An aircraft flies from its base 200km on a bearing 162 degrees, then 350 km on a bearing 260 degrees, and then returns directly to base. Calculate the length and bearing of the return journey!!!

Please help me!!!!!

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- Sep 9th 2008, 10:11 AM<3<3Math Trig. Help!
An aircraft flies from its base 200km on a bearing 162 degrees, then 350 km on a bearing 260 degrees, and then returns directly to base. Calculate the length and bearing of the return journey!!!

Please help me!!!!! - Sep 9th 2008, 10:41 AMSimplicity
Draw a diagram, it does half the question for you.

To work out the length it travels, use the cosine rule. $\displaystyle a^2 = b^2 + c^2 - 2bc\cos A$, where:

- $\displaystyle a,b,c$ are your lengths.
- $\displaystyle A$ is the angle opposite to the side $\displaystyle a$.

- Sep 9th 2008, 10:50 AMshinhidora
As I replied to your other topic!

If you draw that out, you should en up with a triangle...

'A' being your base

'B' being your first point where you change bearing

'C' being the point where you return to your base

â = 180 - 162 = 18 degrees

b = 360 - 260 = 100 degrees

c = 180 - 18 - 100 = 62 degrees

AB = 200km

BC = 350 km

CA = X

you need to find x :p I drew it out in paint but it ended up horribly so nvm :p

You know how to do it now? Or are the formulae what you're looking for?