1. ## Problem Solving

Two hikers set out from the same campsite. One walks 7 km in the direction 043° T and the other walks 10 km in the direction 133° T.

What is the distance between the two hikers?

And I have drawn a diagram. Any help on the above will be appreciated!

2. Luckily, the two had walked at an angle of 90degrees relative to each other or this problem would have been unsolveable. (133-43=90)

So now, thanks to our friend Pythagoras, the problem is quite simple:

a^2+b^2=c^2

If you need any more help on this let me know.

3. Originally Posted by VonNemo19
Luckily, the two had walked at an angle of 90degrees relative to each other or this problem would have been unsolveable. (133-43=90)
Say what?? You're right in that, since they are moving at right angles relative to each other, the simplest solution is the one you describe. However, it can be solved for any angles.

So we have two vectors. We can convert them to rectangular coordinates as such:

$\displaystyle x_1 = 7~\cos 43 = 5.119$
$\displaystyle y_1 = 7~\sin 43 = 4.774$
$\displaystyle x_2 = 10~\cos 133 = -6.820$
$\displaystyle y_2 = 10~\sin 133 = 7.314$

So now we have two vectors in rectangular form: <5.119, 4.774> and <-6.820, 7.314>. All we need to do now is find the distance between them:

$\displaystyle d = \sqrt {(5.119 - (-6.820))^2 + (4.774 - 7.314)^2} = 12.21$

This procedure works for any angles. It is in no way unsolvable if the two vectors aren't at right angles to each other.