Question

the angle of elevation from a viewer to the top of a flagpole is 50°. the viewer is 40 ft away and the viewers eyes are 5.5 ft from the ground. how high is the pole to the nearest tenth of a foot? height = ft

Explanation:

Step1: Use the tangent function

Let \( h \) be the height from the viewer's eyes to the top of the flagpole. We know that \(\tan\theta=\frac{h}{d}\), where \(\theta = 50^{\circ}\) and \(d = 40\) ft. So \(h = 40\times\tan(50^{\circ})\).
Using a calculator, \(\tan(50^{\circ})\approx1.191753592\), then \(h = 40\times1.191753592=47.67014368\) ft.

Step2: Calculate the total height of the flagpole

The total height \(H\) of the flagpole is the sum of the height from the viewer's eyes to the top of the flagpole (\(h\)) and the height of the viewer's eyes from the ground (\(5.5\) ft). So \(H=h + 5.5\).
Substitute \(h = 47.67014368\) ft into the formula: \(H=47.67014368+5.5 = 53.17014368\approx53.2\) ft.

Answer:

\(53.2\)