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Mathematics 12 Online
OpenStudy (osanseviero):

How to know which is the angle from the x-axis of a vector with components (2 3)

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?

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?

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

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):

\[\frac{ -18 }{ -2(3\sqrt{13}) }=A\]

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?

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

OpenStudy (kc_kennylau):

?

OpenStudy (osanseviero):

|dw:1385000320550:dw|

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