A teenager whose eyes are five feet above graund level is looking into a mirror on the ground and can see the top of the building that is 30 feet away from the teenager. The angle of elevation from the center of the mirror to the top of the building is 70º. How tall is the building? How far away from the teenager's feet is the mirror?

Guest Feb 19, 2020

#1**+1 **

I believe that this follows something known as "Snell's Law"

I'm assuming that the building is 30ft from the teenager

Let the distance that the teenager is from the mirror = D

Let the distance that the mirror is from the mirror =30 - D

Let the building height = H

The angle of incidence = the angle of reflection....this means that

tan (70) = 5 / D (1) and tan (70) = H / (30 - D) (2)

Rearranging (1) we get that D = 5 /tan (70)

Sub this into (2) and we have that

tan (70) = H / [ 30 - 5/tan(70) ]

[ 30 -5 / tan (70) ] * tan (70) = H ≈ 77.4 ft =building height

Using similar triangles

D / 5 = (30 - D) / 77.4 cross-multiply

77.4D = 5 (30 - D)

77.4 D =150 - 5D

82.4D =150

D = 150 / 82.4 ≈ 1.82 ft = distance teenager is from the mitrror

CPhill Feb 19, 2020