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what do S30°W and 110° T mean ?

|dw:1315630919336:dw|

like that ^^^ ?

|dw:1315631114054:dw|

Um, idk , it looks right though xDD

How can I find the distance between them ?

|dw:1315631564224:dw|

Um, I don't think that it requires sin and cos, its bearings

|dw:1315631903573:dw|

What? There's not values of x and y, how can you use the distance formula ?

Okay, tyvm (:

does it make sense ?
did you learn this stuff in the past ?

Well , if there is an easier way, I hope someone will post it here - coz I'd like to know it too :)

Hey, you know your diagram what is the angle degrees thing for AB

what do you mean ?

|dw:1315632702671:dw|

yes we can calculate that - it's 240-110 = 130 degrees

and you're right that we can make the problem simpler

|dw:1315633004738:dw|

|dw:1315634036159:dw|

I would need to calculate to find out - why are you asking ?

If that angle can be found, then I can use the sine rule xD

are you trying to use the law of the cosines ?

No, the Sine Rule it says it in my book xD

which sine rule ?
a/sin(A)=b/sin(B)=c/sin(C)
?

Yep that xD

yes, but we only have one angle , that 130 degrees at this point.

I know , but the other angle is 30 right ? the one that I drew

why did you pick 30 degrees ?

is it becasue sin(30)=1/2 ?

Well, the question S30°W so that angle is 30 right ?

|dw:1315634688115:dw|

The 30 degrees would be relative to the black line I drew.

yeah, upsilon says the same thing.

Can it be 60 then, since 90-30 = 60 ?

You can't decide like that you need the angle to be in ratio with the opposite sides...

yep

but the law of the cosines might work here.

http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html

Yea, that will work

|dw:1315635067683:dw|

\[d = \sqrt{ a^2 + b^2 - 2*a*b*cos\alpha}\]

Okay, tyvm (:

Ok, ty (:

and it's much shorter than the first method :)

yup xD. Your method looked so complicated before (: