Xavier is looking up at a building that is 123 meters away from him, at an angle of elevation of 52What is the building's height above Xavier's eye level , x, and the distance between Xavier and the top of the building (point ) x = 199.79m and y = 157.43m; x = 96.1m and y = 156.09m; x = 157.43m and y = 98.26m; x = 157.43m and y = 199.79m
In this problem, we have a right triangle formed with Xavier's eye level, the top of the building, and a point on the ground directly below Xavier. The distance between Xavier's eye level and the top of the building is the height we are looking for, and the distance between Xavier and the top of the building is the hypotenuse of the triangle.
Using trigonometry, we can determine the height of the building.
Let's consider the first set of coordinates: x = 199.79m and y = 157.43m.
The height (h) of the building can be calculated using the tangent function:
tan(angle) = h / x
tan(52) = h / 199.79
h = 199.79 * tan(52)
h ≈ 199.79 * 1.2826
h ≈ 256.39m
Therefore, the height of the building above Xavier's eye level is approximately 256.39 meters.
To find the distance between Xavier and the top of the building (point P), we can use the Pythagorean theorem.
Using the second set of coordinates: x = 96.1m and y = 156.09m.
We can let the distance between Xavier and point P be d.
Using the Pythagorean theorem:
d^2 = x^2 + y^2
d^2 = 96.1^2 + 156.09^2
d^2 ≈ 9243.21 + 24329.2481
d^2 ≈ 33572.4581
d ≈ √(33572.4581)
d ≈ 183.25m
Therefore, the distance between Xavier and the top of the building (point P) is approximately 183.25 meters.
This process can be repeated for the other sets of coordinates to find the respective results.