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)

Question

Responses

x=199.79 m and y=157.43 m
x equals 199.79 m and y equals 157.43 m

x=157.43 m and y=98.26 m
x equals 157.43 m and y equals m

x=96.1 m and y=156.09 m
x equals 96.1 m and y equals 156.09 m

x=157.43 m and y=199.79 m

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 some trigonometric relationships.

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

Using the tangent of the angle of elevation:

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

Here,

• \(\theta = 52°\) (the angle of elevation)
• \(d = 123\) meters (the distance from Xavier to the building)

We can rearrange the equation to solve for \(x\):

\[ x = d \cdot \tan(\theta) \]

Substituting the values:

\[ x = 123 \cdot \tan(52°) \]

Calculating \(\tan(52°)\):

• \(\tan(52°) \approx 1.2799\) (using a calculator)

Now substituting that back:

\[ x \approx 123 \cdot 1.2799 \approx 157.43 \text{ meters} \]

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

To find the distance \(y\) from Xavier to the top of the building, we can use the sine function:

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

Rearranging the equation:

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

Now substituting \(x\) and \(\sin(52°)\) (using a calculator \(\sin(52°) \approx 0.7880\)):

\[ y = \frac{157.43}{\sin(52°)} \approx \frac{157.43}{0.7880} \approx 199.79 \text{ meters} \]

Summary of Results:

• Height above eye level, \(x \approx 157.43\) m
• Distance to the top of the building, \(y \approx 199.79\) m

Thus, the correct answer is:

x=157.43 m and y=199.79 m.