Question

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 ${20}^{\circ }$ and ${22}^{\circ }.$

How high is the balloon?

Answer 1

The height of the balloon:

The angle of depression at a nearer point is always greater than the point farther one.

Thus, the one right angle triangle is formed with angle ${22}^{\circ }$ say the perpendicular or height be h$h$ miles and the base be x miles (say).

$\tan \left({22}^{\circ }\right)=\frac{h}{x}$.

And the other angle of depression ${20}^{\circ }$is at consecutive milepost:

$\tan \left({20}^{\circ }\right)=\frac{h}{x+1}$.

Solve $\tan \left({22}^{\circ }\right)=\frac{h}{x}$ and $\tan \left({20}^{\circ }\right)=\frac{h}{x+1}$:

$$\begin{gathered} \tan \left({20}^{\circ }\right)=\frac{x\tan \left({22}^{\circ }\right)}{x+1} \\ \frac{\tan \left({22}^{\circ }\right)}{\tan \left({20}^{\circ }\right)}=1+\frac{1}{x} \\ x=\frac{\tan \left({20}^{\circ }\right)}{\tan \left({22}^{\circ }\right)-\tan \left({20}^{\circ }\right)} \\ =9.099\text{miles} \end{gathered}$$

The height is:

$$\begin{gathered} h=9.099\times \tan \left({22}^{\circ }\right) \\ =9.099\times 0.4040 \\ =3.6762\text{miles} \end{gathered}$$

Hence, the height of the balloon is $3.6762\text{miles}$.