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Tanker Bearings

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Date: 06/11/2000 at 07:17:59
From: Rachel Crane
Subject: Bearings

I have my end of year exams coming up this week. One of the questions
on the paper will be about bearings, and I don't understand a bit of
it:

From a ship, A, the bearing of an oil tanker, T, is 300 degrees.
From a second ship, B, which is 1000 m due west of A, the bearing of
the oil tanker is 060 degrees. Explain why the oil tanker is the same
distance from A as it is from B. (I also have to draw a rough sketch.)

Thank you so much,
Rachel Crane
```

```
Date: 06/14/2000 at 08:30:18
From: Doctor Rick
Subject: Re: Bearings

Hi, Rachel,

A bearing is a direction, measured as an angle going clockwise from
north. North is 0 degrees, east is 90 degrees, south is 180 degrees,
and west is 270 degrees. In pure math the convention is to measure
directions going counterclockwise from "east" (the positive x axis),
but in practical fields like surveying or navigation, bearings are
used. Don't ask me why there are two different systems; I guess the
people who invented them just never talked with each other.

You can use bearings to draw the geometrical figures requested in your
problem, and then use more familiar geometry to answer the question.
Let's look at the problem:

>A) From a ship, A, the bearing of an oil tanker, T, is 300 degrees.
>From a second ship, B, which is 1000 m due west of A, the bearing of
>the oil tanker is 060 degrees. Explain why the oil tanker is the same
>distance from A as it is from B. (I also have to draw a rough
>sketch.)

Draw a point A. Then draw a ray from A at a bearing of 300 degrees.
Going 300 degrees clockwise from north (up, in your drawing) is the
same as going 60 degrees counterclockwise, because 300 + 60 = 360
degrees, a full circle. We don't know where P is yet, so I'll label a
point on the ray Q just so we can identify the ray. It will look like
this:

Q _           North
|\           ^
\        |
\   60|
\  |
*A

Draw a line from A to B. It goes west (to the right) from A. Make it
any length but label it 1000 m.

Q _           N
|\           ^
\        |
\   60|
30 \  |
B*------------*A

What is the angle between AQ and AB? Since angle NAB is a right angle,
angle NAQ + angle QAB = 90 degrees. Therefore angle QAB = 30 degrees.

We have one more ray to draw, the ray BP. Its bearing is 060, that is,
60 degrees clockwise from north. I'll let you draw it because my
figure would get too complicated. Do the same thing I did above: draw
a line north from B, and mark off an angle of 60 degrees on the right
side of that line. What is the angle between the ray you have drawn
and BA?

Mark P on the figure, at the intersection of AQ and the new ray. Now
you have enough information to answer the question.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/15/2000 at 11:38:43
From: Rachel Crane
Subject: Re: Bearings

Thanks for your message, Dr. Rick. It has helped me a lot. All I need
now is some good luck for tomorrow's math test!

Cheers,
Rachel Crane
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Practical Geometry

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