Chapter 4.7, Problem 54E

Solution

(a)

To prove: $\theta ={\tan }^{-1}\left(\frac{x}{3}\right)$

Given information: $\theta ={\tan }^{-1}\left(\frac{x}{3}\right)$

Formula used: height and distance.

Proof: $\Delta QPL$ is a right triangle, just in a different view. To find $\theta$ , use the tangent ratio:

  $\begin{array}{l}\tan \theta =\frac{opp}{hyp}=\frac{PQ}{PL}\\ \tan \theta =\frac{x}{3}\end{array}$

Taking the inverse tangent of both sides, then we get: $\theta ={\tan }^{-1}\left(\frac{x}{3}\right)$

(b)

To graph: graph $\theta$ in the $[-20,20]$ by $[-90,90]$ viewing window using degree mode.

Given information: region $[-20,20]$ , $[-90,90]$

Graph:

  

Interpretation: graph the equation by entering the right side on Y1 and use the given window:

  $\theta ={\tan }^1\left(\frac{x}{x3}\right)$

Negative values of $x$ mean a distance away from P along the straight shoreline, opposite of $Q$ .

A positive angle means that the beacon rotates counterclockwise with the light towards $Q$ .

A negative angle means that the beacon rotates clockwise with the light away from $Q$ .

(c)

To calculate: find $\theta$

  $\theta \approx {78.69}^{\circ }$

Given information: $x=15$

Formula used: trigonometry

Calculation: substitute $x=15$ and use a calculator in degree mode:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{15}{3}\right)\\ {\tan }^{-1}(5)\approx {78.69}^{\circ }\end{array}$

Therefore, $\theta \approx {78.69}^{\circ }$