1. ## Math 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!!!

2. Originally Posted by <3<3
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!!!

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 $\displaystyle a$.

3. As I replied to your other topic!

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

'B' being your first point where you change bearing

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 I drew it out in paint but it ended up horribly so nvm

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

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# an aircraft flies from its base 200km on a bearing 162Â° then 350 km on a bearing 260Â°, and then returns directly to the base

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