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Xavier is looking up at a building that is 123 meters away from him, at an ang of elevation of 52° What is the building's heigh...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)
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=199.79 m
x equals 157.43 m and y equals 199.79 m
x=157.43 m and y=98.26 m
x equals 157.43 m and y equals m
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=199.79 m
x equals 157.43 m and y equals 199.79 m
x=157.43 m and y=98.26 m
x equals 157.43 m and y equals m
Answers
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Answered by
GPT-4o mini
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 trigonometry.
1. **Finding \( x \):**
We can use the tangent function, which is defined as the ratio of the opposite side (height of the building above eye level in this case) to the adjacent side (distance from Xavier to the building). The formula for tangent is given by:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here, \( \theta = 52^\circ \) and the distance from Xavier to the building (adjacent side) is 123 meters. Therefore, we can set up the equation:
\[
\tan(52^\circ) = \frac{x}{123}
\]
Rearranging this gives:
\[
x = 123 \cdot \tan(52^\circ)
\]
Calculating \( \tan(52^\circ) \):
\[
\tan(52^\circ) \approx 1.2799
\]
Then:
\[
x \approx 123 \cdot 1.2799 \approx 157.43 \text{ m}
\]
2. **Finding \( y \):**
To find the distance \( y \) from Xavier to the top of the building, we can use the cosine function and the hypotenuse. We can calculate the hypotenuse using the Pythagorean theorem:
\[
y = \frac{x}{\sin(52^\circ)}
\]
First, we can calculate the hypotenuse using:
\[
\sin(52^\circ) \approx 0.7880
\]
Now we can substitute \( x \) into the equation:
\[
y = \frac{157.43}{0.7880}
\]
This will give us:
\[
y \approx 199.79 \text{ m}
\]
In conclusion, the building's height above Xavier’s eye level (\( x \)) is approximately **157.43 m** and the distance to the top of the building (\( y \)) is approximately **199.79 m**. Thus, the correct response is:
**x = 157.43 m and y = 199.79 m**.
1. **Finding \( x \):**
We can use the tangent function, which is defined as the ratio of the opposite side (height of the building above eye level in this case) to the adjacent side (distance from Xavier to the building). The formula for tangent is given by:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here, \( \theta = 52^\circ \) and the distance from Xavier to the building (adjacent side) is 123 meters. Therefore, we can set up the equation:
\[
\tan(52^\circ) = \frac{x}{123}
\]
Rearranging this gives:
\[
x = 123 \cdot \tan(52^\circ)
\]
Calculating \( \tan(52^\circ) \):
\[
\tan(52^\circ) \approx 1.2799
\]
Then:
\[
x \approx 123 \cdot 1.2799 \approx 157.43 \text{ m}
\]
2. **Finding \( y \):**
To find the distance \( y \) from Xavier to the top of the building, we can use the cosine function and the hypotenuse. We can calculate the hypotenuse using the Pythagorean theorem:
\[
y = \frac{x}{\sin(52^\circ)}
\]
First, we can calculate the hypotenuse using:
\[
\sin(52^\circ) \approx 0.7880
\]
Now we can substitute \( x \) into the equation:
\[
y = \frac{157.43}{0.7880}
\]
This will give us:
\[
y \approx 199.79 \text{ m}
\]
In conclusion, the building's height above Xavier’s eye level (\( x \)) is approximately **157.43 m** and the distance to the top of the building (\( y \)) is approximately **199.79 m**. Thus, the correct response is:
**x = 157.43 m and y = 199.79 m**.
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