Question

from his eye, which stands 1.55 meters above the ground, jaxon measures the angle of elevation to the top of a prominent skyscraper to be 67°. if he is standing at a horizontal distance of 266 meters from the base of the skyscraper, what is the height of the skyscraper? round your answer to the nearest tenth of a meter if necessary.

Explanation:

Step1: Find the height from eye - level to the top of the skyscraper

We know that in a right - triangle (formed by the line of sight, the horizontal distance, and the vertical distance from eye - level to the top of the skyscraper), the tangent of an angle of elevation $\theta$ is given by $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.
Here, $\theta = 67^{\circ}$ and the adjacent side $a = 266$ meters. Let the opposite side (height from eye - level to the top of the skyscraper) be $x$.
Using the formula $\tan\theta=\frac{x}{266}$, and since $\tan67^{\circ}\approx2.3559$, we have $x = 266\times\tan67^{\circ}$.
$x=266\times2.3559 = 626.6694$ meters.

Step2: Find the total height of the skyscraper

The total height of the skyscraper $H$ is the sum of the height from the ground to the eye ($h_1 = 1.55$ meters) and the height from the eye - level to the top of the skyscraper ($x$).
$H=x + 1.55$.
Substitute $x = 626.6694$ into the formula: $H=626.6694+1.55=628.2194\approx628.2$ meters.

Answer:

$628.2$ meters