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: Identify tangent relationship

$\tan(45^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{271}$, where $h$ is height above eye level.

Step2: Calculate height above eye level

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

Step3: Total skyscraper height

Add eye height: $271 + 1.54 = 272.54$ meters.

Answer:

272.54 meters