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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?

 Feb 19, 2020
 #1
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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

 

 

 

cool cool cool

 Feb 19, 2020
edited by CPhill  Feb 19, 2020
edited by CPhill  Feb 19, 2020

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