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?

Answer 1

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