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.
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.