# Consider two vectors where F1 = 48 N, F2 = 63 N, and = 240o and = 25o, measured from the positive x-axis with counter-clockwise being positive.

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### 1 Answer

You did not state so, but I am going to assume that you need to find the net vector if you combine the two given vectors.

Take each vector separately and find the x- and y- components using sine and cosine functions.

For vector one, it is directed @ 60 degrees clockwise below the negative y-axis, in quadrant three.

The y-vector component is 48N * sin(60) = -41.57 N

The x-vector component is 48N * cos (60) = - 24 N

For vector two, it is directed at 25 degree up in quadrant one.

The y-vector component is 63N * sin (25) = 26.62 N

The x-vector component is 63 N * cos(25) = 57.1 N

Now sum the x-and y-components of the two vectors.

For the y- component you have -41.57+ 26.62 = - 14.95 N

For the x-component you have -24 N + 57.1 N = +33.1 N

So your resultant vector will be located in quadrant four at 33.1N in the x-direction and -14.95N in the y-direction.

Use Pythagoren theorem to find the resultant vector: square root of (33.1^2 + 14.95^2) = 36.32 N

To find the angle use the inverse tangent function:

tan^-1(14.95/33.1) = 24.31 degrees clockwise from the x-axis.