Question

Xavier is looking up at a building that is 123 meters away from him, at an angle of elevation of 52°
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)
Responses

x=157.43 m
𝑥
=
157.43
m
and y=199.79 m
𝑦
=
199.79
m

x equals 157.43 m and y equals 199.79 m

x=96.1 m
𝑥
=
96.1
m
and y=156.09 m
𝑦
=
156.09
m

x equals 96.1 m and y equals 156.09 m

x=157.43 m
𝑥
=
157.43
m
and y=98.26 m
𝑦
=
98.26

m

x equals 157.43 m and y is equal to 98 point 2 6 m

x=199.79 m
𝑥
=
199.79
m
and y=157.43 m

Answer 1

Let the horizontal distance be the adjacent side (123 m) and the angle of elevation θ = 52°.

x = 123·tan(52°) ≈ 123·1.2794 = 157.43 m
y = 123·sec(52°) = 123 / cos(52°) ≈ 123 / 0.6157 = 199.79 m

So x = 157.43 m and y = 199.79 m.