# Thread: Basic trig bearing problem

1. ## Basic trig bearing problem

FYI - I'm doing this not as homework but as revision, I'm not asking for the answer, I want to know the process in which the answer was found.

"A 4WD vehicle leaves a camp site and dravels across a flat, sandy plain in a direction of S 65 degrees E, for a distance of 8.2km. It then heads due south for 6.7km to reach a waterhole."

The two questions are:

a) How far is the waterhole from the camp site?
b) What is the bearing of the waterhole from the camp site?

I don't remember doing this in class and I've tried many things but I can't get the answer. I've got a correct (I think) diagram, but I don't know where to start. Can you help?

2. Originally Posted by Ruscur
FYI - I'm doing this not as homework but as revision, I'm not asking for the answer, I want to know the process in which the answer was found.

"A 4WD vehicle leaves a camp site and dravels across a flat, sandy plain in a direction of S 65 degrees E, for a distance of 8.2km. It then heads due south for 6.7km to reach a waterhole."

The two questions are:

a) How far is the waterhole from the camp site?
b) What is the bearing of the waterhole from the camp site?

I don't remember doing this in class and I've tried many things but I can't get the answer. I've got a correct (I think) diagram, but I don't know where to start. Can you help?

I presume your diagram shows a line extending 65 degrees above the horizontal of length 8.2 and a line downward from that of length 6.7. It should be clear from your diagram that, since the "south" line (vertical) is perpendicular to the "east" line (horizontal), if they did cross you would have a right triangle. That is enough to show that the angle between the two lines you drew is 90- 65= 25 degrees. You have a (non right) triangle with two sides of length 8.2 and 6.7 and angle between them 25 degrees. You can use the cosine law to find the length of the third side (the distance from the camp to the water hole) and then use the sine law to find the angle that third side makes with the vertical (and the bearing will be that many degrees "west of north".

3. 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 $A$ to $B.$
. . $AB = 8.2,\;\angle SAB = 65^o = \angle ABN$

Then it goes 6.7 km south to $C.$
. . $BC = 6.7,\;\angle ABC = 115^o$

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

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

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

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

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

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