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?
+128448 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
