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The angle of elevation from a viewer to the top of a flagpole is 50˚. The viewer is 40 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?

Mar 25, 2020

#1
+1956
+3

Hint:

Tan 50 degrees will represent the angle of elevation. Now, write the equation.

Hope this helped!

Mar 25, 2020
#2
+483
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Tan 50 is not the angle of elevation, that's arctan of that ratio. Tangent is a ratio of sides, not the angle itself. The inverse sine (arcsin)is the angle itself given the sine for example. Consequently, the inverse tangent(arctan) is the angle itself given the tangent.

jfan17  Mar 25, 2020
edited by jfan17  Mar 25, 2020
edited by jfan17  Mar 25, 2020
edited by jfan17  Mar 25, 2020
edited by jfan17  Mar 25, 2020
#3
+23566
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= height / 40        when you figure out THIS height you must add the 5.5 ft observer's height

tan 50 = 1.191754

Mar 25, 2020
#4
+111321
+2

We  have  this

tan (50°)  =  x  / 40

40 * tan (50°)  = x

The  height of the  flagpole  =

5.5   +   x   =

5.5  +  40 * tan (50°)    ≈   53.2  ft

Mar 25, 2020