Originally Posted by
1005 You need to assign a reference for your displacement vectors and then sum them. Most intuitive for me would be to think of the directions as if they were on a map: west being negative x-axis, east being positive x-axis, north being positive y-axis, and south being negative y-axis. From there, you see the first vector is $\displaystyle -200$ and the second vector is $\displaystyle (150 \angle 180^\circ - 60^\circ)$ or $\displaystyle (150 \angle 120^\circ)$
Drawing pictures helps with vector problems such as this one. Do do vector addition, you need to break each vector into rectangular components, and add them.
x: $\displaystyle -200 + 150 \cos (120^\circ)$
y: $\displaystyle 150\sin (120^\circ)$
Your answer then may look like this:
$\displaystyle (-200 + 150 \cos (120^\circ))\hat i \ + (150\sin (120^\circ))\hat j \$
Where $\displaystyle \hat i \$ and $\displaystyle \hat j \$ are unit vectors designating only direction and not magnitude.
If you wanted the answer expressed in phasor form, then do this:
$\displaystyle (\sqrt{(-200 + 150 \cos (120^\circ))^2 + (150\sin (120^\circ))^2}\angle \arctan (\frac{150\sin (120^\circ)}{-200 + 150 \cos (120^\circ)}))$