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January 20th, 2010, 01:03 AM   #1
 
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Finding distance using two angles of elevation

A question in my trigonometry book has bugged me for the longest time. A Chapter 1 level-2 application reads like this: "In the movie Close Encounters of the Third Kind, there was a scene in which the star, Richard Dreyfuss, was approaching Devil's Tower in Wyoming. He could have determined his distance from Devil's Tower by first stopping at a point P and estimating the angle P. After moving 1.000 x 10^2 m toward Devil's Tower, he could have estimated the angle N, as shown in the figure. How far away from Devil's Tower is point N?" So, basically the first angle of elevation is 13.5 degrees and after moving 100 m in the direction toward Devil's Tower the angle of elevation is 14.8 degrees.

What I want to know is how this can be done without advanced techniques in trigonometry. This is in the first chapter of the book and no advanced technique are explained as of yet. Basically, what has been explained thus far in the book in the trigonometric functions, their inverses, and the use of similar triangles to find similar angles and side measures. I would like to know the simplest method that could be used to determine the distance. Thank you in advanced.
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January 20th, 2010, 03:24 AM   #2
 
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Re: Finding distance using two angles of elevation

[attachment=0:3bame9tc]elev.GIF[/attachment:3bame9tc]

By the definition of tan, you have:



But you don't know the value of y. Can you still find x? Yes, if you divide one by the other:



Subtract 1 from both sides, make the left-hand side one fraction, invert both sides and multiply by 100 to get:

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January 20th, 2010, 09:00 PM   #3
 
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Re: Finding distance using two angles of elevation

@aswoods Thank you so much for the quick reply. I can easily see how this works out thanks to your explanation. It is basically using the proportionality of the tangent of the angles to the distances and the height. Would that be a good explanation in word form (just to go a bit further )?
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