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Distance from A to B


Date: 06/17/2001 at 15:06:46
From: Jay
Subject: Trigonometry: bearing problems

Solve with a picture:

The bearing of B from C is 254 degrees. The bearing of A from C is 
344 degrees. The bearing of A from B is 32 degrees. The distance from 
A to C is 780 meters. Find the distance from A to B.

The teacher gave us the hint that it is a right triangle.

Please help me.
- Jay


Date: 06/18/2001 at 12:37:35
From: Doctor Rick
Subject: Re: Trigonometry: bearing problems

Hi, Jay.

Do you know how to find the angles of the triangle ABC? Once you know 
them, I think you'll be on your way. I'll show you how to do this, 
using a different problem so you can do yours yourself. Here's my 
problem:

The bearing of B from C is 215 degrees. The bearing of A from C is 
305 degrees. The bearing of A from B is 338 degrees. The distance from 
A to C is 780 meters. Find the distance from A to B.

The first step is to draw a figure. We can start by drawing BC, which 
is 215 degrees clockwise from north. Draw a vertical line at C (the 
point the bearing is taken FROM). This line represents the direction 
north. Measure 215 degrees clockwise, and draw BC:

                           N
                           |
                           |
                           |--
                           |   \
                       145 *    |
                          / C  / 215
                         /    /
                        /----
                       /
                      /
                     /
                    /
                   /
                  *
                   B

An angle of 215 degrees clockwise is the same as measuring 
360 - 215 = 145 degrees counterclockwise (the sum of these angles 
equals 360 degrees).

Next, add CA. It makes an angle of 305 degrees clockwise from north, 
which is the same as 360 - 55 = 305 degrees counterclockwise:

        A
         *
            \              N
               \           |
                  \     55 |
                     \     |--
                        \  |   \
                           *    |
                          / C  / 215
                         /    /
                        /----
                       /
                      /
                     /
                    /
                   /
                  *
                   B

Now we can find the measure of angle ACB. If the angle NCA = 55 
degrees, and the angle NCB = 145 degrees, then angle ACB = 145 - 55 = 
90 degrees.

We have one more bearing to add to our figure. The bearing of A from B 
is 338 degrees. Draw another north line at B (since this time we're 
measuring from B), and measure 338 degrees clockwise (or 22 degrees 
counterclockwise) to line BA:

        A
         *
            \              N
          \    \           |
                  \     55 |
           \         \     |--
                        \  |   \
             \             *    |
                          / C  / 215
              \   N      /    /
                  |     /----
               \22|    /
                  |   /
                \ |  /
                  | /
                 \|/
                  *
                   B

The next step uses the fact that the two north lines (BN and CN) are 
parallel. Line BC is a transversal, so the interior angles on the same 
side of BC (namely, NBC and BCN) are supplementary: they add to 180 
degrees. Therefore, angle NBC = 180 - angle BCN, or 180 - 145 degrees, 
which equals 35 degrees.

Finally, we can find angle ABC as the sum of angles ABN and NBC: 
22 + 35 = 57 degrees. Now you have two angles of the triangle, so you 
could find the third angle if you needed it. With the length of side 
AC, you have enough information to find the other two sides of the 
triangle. I'll leave that to you. Remember that this is a right 
triangle (and so is the triangle in your problem, as the hint said), 
so you can use the definitions of trig functions in a right triangle 
to solve for side AB.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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