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)

Answer 1

To find the height of the building above Xavier's eye level, x, we can use the trigonometric relationship of tangent.

Tangent(angle) = opposite/adjacent

Tangent(52°) = x/123m

x = tangent(52°) * 123m ≈ 97.71m

Therefore, the building's height above Xavier's eye level is approximately 97.71 meters.

To find the distance between Xavier and the top of the building, y, we can use the trigonometric relationship of cosine.

Cosine(angle) = adjacent/hypotenuse

Cosine(52°) = y/123m

y = cosine(52°) * 123m ≈ 75.32m

Therefore, the distance between Xavier and the top of the building is approximately 75.32 meters.