Chapter 5, Problem 63RE

Solution

To find:The height of the balloon above the ground.

The height of the balloon above the ground is $0.6\text{mile}$ .

Given information:

The included angles of the triangle given in the figure are $33°$ and $37°$ .

Formula used:

The formula for $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$ .

Calculation:

Consider $h$ as the height of the balloon above the ground in miles. Assume $x$ as the distance of the observer from $D$ directly below the balloon such that $1.75-x$ is the distance between the second observer and $D$ . All distances are in limes.

  

Consider the triangle ACD.

Substitute $33°$ for $\theta$, $h$ for opposite side, and $x$ for adjacent side in $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$ .

  $\begin{array}{c}\tan 33=\frac{h}{x}\\ x=\frac{h}{\tan 33}\end{array}$

Consider the triangle BCD.

Substitute $37°$ for $\theta$, $h$ for opposite side, and $1.75-x$ for adjacent side in $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$ .

  $\begin{array}{c}\tan 37=\frac{h}{1.75x}\\ 1.75-x=\frac{h}{\tan 37}\end{array}$

Substitute $\frac{h}{\tan 33}$ for $x$ in $1.75-x=\frac{h}{\tan 37}$ .

  $\begin{array}{c}1.75-\frac{h}{\tan 33}=\frac{h}{\tan 37}\\ h\approx 0.6\text{mile}\end{array}$

Therefore, the height of the balloon above the ground is $0.6\text{mile}$ .