Chapter 10.1, Problem 32PS

Solution

a.

To write an equation that gives the car’s distance $d$ from the base of the bridge as a function of the angle $\theta .$

  $d=140\tan \theta$

Given:

The car position with respect to the person on the bridge is shown as:

  

Calculation:

Now, calculating the distance $d$ using the trigonometric function as:

  $\begin{array}{l}\tan \theta =\frac{\text{Base}}{\text{Height}}\\ \tan \theta =\frac{d}{140}\\ d=140\tan \theta \end{array}$

Hence, the distance $d$ as a function of $\theta$ is $d=140\tan \theta .$

b.

To draw the graph of the function of part(a).

Given:

The function of a car’s distance is given as:

  $d=140\tan \theta$

Graph:

The graph of the given function is shown below:

  

Interpretation:

Here, when the car moves, $\theta$ will move from $0^{\circ }$ to ${90}^{\circ }$ . From the graph, it is observed that when the $\theta =0^{\circ }$ , the distance $d$ is zero. It means that the car is just under the bridge.

If $\theta$ increases, then the distance of car from the base of bridge increases according to the trigonometry tangent function. Here, when the car goes to an infinite distance far away, then it will make an angle of $\theta ={90}^{\circ }$ .

c.

To make a table to find the car’s distance from the bridge for different values of $\theta .$

  $\begin{array}{cccc}\text{Angle}\theta & {20}^{°}& {40}^{°}& {60}^{°}\\ \text{Car's distance }d& 50.96\text{ ft}& 117.46\text{ ft}& 242.48\text{ ft}\end{array}$

Given:

The function of the car’s distance $d$ is given as:

  $d=140\tan \theta$

Calculation:

Calculating the values of $d$ for different values of $\theta$ as:

  $d=140\tan \theta$

For $\theta ={20}^{°}$

  $\begin{array}{l}d=140\tan {20}^{°}\\ d=140\times 0.364\\ d=50.96\text{ ft}\end{array}$

For $\theta ={40}^{°}$

  $\begin{array}{l}d=140\tan {40}^{°}\\ d=140\times 0.839\\ d=117.46\text{ ft}\end{array}$

For $\theta ={60}^{°}$

  $\begin{array}{l}d=140\tan {60}^{°}\\ d=140\times 1.732\\ d=242.48\text{ ft}\end{array}$

Hence, the car’s position at different angles is:

  $\begin{array}{cccc}\text{Angle}\theta & {20}^{°}& {40}^{°}& {60}^{°}\\ \text{Car's distance }d& 50.96\text{ ft}& 117.46\text{ ft}& 242.48\text{ ft}\end{array}$