Question

Movie Theater Screens Suppose that a movie theater has a screen that is $28$feet tall. When you sit down, the bottom of the screen is $6$feet above your eye level. The angle formed by drawing a line from your eye to the bottom of the screen and your eye and the top of the screen is called the viewing angle. In the figure, $\theta$is the viewing angle. Suppose that you sit $x$feet from the screen. The viewing angle $\theta$is given by the function:

$\theta \left[x\right]={\tan }^{-1}\left(\frac{34}{x}\right)-{\tan }^{-1}\left(\frac{6}{x}\right)$

(a) What is your viewing angle if you sit $10$feet from the screen? $15$feet? $20$feet?

(b) If there is $5$feet between the screen and the first row of seats and there is $3$feet between each row, which row results in the largest viewing angle?

(c) Using a graphing utility, graph

$\theta \left[x\right]={\tan }^{-1}\left(\frac{34}{x}\right)-{\tan }^{-1}\left(\frac{6}{x}\right)$.

Short answer

$$\begin{gathered} \mathrm{The}\mathrm{viewing}\mathrm{angle}\mathrm{from} \\ \left(a\right)10\mathrm{ft}-42.65\mathrm{degrees} \\ 15\mathrm{ft}-44.39\mathrm{degrees} \\ 20\mathrm{ft}-42.83\mathrm{degrees} \\ \left(b\right)x=5=31.44 \\ x=8=39.89 \\ x=11=43.46 \\ x=14=44.42 \\ x=17=43.99 \\ \mathrm{The}\mathrm{largest}\mathrm{viewing}\mathrm{angle}\mathrm{is}44.42 \\ \left(c\right)x=13.33333 \end{gathered}$$

The graph of part (c) is

Step-by-step solution

Part (a) Step 1. Given Information.

The given equation is :

$\theta \left[x\right]={\tan }^{-1}\left(\frac{34}{x}\right)-{\tan }^{-1}\left(\frac{6}{x}\right)$

The given figure :

Part (a) Step 2. Calculation.

Find the viewing angle for different given distances :

$10ft;$

localid="1649408471470" $$\begin{gathered} \theta \left[10\right]={\tan }^{-1}\left(\frac{34}{10}\right)-{\tan }^{-1}\left(\frac{6}{10}\right) \\ \theta \left[10\right]=73.6104-30.9637 \\ \theta \left[10\right]=42.6467degrees. \end{gathered}$$

For $x=15ft;$

localid="1649408728574" $$\begin{gathered} \theta \left[15\right]={\tan }^{-1}\left(\frac{34}{15}\right)-{\tan }^{-1}\left(\frac{6}{15}\right) \\ \theta \left[15\right]=66.19405-21.80140 \\ \theta \left[15\right]=44.39265degrees. \end{gathered}$$

For $x=20;$

$$\begin{gathered} \theta \left[20\right]={\tan }^{-1}\left(\frac{34}{20}\right)-{\tan }^{-1}\left(\frac{6}{20}\right) \\ \theta \left[20\right]=59.5344-16.6992 \\ \theta \left[20\right]=42.8352degrees. \end{gathered}$$

Part (b) Step 1. Calculation.

To substitute the values of $x$to determine the maximum value.

Ist row $\left(x=5\right)$;

$$\begin{gathered} \theta \left[5\right]={\tan }^{-1}\left(\frac{34}{5}\right)-{\tan }^{-1}\left(\frac{6}{5}\right) \\ \theta \left[5\right]=81.63411-50.19442 \\ \theta \left[5\right]=31.43969degrees. \end{gathered}$$

IInd row, $\left(x=8\right)$;

$$\begin{gathered} \theta \left[8\right]={\tan }^{-1}\left(\frac{34}{8}\right)-{\tan }^{-1}\left(\frac{6}{8}\right) \\ \theta \left[8\right]=39.88958degrees. \end{gathered}$$

IIIrd row, $\left(x=11\right);$

$$\begin{gathered} \theta \left[11\right]={\tan }^{-1}\left(\frac{34}{11}\right)-{\tan }^{-1}\left(\frac{6}{11}\right) \\ \theta \left[11\right]=43.4616degrees. \end{gathered}$$

IVth row, $\left(x=14\right)$;

$$\begin{gathered} \theta \left[14\right]={\tan }^{-1}\left(\frac{34}{14}\right)-{\tan }^{-1}\left(\frac{6}{14}\right) \\ \theta \left[14\right]=44.4212degrees. \end{gathered}$$

Vth row, $\left(x=17\right)$;

$$\begin{gathered} \theta \left[17\right]={\tan }^{-1}\left(\frac{34}{17}\right)-{\tan }^{-1}\left(\frac{6}{17}\right) \\ \theta \left[17\right]=43.9949degrees. \end{gathered}$$

Since the angle has started to decrease, we see that the maximum viewing angle occurs in the 4th row.

Part (c) Step 1. Graphing Utility.

Using a graphing calculator, graph the function and determine the max $x$value, we get

$x=13.3333$.