4200.25 – Sighting an Airplane

Tags:Problem Set 14Pre-Calculus MathTrigonometry

You are standing outside on a beautiful clear day. An airplane at 3000 feet files over head at 200 mph. You don't notice this until you hear the plane, but the instant that happens you spot the plane at xx degrees above the horizon. The sound you are hearing, however, came from the plane somewhat earlier. At that time it was 20 degrees above the horizon.

Find xx, assuming that 1100 ft/sec is the speed of sound.


Referring to the diagram: You are at point YY. When you first heard the plane it was at BB but the sound arrived from AA. The plane flew from AA to BB while the sound traveled to you, that is, traveled from AA to YY. Our strategy is to find aa, bb and cc, concentrating on ABY\triangle ABY. Once we find ww we can find x=w+20x = w + 20.


3000b=sin20b=3000sin208771 feet.\frac{3000}{b} = \sin 20 \leadsto b = \frac{3000}{\sin 20} \approx 8771 \text{ feet}.

At 1100 ft/sec1100 \text{ ft/sec} the sound took 7.97 sec. to get to YY. In that time, the plane flew from AA to BB at 200 mi\hr. So

c=AB=20036007.97=.443 mi.2339 ft. c = AB = \frac{200}{3600} 7.97 = .443 \text{ mi.} \approx 2339 \text{ ft}.

So much for bb and cc. We now get ww by first finding aa. Using the law of cosines:

a2=23392+877122(2339)(8771)cos20=43845080a6622 feet.a^2 = 2339^2 + 8771^2 - 2 (2339) (8771) \cos 20 = 43845080 \leadsto a \approx 6622 \text{ feet}.

Then, using the law of sines:

sinw=csin20a2339sin206622=0.1208w6.94 degrees. \sin w = \frac{c \sin 20}{a} \approx \frac{2339 \sin 20}{6622} = 0.1208 \leadsto w \approx 6.94 \text{ degrees}.

Thus x26.94 degreesx \approx 26.94 \text{ degrees}.

Note: This problem is presented implausibly as the viewer could note the angle of elevation when spotting the plane but could not know its angle of elevation when the sound that was heard was made. It might be better to put the airplane directly overhead and ask for its position and angle of elevation when emitting that sound.