Chapter 10.2, Problem 53PS

Solution

a.

To determine the model for the distance $d$ as a function of the angle of elevation $\theta .$

  $d=200-300\tan \theta$

Given:

The friend’s position with respect to the person is shown as:

  

Calculation:

Calculating the distance $d$ using the trigonometric function as:

  $\begin{array}{l}\tan \theta =\frac{\text{Height}}{\text{Base}}\\ \tan \theta =\frac{BC}{AB}\\ \tan \theta =\frac{200-d}{300}\\ 200-d=300\tan \theta \\ d=200-300\tan \theta \end{array}$

Hence, the distance $d$ as a function of the angle of elevation $\theta$ is $d=200-300\tan \theta .$

b.

To draw the function of part(a).

Given:

The friend’s position with respect to the person is shown as:

  

Calculation:

The distance function is:

  $d=200-300\tan \theta$

Graph:

The graph of the function is shown below:

  

c.

To determine the angle of elevation $\theta .$

The angle of elevation $\theta$ is ${18.4}^{°}.$

Given:

The friend’s position with respect to the person is shown as:

  

And the friend has repelled halfway down the cliff.

Calculation:

The person has repelled halfway down the cliff, so the new height is given as:

  $\begin{array}{l}Height(h)=\frac{200}{2}\\ h=100\text{ feet}\end{array}$

Now, calculating the angle $\theta$ as:

  $\begin{array}{l}\tan \theta =\frac{\text{Height}}{\text{Base}}\\ \tan \theta =\frac{100}{300}\\ \theta ={\tan }^{-1}\left(0.333\right)\\ \theta ={18.4}^{°}\end{array}$

Hence, the angle of elevation $\theta$ is ${18.4}^{°}.$