## Algebra and Trigonometry 10th Edition

$3$ and $\theta =60^{\circ}$
$\overrightarrow{v} = \dfrac{3}{2 } \ \widehat{i} +\dfrac{3 \sqrt 3}{2} \ \widehat{j}$ Magnitude: $||\overrightarrow{v}|| = \sqrt {(\dfrac{3}{2 })^{2} + (\dfrac{3 \sqrt 3}{2} )^{2}} = 3$ The direction can now be found: $\tan \ \theta = \dfrac{\dfrac{3 \sqrt 3}{2} }{3/2} = \sqrt 3 \implies \theta =60^{\circ}$