If the angle of elevation of the top of a vertical 25 m high building is 46.333o, how far is it to the building?

We can solve this problem using **trigonometry**. Specifically, the **tangent function** is suitable since we are dealing with a right triangle formed by the building, the ground, and the observer’s line of sight to the top of the building.

The tangent function is given by:  
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Where:  
- \(\theta = 46.333^\circ\) (angle of elevation)  
- Opposite side = 25 m (the height of the building)  
- Adjacent side = the distance to the building (let's call it \(d\))

### Step-by-Step Solution

\[
\tan(46.333^\circ) = \frac{25}{d}
\]

Rearranging for \(d\):

\[
d = \frac{25}{\tan(46.333^\circ)}
\]

Using a calculator:

\[
\tan(46.333^\circ) \approx 1.043
\]

\[
d = \frac{25}{1.043} \approx 23.97 \, \text{m}
\]

### Final Answer
The distance to the building is approximately **23.97 meters**.

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Do you want more details or have any further questions?

Here are five related questions to explore:
1. How would the answer change if the building height were 30 m?
2. What is the horizontal distance if the observer is standing on a 5 m platform?
3. What if the angle of elevation is reduced to 30°? How does the distance change?
4. How would you calculate the angle of elevation if the distance is known?
5. How can sine and cosine functions help in different elevation scenarios?

**Tip:** Always ensure your calculator is in **degree mode** when working with angles in degrees.

If the angle of elevation of the top of a vertical 25 m high building is 46.333°, how far is it to the building?

Learn how to calculate the distance to a building using an angle of elevation of 46.333° and the building's height of 25m. This step-by-step trigonometry solution involves applying the tangent function and is suitable for students in Grades 9-10.

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