Chapter 4.8, Problem 41E

a.

To determine

To find: The angle of elevation. And distance if angle of elevation is ${42}^{\circ }$ .

Answer to Problem 41E

The angle of elevation to the top the screen from distance is $\text{15 ft}\text{.}$ and $\text{100 ft}\text{.}$ are ${\tan }^{-1}\left(\frac{28}{15}\right)$ and ${\tan }^{-1}\left(\frac{28}{100}\right)$ .

Explanation of Solution

Given information:

Consider, the figure,

  

Calculation:

From, the above figure, it can be observed that,

  $\begin{array}{l}\text{AC}=22+6=28\text{ ft}.\\ \text{BC}=100\text{ ft}\text{.}\\ \text{CD}=15\text{ ft}\text{.}\end{array}$

From $\Delta \text{ACD}$

Let, $\angle \text{CAD}=\theta$

  $\begin{array}{l}\tan \theta =\frac{\text{AC}}{\text{CD}}\\ \Rightarrow \tan \theta =\frac{28}{15}\\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{28}{15}\right)\end{array}$

And from $\Delta \text{ACB}$

  $\angle \text{CAB}=\alpha \left(\text{say}\right)$

  $\begin{array}{l}\tan \alpha =\frac{28}{100}\\ \Rightarrow \alpha ={\tan }^{-1}\left(\frac{28}{100}\right)\end{array}$

Hence, the angle of elevation to the top the screen from distance is $15\text{ ft}.$ and $100\text{ ft}.$ are ${\tan }^{-1}\left(\frac{28}{15}\right)$ and ${\tan }^{-1}\left(\frac{28}{100}\right)$

b.

To determine

To find: The angle of elevation. And distance if angle of elevation is ${42}^{\circ }$ .

Answer to Problem 41E

Distance $\simeq 31.1\text{ ft}.$

Explanation of Solution

Given information:

Consider, the figure,

  

Calculation:

From, the above figure, it can be observed that,

  $\begin{array}{l}AC=22+6=28\text{ ft}.\\ BC=100\text{ ft}\text{.}\\ CD=15\text{ ft}\text{.}\end{array}$

If, $\theta ={42}^{\circ }$

Let, distance $=d$

Then,

  $\begin{array}{l}\tan {42}^{\circ }=\frac{28}{d}\\ \Rightarrow d=\frac{28}{\tan {42}^{\circ }}\\ \Rightarrow d=\frac{28}{0.9}\\ \Rightarrow d\simeq 31.1\text{ ft}\text{.}\end{array}$

Hence, Distance $\simeq 31.1\text{ ft}.$