Question

from his eye, which stands 1.67 meters above the ground, jeriel measures the angle of elevation to the top of a prominent skyscraper to be 61°. if he is standing at a horizontal distance of 258 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: Identify the trigonometric relationship

We have a right triangle where the horizontal distance (adjacent side) is 258 meters, the angle of elevation is \(61^\circ\), and we need to find the vertical height from the eye level to the top of the skyscraper (opposite side). We use the tangent function: \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). So \(\tan(61^\circ)=\frac{h}{258}\), where \(h\) is the height from eye level to the top.

Step2: Calculate the height from eye level

Solve for \(h\): \(h = 258\times\tan(61^\circ)\). Using a calculator, \(\tan(61^\circ)\approx1.8040\), so \(h\approx258\times1.8040 = 465.432\) meters.

Step3: Add the eye - level height to the ground

The total height of the skyscraper is the height from eye level to the top plus the height of the eye above the ground. So total height \(H=465.432 + 1.67=467.102\) meters. Rounding to the nearest tenth, we get \(467.1\) meters.

Answer:

The height of the skyscraper is approximately \(\boldsymbol{467.1}\) meters.