Question

question
from his eye, which stands 1.64 meters above the ground, keshawn measures the angle of elevation to the top of a prominent skyscraper to be 33°. if he is standing at a horizontal distance of 385 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: Define known values

Let $h_1$ = height from eye to ground = $1.64$ m, $\theta = 33^\circ$, horizontal distance $d = 385$ m. Let $h_2$ = height from eye level to top of skyscraper.

Step2: Use tangent function for $h_2$

$\tan(\theta) = \frac{h_2}{d}$
Rearrange to solve for $h_2$:
$h_2 = d \times \tan(\theta)$
Substitute values:
$h_2 = 385 \times \tan(33^\circ)$
Calculate $\tan(33^\circ) \approx 0.6494$, so:
$h_2 \approx 385 \times 0.6494 \approx 250.019$ m

Step3: Total height of skyscraper

Total height $H = h_1 + h_2$
$H \approx 1.64 + 250.019$

Answer:

$251.7$ meters (rounded to the nearest tenth)