# Find y-component of vector A

• Sep 28th 2012, 06:43 PM
icelated
Find y-component of vector A
Vector A is in the direction 34.0 degrees clockwise from the -y axis. The x-component of vector A is Ax = - 16.0m

What is the y-component of vector A?

Im not sure how to solve it since the magnitude is not known. CI drew a picture of how i think its suppose to be.
Can someone help me set it up and how they got that answer?

I think since the hypotenuse is not known i use tan somehow.

Thank you
• Sep 28th 2012, 06:56 PM
MarkFL
Re: Find y-component of vector A
Is A supposed to be counter-clockwise from the -y-axis? This is the way you have drawn it and is the only way the x-component can be negative.

Otherwise, your hunch about using the tangent function is a good one...
• Sep 28th 2012, 07:25 PM
icelated
Re: Find y-component of vector A
Quote:

Originally Posted by MarkFL2
Is A supposed to be counter-clockwise from the -y-axis?

No, Vector A is clockwise from the -y axis.
• Sep 28th 2012, 07:42 PM
MarkFL
Re: Find y-component of vector A
D'oh...it seems I forgot how clocks move...:)
• Sep 29th 2012, 03:55 AM
Plato
Re: Find y-component of vector A
Quote:

Originally Posted by icelated
Vector A is in the direction 34.0 degrees clockwise from the -y axis. The x-component of vector A is Ax = - 16.0m
What is the y-component of vector A?
Im not sure how to solve it since the magnitude is not known.

$\displaystyle \|A\|=16\cos(56^o)$. From that you can find $\displaystyle A_y$.
• Sep 29th 2012, 08:38 AM
icelated
Re: Find y-component of vector A
90 - 34 ?
• Sep 29th 2012, 08:45 AM
HallsofIvy
Re: Find y-component of vector A
Quote:

Originally Posted by icelated
90 - 34 ?

I have no clue what you mean by this. You have a right triangle with angle 34 degrees and "opposite side" of length 16. Since you mention the tangent function, yes, if you call the length of the near side "y" then you have $tan(34)= 16/y$. You can solve that for the absolute value of y (since lengths are always positive).