Question

Xavier is looking up at a building that is 123 meters away from him, at an angle of elevation of 52 degrees. What is the buildings height above Xavier’s eye level, x, and the distance between Xavier and the top of the building, y?

Answer 1

To solve for the height of the building above Xavier's eye level (x) and the distance between Xavier and the top of the building (y), we can use some trigonometry.

1. **Finding the height of the building above Xavier's eye level (x):**

Given:
- The distance from Xavier to the building (adjacent side, a) = 123 meters
- The angle of elevation (θ) = 52 degrees

We can use the tangent function, which relates the angle to the opposite and adjacent sides:

\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]

Here, the opposite side represents the height above eye level (x), so we can write:

\[
\tan(52^\circ) = \frac{x}{123}
\]

To isolate x, we rearrange the equation:

\[
x = 123 \cdot \tan(52^\circ)
\]

Now, we compute \( \tan(52^\circ) \):

Using a calculator:

\[
\tan(52^\circ) \approx 1.2799
\]

Now plug this value in:

\[
x = 123 \cdot 1.2799 \approx 157.4 \text{ meters}
\]

So, the height of the building above Xavier's eye level (x) is approximately **157.4 meters**.

2. **Finding the distance between Xavier and the top of the building (y):**

To find y, we can use the Pythagorean theorem since we now have one side (adjacent) and the opposite side (height).

\[
y^2 = 123^2 + x^2
\]

Substituting x:

\[
y^2 = 123^2 + 157.4^2
\]

Calculating:

\[
123^2 = 15129
\]
\[
157.4^2 \approx 24855.76
\]

Now add them:

\[
y^2 = 15129 + 24855.76 \approx 39984.76
\]

Taking the square root of both sides:

\[
y \approx \sqrt{39984.76} \approx 199.96 \text{ meters}
\]

Thus, the distance between Xavier and the top of the building (y) is approximately **200.0 meters**.

In summary:
- Height above eye level (x): **157.4 meters**
- Distance to the top of the building (y): **200.0 meters**