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## What is the height of the building?

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?

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
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.