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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,

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 

             \ B
                    \                         A
          / 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 **
                 *****           *****

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   
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Practical Geometry

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