Question

The angle of elevation from a viewer to the top of a flagpole is 50˚. The viewer is 80 ft away and the viewer’s eyes are 5.5 ft from the ground. How high is the pole to the nearest tenth of a foot?

Answer 1

Step-by-step explanation:

frist draw the diagram

here x is the distance from the top to the height of the viewer. therefore the height of pole is 95.3+5.5=100.8 ft

Answer 2

Flagpole length  from the ground to top of a flagpole  is 100.78 ft.    

Step-by-step explanation:

Given :

Viewer angle of elevation  to the top of a flagpole is 50°.

The viewer is 80 ft away and  viewer's eyes are 5.5 ft from the ground.

To Find:

The height of the flag pole.

Formula Used:

In  a right   angle triangle-

$tan\theta = \frac{perpendicular}{base}$

Solution:

As given- Viewer angle of elevation  to the top of a flagpole is 50°.

 Let viewer angle of elevation  to the top of a flagpole is  $= \theta$

 Let the viewer   distance from flagpole = 80 ft.

Let  Top of flagpole height for viewer’s eyes level = H.

$tan\theta$  = flagpole height for viewer’s eyes level/ the viewer   distance from flagpole

        $tan 50 \textdegree =\frac{H}{80}$  

     $H= 80\times tan 50\textdegree$

Putting value of $tan 50 \textdegree = 1.191$

$H= 80\times 1.191$

$H= 95.28 ft$

Flagpole length  from the ground to top of a flagpole = the viewer's eyes are 5.5 ft from the ground+ Top of flagpole height for viewer’s eyes level.

  Flagpole length  from the ground to top of a flagpole  $= 5.5 + 95.28$                                        

$= 100.78 ft$

Thus,   Flagpole length  from the ground to top of a flagpole  is 100.78 ft.