1. ## Bearing problem jd

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?

2. Hello, giles84!

I assume you understand "bearings."

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.