Question

question
from her eye, which stands 1.68 meters above the ground, hannah measures the angle
of elevation to the top of a prominent skyscraper to be 31°. if she is standing at a
horizontal distance of 194 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.
answer attempt 1 out of 2

Explanation:

Step1: Use the tangent function

We know that \(\tan\theta=\frac{opposite}{adjacent}\). Here, \(\theta = 31^{\circ}\), and the adjacent side (horizontal distance) \(x = 194\) meters. Let \(h\) be the height from the eye - level to the top of the skyscraper. Then \(\tan(31^{\circ})=\frac{h}{194}\).
So, \(h = 194\times\tan(31^{\circ})\).
Since \(\tan(31^{\circ})\approx0.6009\), then \(h=194\times0.6009 = 194\times\frac{6009}{10000}=194\times0.6009 = 116.5746\) meters.

Step2: Add the eye - level height

The total height \(H\) of the skyscraper is the sum of the height from the eye - level to the top of the skyscraper (\(h\)) and the eye - level height (\(1.68\) meters).
\(H=h + 1.68\)
Substitute \(h = 116.5746\) into the formula: \(H=116.5746+1.68=118.2546\approx118.25\) meters

Answer:

\(118.25\) meters