1. ## bearings

a woman walks 120m on a bearing of 312, then turns and walks for a further 96m on a bearing of 056.

how far is she from her starting point??
can you use the cosine rule
a^2 = b^2 + c^2 - 2bccos(A)???
it gives 18042m
and sqrt gives 134.32m
is that right??

find the bearing of the woman from her starting point.
is it 044???

2. Originally Posted by deej813
a woman walks 120m on a bearing of 312, then turns and walks for a further 96m on a bearing of 056.

how far is she from her starting point??
can you use the cosine rule
a^2 = b^2 + c^2 - 2bccos(A)???
it gives 18042m
and sqrt gives 134.32m
is that right??

find the bearing of the woman from her starting point.
is it 044???
Did you make a (more or less) exact drawing?

I've attached a sketch where 1 cm of the drawing correspond with 10 m in reality(?).

The resulting distance and the resulting bearing are drawn in red.

3. yeah i tried
but i think we are supposed to work it out without a drawing
was my working out right??
or close?

4. Originally Posted by deej813
yeah i tried
but i think we are supposed to work it out without a drawing
was my working out right??
or close?
Unfortunately there aren't "nearly correct" solutions.

You got the final distance correctly.

You calculated the angle at the starting point correctly. You forgot to convert the included angle into a bearing:

The included angle has a value of 43.906°. Add this value to the initial bearing and you'll get the bearing of the final distance:

312° + 43.906° = 355.906°

5. oh ok
great
i'm glad i got the other bits right
thanks heaps