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If vector A⃗ has components Ax and Ay and makes an angle θ with the +x axis, then...
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tanθ =Ax/Ay ?
Yup
isnt it supposed to be y/x instead?
wait crap. Yea. It's y/x.
A=Ax+Ay (where Ais the magnitude of A⃗ ) θ =AyAx cosθ = AxA2x+A2y√ tanθ =AxAy which of these is true? the first one?
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Hmmm not quite sure...might want to look that up or ask someone else. I don't want to give any false info.
k cool thanks
yes the tangent of the angle is ay/ax
The vector is F = (Ax,Ay) and it has direction @ such that tan@ = Ay/Ax
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