Question

from her eye, which stands 1.66 meters above the ground, deondra measures the angle of elevation to the top of a prominent skyscraper to be 70°. if she is standing at a horizontal distance of 276 meters from the base of the skyscraper, what is the height of the skyscraper? round your answer to the nearest tenth of a meter if necessary.

Explanation:

Step1: Find the height from eye level to the top of the skyscraper

We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 70^{\circ}\) and the adjacent side (horizontal distance) \(x = 276\) meters. Let \(h\) be the height from eye - level to the top of the skyscraper. Then \(\tan70^{\circ}=\frac{h}{276}\).
Since \(\tan70^{\circ}\approx2.747\), we have \(h = 276\times\tan70^{\circ}\).

$$h=276\times2.747 = 758.172$$

Step2: Find the total height of the skyscraper

The total height \(H\) of the skyscraper is the sum of the height from the ground to the eye - level (\(h_0=1.66\) meters) and \(h\).
\(H=h + 1.66\)
Substitute \(h = 758.172\) into the formula: \(H=758.172+1.66=759.832\)

Answer:

\(759.8\) meters