Height of the Plane

Date: 03/29/2001 at 13:54:29
From: Peter Komorowski
Subject: Pre-Calculus

Question: Towers A and B are known to be 4.1 mi. apart on level 
ground. A pilot measures the angles of depression to the towers to be 
36.5 degrees and 25 degrees, respectively. Find the height of the 
airplane.

I tried to answer this question and simply did not have any idea how 
to do it.

Date: 03/29/2001 at 17:02:06
From: Doctor Jaffee
Subject: Re: Pre-Calculus

Hi Peter,

A good picture can be very helpful in solving a problem like this one. 

Draw line PQ, where P and Q represent the bases of the two towers and 
the length of the segment PQ is 4.1 miles.

Then draw line AB parallel to line PQ. Point A represents the location 
of the airplane. Point A should be situated so that the measure of 
angle BAQ is 25, and the measure of angle BAP is 36.5.

Finally, locate point C on line PQ so that AC is perpendicular to PQ.

Now you should be able to calculate the measures of all the angles in 
the drawing fairly easily, and then use the Law of Sines in triangle 
APQ to calculate the length of AP. Once you know that, you should be 
able to work in triangle ACP and calculate AC, the height of the 
airplane.

Give it a try and if you want to check your answer with me or if you 
need more help, write back.

- Doctor Jaffee, The Math Forum
  http://mathforum.org/dr.math/