Question

from her eye, which stands 1.69 meters above the ground, sadie measures the angle of elevation to the top of a prominent skyscraper to be $36^{\circ}$. if she is standing at a horizontal distance of 275 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: Define known values

Let $h_1$ = height from eye to top, $d=275$ m, $\theta=36^\circ$, eye height $h_2=1.69$ m.

Step2: Calculate $h_1$ via tangent

$\tan\theta = \frac{h_1}{d}$
$h_1 = d \times \tan(36^\circ)$
$h_1 = 275 \times \tan(36^\circ) \approx 275 \times 0.7265 = 200.7875$

Step3: Total height of skyscraper

Total height $H = h_1 + h_2$
$H = 200.7875 + 1.69$

Answer:

202.48 meters