## College Physics (4th Edition)

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.
$A = \sqrt{3.0}~\hat{j}$ $B = -1.0~\hat{i}$ $A+B = -1.0~\hat{i} + \sqrt{3.0}~\hat{j}$ 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.