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?

Answer 1

Answer:

53.2 feet

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

In the right triangle ABC

$tan(50^o)=\frac{AC}{BC}$ -----> by TOA (opposite side divided by the adjacent side)

substitute the values

$tan(50^o)=\frac{h}{40}$

solve for h

$h=(40)tan(50^o)$

$h=47.7\ ft$

therefore

The height of the pole is

$h+5.5=47.7+5.5=53.2\ ft$