Chapter 4.7, Problem 53E

Solution

(a)

To prove: $\theta ={\tan }^{-1}\left(\frac{14}{x}\right)-{\tan }^{-1}\left(\frac{2}{x}\right)$

Given information: $\theta ={\tan }^{-1}\left(\frac{14}{x}\right)-{\tan }^{-1}\left(\frac{2}{x}\right)$

Formula used: height and distance.

Proof: let $A$ be the acute of the larger right triangle with respect to the horizontal and $B$ be the acute angle of the smaller right triangle with respect to the horizontal so that:

  $\theta +B=A$ or $\theta =A-B$

To find $A$ , use the tangent ratio on the larger right triangle:

  $\begin{array}{c}\tan A=\frac{opp}{adj}\\ =\frac{12+2}{x}\\ =\frac{14}{x}\\ A={\tan }^{-1}\frac{14}{x}\end{array}$

To find $B$ , use the tangent ratio on the smaller right triangle:

  $\begin{array}{c}\tan B=\frac{opp}{adj}\\ =\frac{2}{x}\\ B={\tan }^{-1}\frac{2}{x}\end{array}$

Hence,

  $\theta ={\tan }^{-1}\left(\frac{14}{x}\right)-{\tan }^{-1}\left(\frac{2}{x}\right)$

(b)

To graph: graph $\theta$ in the $[0.25]$ by $[0,55]$ viewing window using degree mode.

Given information: region $[0.25]$ , $[0,55]$

Graph:

  

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

  $\theta ={\tan }^1\left(\frac{14}{x}\right)-{\tan }^{-1}\left(\frac{2}{x}\right)$

Then use the maximum feature to obtain $x\approx 5.3$ ft.

(c)

To calculate: distance from wall.

About $2$ ft. or $15$ ft.

Given information: $\theta ={35}^{\circ }$

Formula used: trigonometryCalculation: graph the line $y=35$ representing $\theta ={35}^{\circ }$

on Y2 together with the graph from part (b) then use the intersect feature to find the x-values:

  

To the nearest foot, you are standing from the wall about $2$ ft. or $15$ ft. if $\theta ={35}^{\circ }$ .