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 above Xavier’s eye level, x, and the distance between Xavier and the top of the building, y? (1 point)

Answer 1

To find 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 trigonometry.

### Given:
- Distance from Xavier to the building (base): \( d = 123 \) meters
- Angle of elevation: \( \theta = 52^\circ \)

### Step 1: Calculate the height of the building above Xavier's eye level (x)

Using the tangent function, which relates the angle of elevation to the opposite side (height of the building) and the adjacent side (distance to the building):

\[
\tan(\theta) = \frac{x}{d}
\]

Substituting the values we have:

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

Now, solving for \( x \):

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

Using a calculator to find \( \tan(52^\circ) \):

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

So we have:

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

### Step 2: Calculate the distance between Xavier and the top of the building (y)

To find \( y \), we can use the sine function. The hypotenuse in this case is the distance from Xavier to the top of the building:

\[
\sin(\theta) = \frac{x}{y}
\]

Rearranging this gives:

\[
y = \frac{x}{\sin(\theta)}
\]

Now substituting in the value of \( x \) and \( \sin(52^\circ) \):

Calculating \( \sin(52^\circ) \):

\[
\sin(52^\circ) \approx 0.7880
\]

Now, substituting the values:

\[
y \approx \frac{157.4}{0.7880} \approx 199.0 \text{ meters}
\]

### Final Results:
- The height of the building above Xavier's eye level, \( x \approx 157.4 \) meters.
- The distance between Xavier and the top of the building, \( y \approx 199.0 \) meters.