What is the height of the building?

Question

Jay is 120 meters from the base of a building and Mary is 40 meters from the base of the building. Jay's angle of elevation when looking at the top of the building is X and Mary's is 2X. What is the height of the building?

Answer 1

tangent=opposite/adjacent, H=height of building

tan(x)=H/120

tan(2x)=H/40

120tan(x)=H

40tan(2x)=H

120tan(x)=40tan(2x)

3tan(x)=tan(2x)

tan(2x)=3tan(x)

tan(2x)=2tan(x)/[1-tan^2(x)]

2tan(x)/[1-tan^2(x)]=3tan(x)

2tan(x)=3tan(x)[1-tan^2(x)]

2=3[1-tan^2(x)]

2/3=1-tan^2(x)

-1/3=-tan^2(x)

1/3=tan^2(x)

find the square root of 1/3

square root of 1/3=0.57735

tan of what angle is 0.57735 from the table of tangents ?

the angle x=30 degrees

tan(30)=H/120

120*0.57735=69.28 meters

2x=60 degrees

tan(60)=H/40

tan(60)=1.7321

40*1.7321=69.28 meters

Answer 2

Use the definition of tangent (opposite/adjacent):

 

From Jay's point of view, tan (X) = h/120

From Mary's point of view, tan (2X) = h/40

 

Divide the two equations:

 

tan(2X)/tan(X) = (h/40)/(h/120) = 3

tan(2X) = 3 tan(X)

 

Use the double-angle formula

 

tan(2X) = 2 tan(X) /(1 - tan²(X) )

 

Get

 

2 tan(X) /(1 - tan²(X) ) = 3 tan(X)

 

2  = 3 (1 - tan²(X) )

 

tan²(X) = 1/3

 

X= 30°

 

Therefore,

 

h=120 tan (30°) ≈ 69 meters

 

is the height.