Question

taub is trying to find the height of a radio antenna on the roof of a local building. she stands at a horizontal distance of 20 meters from the building. the angle of elevation from her eyes to the roof (point a) is 43°, and the angle of elevation from her eyes to the top of the antenna (point b) is 47°. if her eyes are 1.57 meters from the ground, find the height of the antenna (the distance from point a to point b). round your answer to the nearest meter if necessary.

Explanation:

Step1: Find the height from eye - level to point A

Let the height from eye - level to point A be \(h_1\).
We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For \(\theta = 43^{\circ}\) and adjacent side \(x = 20\) meters.
\(\tan43^{\circ}=\frac{h_1}{20}\), so \(h_1 = 20\times\tan43^{\circ}\).
Using a calculator, \(\tan43^{\circ}\approx0.9325\), then \(h_1=20\times0.9325 = 18.65\) meters.

Step2: Find the height from eye - level to point B

Let the height from eye - level to point B be \(h_2\).
For \(\theta = 47^{\circ}\) and adjacent side \(x = 20\) meters.
\(\tan47^{\circ}=\frac{h_2}{20}\), so \(h_2 = 20\times\tan47^{\circ}\).
Using a calculator, \(\tan47^{\circ}\approx1.0724\), then \(h_2=20\times1.0724 = 21.448\) meters.

Step3: Calculate the height of the antenna (AB)

The height of the antenna \(AB=h_2 - h_1\).
\(AB=21.448-18.65=2.798\approx3\) meters.

Answer:

The height of the antenna (from point A to point B) is approximately \(3\) meters.