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

Answer 1

To find the height of the building above Xavier's eye level (x), we can use trigonometry. The tangent function relates the angle of elevation to the opposite (x) and adjacent (123 meters) sides of the triangle formed.

We can use the formula: tangent(angle of elevation) = opposite / adjacent

Tangent(52°) = x / 123 meters

To find x, we can rearrange the equation:

x = tangent(52°) * 123 meters

Using a calculator, we can find:

x ≈ 101.15 meters

So, the building's height above Xavier's eye level is approximately 101.15 meters.

To find the distance between Xavier and the top of the building (y), we can use trigonometry again. The sine function relates the angle of elevation to the opposite (x) and hypotenuse (y) sides of the triangle formed.

We can use the formula: sine(angle of elevation) = opposite / hypotenuse

Sine(52°) = x / y

To find y, we can rearrange the equation:

y = x / sine(52°)

Using a calculator, we can find:

y ≈ 130.66 meters

So, the distance between Xavier and the top of the building is approximately 130.66 meters.