Question

lillian is trying to find the height of a radio antenna on the roof of a local building. she stands at a horizontal distance of 15 meters from the building. the angle of elevation from her eyes to the roof (point a) is 32°, and the angle of elevation from her eyes to the top of the antenna (point b) is 54°. if her eyes are 1.51 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 antenna top at point A

Use the tangent function in the right - triangle for point A. The formula for tangent is \(\tan\theta=\frac{opposite}{adjacent}\).
Given \(\theta = 32^{\circ}\) and \(adjacent = 15\) meters. Let \(h_1\) be the height from eye - level to the top of the antenna at point A. Then \(\tan32^{\circ}=\frac{h_1}{15}\).
So, \(h_1 = 15\times\tan32^{\circ}\).
Using a calculator, \(\tan32^{\circ}\approx0.6249\), then \(h_1=15\times0.6249 = 9.3735\) meters.

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

Use the tangent function in the right - triangle for point B. Let \(h_2\) be the height from eye - level to the top of the antenna at point B. Given \(\theta = 54^{\circ}\) and \(adjacent\) (distance from B to the building) be \(x\).
First, find \(x\) from the first triangle (point A): we know \(h_1 = 9.3735\) and \(\tan32^{\circ}=\frac{h_1}{15}\).
For the second triangle (point B), \(\tan54^{\circ}=\frac{h_2}{x}\). But we can also use the fact that we can find the height of the antenna.
Another way: Let the total height of the antenna (from eye - level) be \(H\).
We know that \(\tan32^{\circ}=\frac{H}{15}\) and \(\tan54^{\circ}=\frac{H}{d}\) (where \(d\) is the distance from B to the building). But we can also use the difference in angles.
Let the height of the antenna from eye - level be \(h\).
We know that \(\tan32^{\circ}=\frac{h}{15}\) and \(\tan54^{\circ}=\frac{h}{d}\). But we can use the formula \(h = 15\times\tan32^{\circ}\) (from step 1) and also, if we consider the two - triangle situation.
Let's use the formula for the height of the antenna (from eye - level):
We know that \(\tan32^{\circ}=\frac{h}{15}\) and \(\tan54^{\circ}=\frac{h}{d}\). But we can also use the formula \(h=\frac{15\times\tan32^{\circ}\times\tan54^{\circ}}{\tan54^{\circ}-\tan32^{\circ}}\)
\(\tan32^{\circ}\approx0.6249\), \(\tan54^{\circ}\approx1.3764\)
\(h=\frac{15\times0.6249\times1.3764}{1.3764 - 0.6249}=\frac{15\times0.6249\times1.3764}{0.7515}\)
\(h=\frac{15\times0.8609}{0.7515}=\frac{12.9135}{0.7515}=17.18\) meters

Step3: Calculate the total height of the antenna

The height of the antenna \(H=h + 1.51\) (adding the height of the eyes from the ground)
\(H=17.18+1.51=18.69\approx19\) meters

Answer:

\(19\) meters