xavier is looking up at a building that is 123 meters away from him, at an angle of elevaton of 52 degrees. what is the buildings height about xaviers eye level,x, and the distance between xavier and the top of the building, y
To determine the height of the building above Xavier's eye level (x), we can use the trigonometric relationship of tangent.
Let h be the height of the building.
tan(52 degrees) = h / 123 meters
Solving for h:
h = tan(52 degrees) * 123 meters
Using a scientific calculator, we find:
h ≈ 141.71 meters
Therefore, the height of the building above Xavier's eye level is approximately 141.71 meters.
To determine the distance between Xavier and the top of the building (y), we can use the trigonometric relationship of cosine.
cos(52 degrees) = y / 123 meters
Solving for y:
y = cos(52 degrees) * 123 meters
Using a scientific calculator, we find:
y ≈ 76.57 meters
Therefore, the distance between Xavier and the top of the building is approximately 76.57 meters.