## Algebra and Trigonometry 10th Edition

$\lt \sqrt 6, \sqrt 6 \gt$
We are given that $||\overrightarrow{v}|| = 2 \sqrt 3$ and $\theta = 45^{\circ}$ We know that $v=||v|| \lt \cos \theta , \sin \theta \gt$ Thus, $v=2 \sqrt 3 \lt \cos 45^{\circ} , \sin 45^{\circ} \gt \\=2 \sqrt 3 \lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2} \gt \\= \lt \sqrt 6, \sqrt 6 \gt$