Question

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.

Answer 1

Answer:

  a)  x = 54/tan(theta)

  b)  theta = arctan(54/x)

Step-by-step explanation:

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)

Answer 2

The expressions for both questions are :

$x = \frac{54}{tan\theta}$

$\theta = tan^{-1}(\frac{54}{x})$

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

$tan\theta = \frac{opposite}{adjacent}\\\\tan \theta = \frac{54}{x}$

Express x as the subject :

$x = \frac{54}{tan\theta}$

B.)

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

Express $\theta$ as the subject :

Take the Tan inverse of both sides :

$\theta = tan^{-1}(\frac{54}{x} )$

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