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This problem involves trigonometry, specifically dealing with the tangent of the angle of elevation.

### Problem Summary:
- Neil is standing 250 m from a building.
- The angle of elevation to the top of the building is 31°.
- We are asked to find the height difference \( h \) between the top of the building and Neil's eye level.

### Solution:
We can use the formula for the tangent of the angle of elevation:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Where:
- \( \theta = 31^\circ \) (the angle of elevation),
- \( \text{opposite} = h \) (the height difference we want to find),
- \( \text{adjacent} = 250 \, \text{m} \) (the distance Neil is from the building).

Thus, the equation becomes:
\[
\tan(31^\circ) = \frac{h}{250}
\]
To solve for \( h \):
\[
h = 250 \times \tan(31^\circ)
\]

I will calculate the value of \( h \).The height difference between the top of the building and Neil's eye level is approximately \( 150.22 \, \text{m} \), correct to two decimal places.

Would you like more details or have any questions? Here are some related questions to expand on this problem:

1. How would the height difference change if Neil stood further from the building?
2. What would happen to the height difference if the angle of elevation was smaller?
3. How would you calculate the distance if the height and angle of elevation were given?
4. Can you explain how the tangent function relates to right-angle triangles?
5. What is the significance of using radians vs degrees in trigonometric calculations?

**Tip:** Always ensure your calculator is set to the correct mode (degrees or radians) based on the angle units used in the problem.

Neil is standing 250 m from a building and measures the angle of elevation of the top of the building to be 31°. If the difference in height between the top of the building and Neil’s eye-level is h, find h correct to two decimal places.

Calculate the height difference between the top of a building and Neil’s eye-level using trigonometric principles. Given the angle of elevation (31°) and distance (250 m), this problem demonstrates the use of the tangent function. Suitable for students in Grades 10-12.

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