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From her eye, which stands 1.66 metersabove the ground, Deondra measures theangle of elevation to the top of aprominent skyscraper to be 70 degrees.If she is standing at a horizontal distanceof 276 meters from the base of theskyscraper, what is the height of theskyscraper? Round your answer to thenearest tenth of a meter if necessary.

From her eye, which stands 1.66 metersabove the ground, Deondra measures theangle-example-1
User Ruario
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Given:

Angle of elevation = 70 degrees.

Distance from her eyes to the ground = 1.66 meters

Horizontal distance from the base = 276 meters.

Let's find the height of the skyscraper.

Let's first sketch a figure which represents this situation:

To find the height of the skyscraper, apply the trigonometric ratio formula for tangent:


tan\theta=\frac{\text{ opposite}}{adjacent}

Where:

θ is the angle of elevation = 70 degrees

opposite side is the side opposite the angle = h

Adjacent side is the side adjacent the angle = 276 m

Thus, we have:


\begin{gathered} tan70=(h)/(276) \\ \\ h=276tan70 \\ \\ h=758.3 \end{gathered}

Now, the total height of the skyscraper will be:

758.3 + 1.66 = 759.96 ≈ 760 m

Therefore, the height of the skyscraper is 760 meters.

ANSWER:

From her eye, which stands 1.66 metersabove the ground, Deondra measures theangle-example-1
User Ahmad Shahwan
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