Finding equilibrant vector and x-y components from magnitude and angle

[buffgilville]

Can someone please help me with this?
1) A resultant vector is 5 units long and makes an angle of 23 degrees measured counter-clockwise with respect to the positive x-axis. What are the magnitude and angle (measured counter-clockwise with respect to the positive x-axis) of the equilibrant vector?

2) Find the x- and y-components of the vector whose magnitude is 8.73 units making an angle of 155 degrees measured counter-clockwise with respect to the positive x-axis.

 

[marlon]

These kinds of questions have already been answered, look at some other posts.

Think of the fact that the x-component of a vector of magnitude F and angle x with the x-axis is
equal to F*cos (x). The y-component is F*sin(x)

Once the components are given you can work the other way around and determin the magnitude of the vector as F = sqrt(x² + y²)

regards
marlon

 

[buffgilville]

But what does the counter-clockwise mean? Does it mean that the angle is negative? I'm confused.

 

[Pyrrhus]

Clockwise means a negative angle, and counter-clockwise means a positive angle.