I'll presume you're looking for the angle which the vector makes with the #x#-axis.
We use our old friend #tan# for this.
#tantheta=y/x#
#=>theta=tan^-1(y/x)#
#color(white)(=>theta)=tan^-1((-3)/4)=tan^-1(-0.75)#
#color(white)(=>theta)=-0.64#
#color(white)(=>theta)approx-36.67°#
Because #tan^-1# can only return angles between -90° and +90° (that is, in #Q_"IV"# or #Q_"I"#), we then check which quadrant the point #(4,"-"3)# is in, to confirm whether we can keep this answer or if we need to add 180°.
#(4,"-"3)# is in quadrant 4, and so the value given by #tan^-1# is fine as it is.
The vector #V=(4,"-"3)# is at an angle of #-36.67°# to the positive #x#-axis.
