What is the height of a building when the distance between its angles of elevation which are 29 degrees and 37 degrees is 30 meters on level ground?
Using trigonometry its height works out as 63 meters to the
nearest meter.
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let:
h = height building
α, β be the angles of elevation (29° and 37° in some order)
d be the distance between the elevations (30 m).
x = distance from building where the elevation of angle α is
measured.
Then:
angle α is an exterior angle to the triangle which contains the
position from which angle α is measured, the position from which
angle β is measured and the point of the top of the building. Thus
angle α = angle β + angle at top of building of this triangle →
angle α > angle β as the angle at the top of the building is
> 0
→ α = 37°, β = 29°
Using the tangent trigonometric ratio we can form two equations,
one with angle α, one with angle β:
tan α = h/x → x = h/tan α
tan β = h/(x + d) → x = h/tan β - d
→ h/tan α = h/tan β - d
→ h/tan β - 1/tan α = d
→ h(1/tan β - 1/tan α) = d
→ h(tan α - tan β)/(tan α tan β) = d
→ h = (d tan α tan β)/(tan α - tan β)
We can now substitute the values of α, β and x in and find the
height:
h = (30 m × tan 37° × tan 29°)/(tan 37° - tan 29°) ≈ 63 m