# Bearing problem jd

• Apr 20th 2009, 12:50 PM
giles84@hotmail.com
Bearing problem jd
This is a question i have been stuck on for hours. please help!

The following is all that is given in the question:

An airplane leaves an airport on a bearing of 45degrees.
After flying for 65km, the airplane turns and flies on a bearing of 240 degrees for 62 km. To the nearest km, how far is the airplane fom the airport?
• Apr 21st 2009, 06:24 AM
Soroban
Hello, giles84!

I assume you understand "bearings."

Quote:

An airplane leaves an airport on a bearing of 45°.
After flying for 65 km, the airplane turns and flies on a bearing of 240° for 62 km.
To the nearest km, how far is the airplane fom the airport?

It leaves the airport $A$ on a bearing of 45° and flies 65 km to $B.$
Code:

                                    o B                                   * :                                 *45°:                               *    :                             *                           *                         *             :        * 65             :      *             :    *             :45°*             : *             o             A

Then it flies on a heading of 240° for 62 km to $P.$
Code:

                                    :                               120° :                                     o B                               *    240°                   62      *                     *                 *           *     P o

The final diagram looks like this:
Code:

                                    :                               120° :                                     o B                               *  * :                   62      *    *  :                     *  15°  * 45° :                 *          *      :           *              *     P o                *       *              * 65         *  :      *         *  :    *           * :45°*           *: *             o             A

We find that $\angle PBA = 15^o$
And we can use the Law of Cosines.

$PA^2 \;=\;65^2 + 62^2 - 2(65)(62)\cos15^o \;=\;283.6378401$

Therefore: . $PA \;=\;16.841552 \;\approx\;17$ km.