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each vector has a y and x component

i and j :) ?

well you're given a magnitude

|dw:1359604762762:dw|

you there?

|dw:1359605259789:dw|

yeah, im here.. trying to understand

|dw:1359605476238:dw|

now using trig functions you can find the components correct ? can you do that?

using sohcahtoa right?

A_x=sin(16.6)*A?

yes

i got 3.16 as side A_y..

so now we have
\[A_x=Acos(16.6)\]
\[A_y=Asin(16.6)\]

you had them switched around =]

and 0.95 as a_y.. A=3.30 right?

yep sounds correct

atleast according to my diagram lol

alright so the B is easy since there is only one component to it... which is the \[B_y\]

B=3.30 but theres no theta?

yes but it's a simple line on the y axis so it only has one component an x is what i meant =/

so you have
\[B_x=3.30\]

|dw:1359606759899:dw|

"like components" ? 8.1?

so B is 3.30 and for A do I add a_x and a_y and A?

well lets do it like my book says
\[R_x=A_x+B_x\]
\[R_y=A_y+B_y\]

\[|A+B|=\sqrt{R_x^2+R_y^2}\]

R_x=4.25
R_y= 3.16

|dw:1359607102751:dw|

5.3 m?

\[B_x=3.30\]
\[A_x=3.30cos(16.6)\]
\[A_y=3.30sin(16.6)\]

6.6 m?

\[R_x=6.46\]
\[R_y=.943\]
\[|R|=\sqrt{6.46^2+.943^2}\]

i'm getting 6.5

6.53 to be more exact

yeah i got that too now. i rounded wrong earlier.

so for A-B would i just subtract?

yes to do it easily i'm trying to remember what it looks like lol|dw:1359607683207:dw|

|dw:1359607791579:dw|

I'll go back later when i have more imte tomorrow and try to figure it out lol.