) = \sqrt{3} \quad \text{and} \quad \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Therefore, we can write the equations as:
\[ \sqrt{3} = \frac{h}{x} \quad \text{(1)} \]
\[ \frac{1}{\sqrt{3}} = \frac{h}{x + 45} \quad \text{(2)} \]

4. **Express \( h \) in terms of \( x \):**
From equation (1):
\[ h = \sqrt{3} x \quad \text{(3)} \]
From equation (2):
\[ h = \frac{1}{\sqrt{3}}(x + 45) \quad \text{(4)} \]

5. **Set Equations (3) and (4) Equal:**
Now, we can set the two expressions for \( h \) equal to each other:
\[ \sqrt{3} x = \frac{1}{\sqrt{3}}(x + 45) \]

6. **Clear the Fraction:**
Multiply both sides by \( \sqrt{3} \) to eliminate the fraction:
\[ 3x = x + 45 \]

7. **Solve for \( x \):**
Rearranging gives:
\[ 3x - x = 45 \]
\[ 2x = 45 \]
\[ x = 22.5 \text{ m} \]

8. **Find the Height \( h \):**
Substitute \( x \) back into equation (3) to find \( h \):
\[ h = \sqrt{3} \times 22.5 \]
\[ h = 22.5 \sqrt{3} \approx 39.0 \text{ m} \quad (\text{using } \sqrt{3} \approx 1.732) \]

### Final Answer:
The height of the pole is approximately \( 39.0 \) meters.