  A man is 250m away from the base of the pole of height 'h' & making an angle of elevation Ø, after travelling 150m towards the pole the angle of elevation becomes 2Ø. Find the height of the pole.

To solve the problem, we will use trigonometry. Let the height of the pole be \( h \) meters, and the angle of elevation be \( \theta \) (given as \( \theta \) and \( 2\theta \) at two different distances).

### Step-by-Step Solution:

1. **Initial condition**: When the man is 250 meters away from the base of the pole, the angle of elevation is \( \theta \).
   - Using the tangent function:
   \[
   \tan(\theta) = \frac{h}{250}
   \]
   Therefore, we can write the equation:
   \[
   h = 250 \cdot \tan(\theta)
   \tag{1}
   \]

2. **After travelling 150 meters**: The man is now \( 250 - 150 = 100 \) meters away from the pole, and the angle of elevation becomes \( 2\theta \).
   - Using the tangent function again:
   \[
   \tan(2\theta) = \frac{h}{100}
   \]
   Therefore, we can write:
   \[
   h = 100 \cdot \tan(2\theta)
   \tag{2}
   \]

3. **Relationship between \( \tan(2\theta) \) and \( \tan(\theta) \)**: 
   Using the double-angle identity for tangent:
   \[
   \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}
   \]

4. **Substitute this into equation (2)**:
   \[
   h = 100 \cdot \frac{2\tan(\theta)}{1 - \tan^2(\theta)}
   \tag{3}
   \]

5. **Equating equations (1) and (3)**:
   Since both expressions represent \( h \), we set them equal to each other:
   \[
   250 \cdot \tan(\theta) = 100 \cdot \frac{2\tan(\theta)}{1 - \tan^2(\theta)}
   \]

6. **Simplify the equation**:
   Cancel out \( \tan(\theta) \) (assuming \( \tan(\theta) \neq 0 \)):
   \[
   250 = 100 \cdot \frac{2}{1 - \tan^2(\theta)}
   \]

7. **Solve for \( \tan^2(\theta) \)**:
   \[
   \frac{250}{100} = \frac{2}{1 - \tan^2(\theta)}
   \]
   \[
   2.5 = \frac{2}{1 - \tan^2(\theta)}
   \]
   Multiply both sides by \( 1 - \tan^2(\theta) \):
   \[
   2.5(1 - \tan^2(\theta)) = 2
   \]
   Expand and solve for \( \tan^2(\theta) \):
   \[
   2.5 - 2.5\tan^2(\theta) = 2
   \]
   \[
   2.5 - 2 = 2.5\tan^2(\theta)
   \]
   \[
   0.5 = 2.5\tan^2(\theta)
   \]
   \[
   \tan^2(\theta) = \frac{0.5}{2.5} = 0.2
   \]
   \[
   \tan(\theta) = \sqrt{0.2}
   \]
   \[
   \tan(\theta) = \frac{\sqrt{2}}{5}
   \]

8. **Find the height \( h \)**:
   Now, substitute \( \tan(\theta) = \frac{\sqrt{2}}{5} \) into equation (1):
   \[
   h = 250 \cdot \frac{\sqrt{2}}{5} = 50\sqrt{2} \, \text{meters}
   \]

### Final Answer:
The height of the pole is \( 50\sqrt{2} \approx 70.71 \) meters.

Would you like more details or have any questions?

Here are five related questions you could explore:
1. What if the man starts 300 meters away from the pole?
2. How would the height change if the angle of elevation doubles again after travelling 100 meters more?
3. How is the double-angle formula for tangent derived?
4. What is the significance of the tangent function in trigonometry?
5. How does the angle of elevation affect the height of objects in real-life scenarios?

**Tip**: Always verify the conditions for using trigonometric identities, especially the restrictions on angles and domains.

A man is 250m away from the base of the pole of height 'h' & making an angle of elevation Ø, after travelling 150m towards the pole the angle of elevation becomes 2Ø. Find the height of the pole.

Find the height of a pole by solving a real-world problem using trigonometric principles. Given the initial and changed angles of elevation, we apply tangent functions and the double-angle identity. Suitable for students in Grades 10-12.

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