Question

from her eye, which stands 1.75 meters above the ground, myesha measures the angle of elevation to the top of a prominent skyscraper to be 19°. if she is standing at a horizontal distance of 337 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: Find the height from eye - level to the top of the skyscraper

We know that in a right - triangle (formed by the line of sight, the horizontal distance, and the vertical distance from eye - level to the top of the skyscraper), the tangent of the angle of elevation $\theta$ is given by $\tan\theta=\frac{opposite}{adjacent}$.
Here, $\theta = 19^{\circ}$ and the adjacent side (horizontal distance) $x = 337$ meters. Let $h$ be the height from eye - level to the top of the skyscraper.
So, $\tan(19^{\circ})=\frac{h}{337}$.
Then $h = 337\times\tan(19^{\circ})$.
Since $\tan(19^{\circ})\approx0.3443$, we have $h = 337\times0.3443=337\times\frac{3443}{10000}=\frac{337\times3443}{10000}$.
$337\times3443=(300 + 30+7)\times3443=300\times3443+30\times3443 + 7\times3443=1032900+103290+24101=1160291$.
So $h=\frac{1160291}{10000}=116.0291$ meters.

Step2: Find the total height of the skyscraper

The total height of the skyscraper $H$ is the sum of the height from the ground to the eye - level ($h_0 = 1.75$ meters) and the height from eye - level to the top of the skyscraper ($h$).
$H=h + 1.75$.
Substitute $h = 116.0291$ into the formula: $H=116.0291+1.75=117.7791\approx117.78$ meters.

Answer:

$117.78$ meters