Question

Xavier is looking up at a building that is 123 meters away from him, at an angle of elevation of $52^{\circ}$. What is the building’s height above Xavier’s eye level, $x$, and the distance between Xavier and the top of the building, $y$?Answers$x = 199.79\text{ m and } y = 157.43\text{ m}$$x = 96.1\text{ m and } y = 156.09\text{ m}$$x = 157.43\text{ m and } y = 199.79\text{ m}$$x = 157.43\text{ m and } y = 98.26\text{ m}$

Answer 1

Let the horizontal distance be 123 m and the angle of elevation be 52°. Then

x = 123 · tan(52°) ≈ 123·1.2802 ≈ 157.43 m,
y = 123 / cos(52°) ≈ 123/0.6157 ≈ 199.79 m.

So x = 157.43 m and y = 199.79 m.