Answer
The vector $A+B$ has a magnitude of 2.0 units and it is directed at an angle of $30.0^{\circ}$ counter-clockwise from the positive y-axis.
Work Step by Step
$A = \sqrt{3.0}~\hat{j}$
$B = -1.0~\hat{i}$
$A+B = -1.0~\hat{i} + \sqrt{3.0}~\hat{j}$
We can find the magnitude of A+B:
$\vert A+B \vert = \sqrt{(-1.0)^2+(\sqrt{3.0})^2}$
$\vert A+B \vert = 2.0$
We can find the angle $\theta$ counter-clockwise from the positive y-axis:
$tan(\theta) = \frac{1.0}{\sqrt{3.0}}$
$\theta = arctan(\frac{1.0}{\sqrt{3.0}})$
$\theta = 30.0^{\circ}$
The vector $A+B$ has a magnitude of 2.0 units and it is directed at an angle of $30.0^{\circ}$ counter-clockwise from the positive y-axis.
