Question

Please Help! Geometry!

Answer 1

Explanation:

Suppose the height of the pole is AB, the viewer is standing at point C and the viewer's eye is at point D i.e 5ft above the ground.

BCDE is a rectangle, so BC=DE= 20ft

The angle of elevation from the viewer's eye at D to the top of the flagpole A = 50°

So, in ∆ AED

`(AE)/(ED) = \tan(50)`

`(AE)/(20) = \tan(50)`

`AE = 20 * \tan(50)`

`AE = 20 * 1.19 = 23.8`

Total height of the pole = 23.8+5 = 28.8 ft

Answer 2

Answer: the height of the flagpole is 28.8 feet

Explanation:

Considering the situation, a right angle triangle is formed. The height, h of the flagpole from the ground represents the opposite side of the right angle triangle.

The horizontal distance between the viewer and the flagpole represents the adjacent side of the right angle triangle.

To determine h, we would apply

the tangent trigonometric ratio.

Tan θ = opposite side/adjacent side. Therefore,

Tan 50 = h/20

h = 20tan50 = 20 × 1.192

h = 23.84

Since the viewer's eyes is 5 feet from the ground, then the total height of the flagpole is 23.84 + 5 = 28.8 feet