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 $\displaystyle A$ on a bearing of 45° and flies 65 km to $\displaystyle B.$ Code:

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

Then it flies on a heading of 240° for 62 km to $\displaystyle P.$ 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 $\displaystyle \angle PBA = 15^o$

And we can use the Law of Cosines.

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

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