Question

from his eye, which stands 1.52 meters above the ground, wyatt measures the angle of elevation to the top of a prominent skyscraper to be 28°. if he is standing at a horizontal distance of 331 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 $h_1 = 1.52$ m (eye height), $d = 331$ m (horizontal distance), $\theta = 28^\circ$ (angle of elevation). Let $h_2$ = vertical height from eye to skyscraper top.

Step2: Use tangent to find $h_2$

$\tan\theta = \frac{h_2}{d}$
$h_2 = d \times \tan(\theta)$
$h_2 = 331 \times \tan(28^\circ)$
Calculate $\tan(28^\circ) \approx 0.5317$, so $h_2 \approx 331 \times 0.5317 \approx 175.99$ m

Step3: Total skyscraper height

Add eye height to $h_2$:
$H = h_1 + h_2$
$H \approx 1.52 + 175.99 = 177.51$ m

Step4: Round to nearest tenth

$H \approx 177.5$ m

Answer:

177.5 meters