Mathematics
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OpenStudy (osanseviero):
How to know which is the angle from the x-axis of a vector with components (2 3)
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OpenStudy (kc_kennylau):
Have you learnt the cosine rule?
OpenStudy (osanseviero):
nope
OpenStudy (osanseviero):
wait a sec :) please
OpenStudy (osanseviero):
|dw:1384999144716:dw|
OpenStudy (osanseviero):
I can say that, right?
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OpenStudy (kc_kennylau):
Yep you can say that, but you really haven't learnt the law of cosines/cosine rule/cosine law?
(it has many names kay)
OpenStudy (osanseviero):
Not yet
OpenStudy (kc_kennylau):
Then you wouldn't be given this problem...
OpenStudy (kc_kennylau):
Here is the cosine rule: \(a^2=b^2+c^2-2bc\cos(A)\)
(a, b, c are sides of a triangle, and A, B, C are angles of a triangle)
OpenStudy (kc_kennylau):
does that help you?
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OpenStudy (osanseviero):
Yep, thanks!
OpenStudy (kc_kennylau):
no problem :)
OpenStudy (osanseviero):
I dont know if my answer is correct, can you do it please?
OpenStudy (kc_kennylau):
what's your answer? :)
OpenStudy (osanseviero):
1.090
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OpenStudy (osanseviero):
With this method I did this
OpenStudy (osanseviero):
4=9+13-2(3*sqrt13)(cosA)
OpenStudy (osanseviero):
oops, I did a mistake, give me a sec
OpenStudy (kc_kennylau):
ok :)
OpenStudy (osanseviero):
\[\frac{ -18 }{ -2(3\sqrt{13}) }=A\]
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OpenStudy (osanseviero):
cos-1 of that
OpenStudy (kc_kennylau):
|dw:1384999945508:dw|
OpenStudy (osanseviero):
oh :(
OpenStudy (kc_kennylau):
don't give up :) try again :D
OpenStudy (osanseviero):
2/sqrt16 = cosA?
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OpenStudy (kc_kennylau):
you mean 13
OpenStudy (osanseviero):
yep
OpenStudy (kc_kennylau):
and yes :)
OpenStudy (osanseviero):
Oh! got the same than with the first method!
OpenStudy (osanseviero):
So that is the angle x, and I need to do 2pi - this
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OpenStudy (kc_kennylau):
?
OpenStudy (osanseviero):
|dw:1385000320550:dw|