A 6-foot man walks on level ground toward a 60 foot flag pole. He notes different angles of elevation θ to the top of the pole when he is different horizontal distances x away from the pole.
a) Express the man's horizontal distance from the base of the pole x (in feet) as a function of the angle of elevation θ in degrees. Type theta for θ.
b) Express the angle of elevation θ as a function of the man's horizontal distance to the base of the pole x.

a) The tangent ratio is the ratio of the side opposite an acute angle in a right triangle to the adjacent side. For the angle of elevation, the adjacent side is the distance x to the flagpole. The side opposite is the 54-foot difference between the top of the man's head and the top of the flagpole. So, we have ...

tan(theta) = 54/x

Solving for x gives ...

x = 54/tan(theta)

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b) Solving the first equation in the previous part for theta involves the inverse tangent function:

theta = arctan(54/x)

fichoh fichoh · Answer 2 · 2021-08-19T14:32:25+02:00

The expressions for both questions are :

[tex]x = \frac{54}{tan\theta}[/tex]

[tex]\theta = tan^{-1}(\frac{54}{x})[/tex]

The solution triangle is given below ;

Using trigonometry ; SOHCAHTOA

The man's horizontal distance, x expressed as a function of the angle of elevation (from the top of the man's head) , θ;

This means, the opposite side = (height of flagpole - man's height) = (60 - 6) = 54 feets

[tex]tan\theta = \frac{opposite}{adjacent}\\\\tan \theta = \frac{54}{x}[/tex]

Express x as the subject :

[tex]x = \frac{54}{tan\theta}[/tex]

B.)

[tex]tan\theta[/tex] = \frac{54}{x}[/tex]

Express [tex]\theta[/tex] as the subject :

Take the Tan inverse of both sides :

[tex]\theta = tan^{-1}(\frac{54}{x} )[/tex]

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