Chapter 6.2, Problem 58E

Height of a Balloon A hot-air balloon is floating above a straight road. To estimate their height above the ground, the balloonists simultaneously measure the angle of depression to two consecutive mileposts on the road on the same side of the balloon. The angles of depression are found to be 20° and 22°. How high is the balloon?

To determine

To find: The height of the balloon.

Answer to Problem 58E

The height of the balloon is $3.7\text{mi}$.

Explanation of Solution

Given:

A balloon floats on a straight Balloonists measure the angle of depression to two consecutive mileposts on the same side of the road with $20°$ and $22°$.

Formula used:

The formula to calculate $\theta$ with tangent is,

$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$ (1)

Calculation:

Assume the distance between the balloons is $x$.

s

Figure (1)

Figure (1) shows the angle of depression are $20°$ and $22°$, $AB=x$, $CD=1\text{mi}$ and $CB=y$.

As $AB\perp L$ sum of the angles at point $A$ is $90°$.

$\begin{array}{c}20°+a=90°\\ a=90°-20°\\ a=70°\end{array}$

The value of a is $70°$.

And the value of b is calculated as,

$\begin{array}{c}22°+b=90°\\ b=90°-22°\\ b=68°\end{array}$

Substitute $68°$ for $\theta$, $y$ for $\text{opposite}$ and $x$ for $\text{adjacent}$ in equation (1).

$\begin{array}{c}\tan 68°=\frac{y}{x}\\ y=x\tan 68°\end{array}$ (2)

Substitute $38°$ for $\theta$, $y+1$ for $\text{opposite}$ and $x$ for $\text{adjacent}$ in equation (1).

$\begin{array}{c}\tan 70°=\frac{y+1\text{mi}}{x}\\ y+1=x\tan 70°\\ y=x\tan 70°-1\end{array}$ (3)

Compare equation (2) with equation (3) to calculate $x$.

$\begin{array}{c}x\tan 68°=x\tan 70°-1\\ 1=x\left(\tan 70°-\tan 68°\right)\end{array}$

Divide both sides by $\left(\tan 70°-\tan 68°\right)$ in the above equation.

$\begin{array}{c}x=\frac{1}{\left(\tan 70°-\tan 68°\right)}\\ =\frac{1}{2.7474-2.4751}\\ =\frac{1}{0.2723}\\ \approx 3.7\text{mi}\end{array}$

Hence, the height of the balloon is $3.7\text{mi}$.