91Ó°ÊÓ

Given that a blimp, suspended in the air at a height of 500 feet, lies directly over a line from Soldier Field to the Adler Planetarium on Lake Michigan (see the figure). If the angle of depression from the blimp to the stadium is 32° and from the blimp to the planetarium is 23°.

We can draw a figure of the given problem as:-

Suppose the distance between soldier field and point C is $x$and the distance between Adler Planetarium and point C is $y.$

Then, in $△ABC,∠C=90°$

$$\begin{gathered} \tan (32°)=\frac{AC}{BC} \\ \tan (32°)×BC=AC \\ BC=\frac{500}{\tan (32°)} \\ BC=\frac{500}{0.624} \\ BC=x=801.28feet \end{gathered}$$

Now in $△ACD,∠C=90°$

width="163">tan(23°)=ACCDtan(23°)×CD=ACCD=500tan(23°)CD=5000.424CD=y=1179.24feet

So, the distance between Soldier Field and the Adler Planetarium is$x+y.$

$$\begin{gathered} x+y=801.28+1179.24 \\ x+y=1980.52feet \end{gathered}$$