Question

from his eye, which stands 1.54 meters above the ground, brody measures the angle of elevation to the top of a prominent skyscraper to be $45^\circ$. if he is standing at a horizontal distance of 271 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 unknown height segment

Let $h$ = height from eye level to skyscraper top.

Step2: Use tangent of elevation angle

$\tan(45^\circ) = \frac{h}{271}$
Since $\tan(45^\circ)=1$, $h = 271 \times 1 = 271$ meters.

Step3: Total skyscraper height

Add eye height to $h$:
$\text{Total Height} = 271 + 1.54$

Answer:

272.54 meters