Answer
$v=\dfrac{5}{2}i+\dfrac{5\sqrt3}{2}j$
Work Step by Step
Let us consider that a vector $v$ is given by $v=pi+qj$ and makes an angle of $\alpha$ with the positive $x$-axis. The magnitude of the vector can be determined using the formula
$||v||=\sqrt{p^2+q^2} $
Then we find $p=||v||cos\alpha$ and $v=||v||sin\alpha$.
Here, we have: $p=||5|| \cos \ (60^\circ)=\dfrac{5}{2}$
and $v=||5 | \sin \ (60^\circ)=\dfrac{5\sqrt3}{2}$
Therefore $v=\dfrac{5}{2}i+\dfrac{5\sqrt3}{2}j$
