Height of the PlaneDate: 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/ |
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