Question

from his eye, which stands 1.63 meters above the ground, isaac measures the angle of elevation to the top of a prominent skyscraper to be 12°. if he is standing at a horizontal distance of 294 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 $h_1 = 1.63$ m (eye height), $d = 294$ m (horizontal distance), $\theta = 17^\circ$ (angle of elevation).

Step2: Calculate vertical height to top

Use tangent function: $\tan\theta = \frac{h_2}{d}$
$h_2 = d \times \tan\theta = 294 \times \tan(17^\circ)$
$\tan(17^\circ) \approx 0.3057$, so $h_2 \approx 294 \times 0.3057 \approx 89.8758$ m

Step3: Total skyscraper height

Add eye height to vertical height: $H = h_1 + h_2$
$H \approx 1.63 + 89.8758$

Answer:

$91.51$ meters