Locating a Ship Using Three Angles
Date: 07/07/99 at 18:19:28 From: Thaddeus Subject: Geometry Angles Dr. Math, The navigator of a ship S saw landmarks in the distance at three points A, B, and C. Taking sightings from the deck of the ship she found that angle ASB = 100, angle BSC = 125 and angle CSA = 135. Then she located the three points A, B and C on a map and used the fact that an inscribed angle is half the measure of the central angle to find the exact position of her ship. How did she do it? I drew an angle ASB and then drew the perpendicular bisectors of each side to draw a circle with points A, S and B on the circle. Then I created a point C such that angle BCS was 125 degrees and drew the perpendicular bisectors of BC and CS to draw a circle with points B, C, and S in common. Then I did it one last time with the last angle. All three circles intersected at S, but I do not understand how this helps me. I never used the inscribed angle theorem and I started with S already located on my paper. I do not believe this is right. Please tell me where I went wrong. Thank you, Thaddeus
Date: 07/08/99 at 12:10:16 From: Doctor Peterson Subject: Re: Geometry Angles Hi, Thaddeus. This is a nice problem. I'll try to give you a hint to get you started in the right direction. As you recognized, you can't start with point S and claim to have found where it is. But you can work backward in another sense to solve a problem like this. Let's start by drawing a point S and three rays from it at the appropriate angles, then pick a point on each ray to be A, B, and C. That sets things up for us; if we can find a method that doesn't use point S, but ends up putting a dot where S is, we'll know we've solved it. \ \ \ B + \ \ \ \ \ \ A S+------------------------+- / / / / + / C / / Now how can we find S? Let's concentrate on points A, B, and S. All we know about them is that angle ASB is 100 degrees. Suppose they were all on a circle. Then this would be an inscribed angle, and it would be half the central angle: *********** ***** ***** *** *** ** ** * * ** ** \ * 200 * \ * * \*B O * +-------------------+ * *\ \ * *\ \ * * \ \ * **\ \ ** *\ \ * **100 \A ** S+**---------------------*+*----- ***** ***** *********** In fact, if we could find that point O to use for the center of the circle, then the circle would be the locus of all points S for which angle ASB = 100 degrees. That's because the central angle, and the arc intercepted by ASB, are the same no matter where S is on the circle. Now you just have to figure out how you can locate point O, using only points A and B and the angle. If you do the same thing for BC and CA, you'll have drawn three circles that intersect at S. (Actually you'll only have to draw two of them, unless you want to check the accuracy of your drawing.) - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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