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 34°. 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), $\theta = 67^\circ$ (angle of elevation). Let $h_2$ be the vertical height from eye level to skyscraper top.

Step2: Calculate $h_2$ using tangent

$\tan\theta = \frac{h_2}{d}$
$h_2 = d \times \tan\theta = 294 \times \tan(67^\circ)$
Calculate $\tan(67^\circ) \approx 2.3559$, so $h_2 \approx 294 \times 2.3559 \approx 692.6346$ m

Step3: Total skyscraper height

Total height $H = h_1 + h_2$
$H \approx 1.63 + 692.6346$

Answer:

$694.26$ meters