A surveyor is standing 25 feet from a building and is looking at the top with an angle of elevation of 65 degrees. How tall was the building?

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A surveyor is standing 25 feet from a building and is looking at the top with an angle of elevation of 65 degrees. How tall was the building?

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Answer 1 · 2016-12-04T00:15:04

The height of the building is $53.6$ $53.6$ feet.

Explanation:

my drawing

Please excuse the crude drawing.

Let $x =$ $x=$ the height of the building.

The base of the triangle is labeled 25 ft, and represents the distance from the building to the surveyor.

The surveyor is standing at the lower right vertex of the triangle.

The angle of elevation is represented by the angle labeled $65$ $65$ degrees. An angle of elevation is the angle between the observer (the surveyor) and the line of sight to the object being observed (the top of the building).

To find the height of the building $x$ $x$ , use the tangent of the $65$ $65$ degree angle. In a right triangle, the tangent of an angle $θ$ $theta$ is the ratio between the side opposite the angle ( $x$ $x$ ) and adjacent to the angle ( $25$ $25$ ).

$tan θ = \frac{opposite}{adjacent}$ $tantheta=frac{"opposite"}{"adjacent"]$

$tan 65 = \frac{x}{25}$ $tan 65=x/25$

Multiply both sides by $25$ $25$

$25 \cdot tan 65 = \frac{x}{25} \cdot 25$ $25* tan 65 = x/25 * 25$

$25* tan 65 = x/cancel25 *cancel25$

$25 * tan65=xcolor(white)(aaa)$ Use a calculator

$53.6=x$

The height of the building is $53.6$ feet.

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A surveyor is standing 25 feet from a building and is looking at the top with an angle of elevation of 65 degrees. How tall was the building?

1 Answer

Explanation:

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