Hello, Ruscur!

A car leaves a camp site and travels in the direction of S 65° E for 8.2 km.

It then heads due south for 6.7 km to reach a waterhole.

a) How far is the waterhole from the camp site?

b) What is the bearing of the waterhole from the camp site? First, we need a sketch . . . Code:

A N
* 8.2 :
:. * 65°:
: 65° * :
: . * B
: . 115°|
: . |
: . |
S . | 6.7
. |
. |
. |
.|
* C

The car goes 8.2 km from $\displaystyle A$ to $\displaystyle B.$

. . $\displaystyle AB = 8.2,\;\angle SAB = 65^o = \angle ABN$

Then it goes 6.7 km south to $\displaystyle C.$

. . $\displaystyle BC = 6.7,\;\angle ABC = 115^o$

Law of Cosines: .$\displaystyle AC^2 \:=\:8.2^2 + 6.7^2 - 2(8.2)(6.7)\cos115^o \:=\:158.5672946$

(a) Therefore: .$\displaystyle \boxed{ AC \;\approx\;13.6\text{ km}}$

Law of Cosines: .$\displaystyle \cos(\angle BAC) \;=\;\frac{8.2^2 _ 13.6^2 - 6.7^2}{2(8.2)(13.6)} \;=\;0.929474534$

Hence: .$\displaystyle \angle BAC \;\approx\;21.6^o$

Therefore: .$\displaystyle \angle SAC \:=\:65^o - 21.6^o \:=\:43.4^o$

The bearing is: .$\displaystyle \boxed{S\,43.4^o\,E}$