Answer:
Explanation:
You first solve point a by finding the components [tex]< R_x; R_y >[/tex] of the resulting vector, which will be
[tex]< 0 + 40 \times cos 30\° + 30 \times cos (- 50\°); 20 + 40 \times sin 30\° + 30 \times sin (-50\°) >[/tex]
Notice that since the 50° angle is taken clockwise, we consider it negative, while the 30° is considered positive.
Plugging the values in a calculator or a spreadsheet (if you are using excel or similar programs, don't forget to convert the angle in radians!) you will get the components, namely [tex]< 53.92;17.02 >[/tex]
At this point magnitude is defined as [tex]R = \sqrt{R_x^2 + R_y^2}[/tex] while for direction you first take the ratio [tex]R_y/R_x[/tex] and again, with a calculator, you get the inverse tangent of it. The result will be a number between [tex]-\frac{\pi}2[/tex] and [tex]\frac\pi2[/tex] (-90° and 90°) which tells you the angle the line containing your vector forms with the positive x axis, in our case it's [tex]tan^-^1 (53.92/17.02) \approx 17.52[/tex] (ie, it sits on the red line)
At this point, based on the fact that both are positive, the end point of the vector is in the first quadrant.
