Question

from his eye, which stands 1.62 meters above the ground, montraie measures the angle of elevation to the top of a prominent skyscraper to be 69°. if he is standing at a horizontal distance of 223 meters from the base of the skyscraper, what is the height of the skyscraper? round your answer to the nearest hundredth of a meter if necessary.

Explanation:

Step1: Use the tangent function

In a right - triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 69^{\circ}\), the adjacent side \(a = 223\) meters. Let the height from the eye - level to the top of the skyscraper be \(x\). Then \(\tan(69^{\circ})=\frac{x}{223}\).
So, \(x = 223\times\tan(69^{\circ})\).
We know that \(\tan(69^{\circ})\approx2.605089\).
Then \(x=223\times2.605089\).
\(x = 223\times2.605089=581.934847\) meters.

Step2: Calculate 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 (\(1.62\) meters) and \(x\).
\(h=x + 1.62\).
Substitute \(x = 581.934847\) into the formula: \(h=581.934847+1.62\).
\(h=583.554847\approx583.55\) meters.

Answer:

\(583.55\) meters