Question

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

x=96.1 m and y=156.09 m
x=157.43 m and y= 199.79 m
x=157.43 m and y= 98.26 m
x=199.79 m and y= 157.43 m

Answer 1

To find the height of the building above Xavier's eye level (x) and the distance from Xavier to the top of the building (y), we can use trigonometry.

1. **Height Above Eye Level (x):**
We can calculate the height (h) of the building above Xavier's eye level using the tangent of the angle of elevation:

\[
\tan(\theta) = \frac{h}{\text{distance}}
\]

Given that the distance is 123 meters and the angle of elevation (θ) is 52 degrees:

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

Calculating:

\[
h \approx 123 \cdot 1.2799 \approx 157.43 \text{ m}
\]

So, the height of the building above Xavier's eye level, x, is approximately 157.43 m.

2. **Distance to the Top of the Building (y):**
To find the distance from Xavier to the top of the building (y), we can use the cosine function in a right triangle where:

\[
y = \frac{\text{distance}}{\cos(\theta)}
\]

Calculating:

\[
y = \frac{123}{\cos(52^\circ)}
\]

Since \(\cos(52^\circ) \approx 0.6157\),

\[
y \approx \frac{123}{0.6157} \approx 199.79 \text{ m}
\]

Putting both results together:

- Height above eye level (x) = 157.43 m
- Distance to the top of the building (y) = 199.79 m

Thus, the correct answer is **x = 157.43 m and y = 199.79 m**.