How do you find a vector in the form  when only the angle and magnitude are given? How do you find a vector in the form  when only the angle and magnitude are given?
Here is an example where an angle of 80 degrees is given along with a magnitude of 3.

 A: Note that the angle the vector $\vec{u}$ makes with the $X$-axis is $80^{\circ}$. Hence, the $x$ component of the vector is $\lvert \vec{u} \rvert \cos(80^{\circ})$.
Similarly, the angle the vector $\vec{u}$ makes with the $Y$-axis is $10^{\circ}$. Hence, the $y$ component of the vector is $\lvert \vec{u} \rvert \cos(10^{\circ})$.
Hence, if you want write the vector as $(x,y)$, then it should be $\left(3 \cos(80^{\circ}), 3 \cos(10^{\circ})\right)$.
A: Polar Coordinate System and Cartesian Coordinate system:
You need only the change of variables from polar coordinate system to cartesian coordinates $(r,\theta)\to (x,y)$ given by $x=r\cos\theta$ and $y=r\sin\theta$. 
The inverse change of variables from cartesian coordinate system to polar coordinate system $(x,y)\to (r,\theta)$ is given by
$$\theta = \begin{cases}
\arctan\left(\left|\frac{y}{x}\right|\right) & \mbox{if } x > 0\\
\arctan\left(\left|\frac{y}{x}\right|\right) + \pi & \mbox{if } x < 0 \mbox{ and } y \ge 0\\
\arctan\left(\left|\frac{y}{x}\right|\right) - \pi & \mbox{if } x < 0 \mbox{ and } y < 0\\
\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y > 0\\
-\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y < 0\\
0 & \mbox{if } x = 0 \mbox{ and } y = 0
\end{cases}$$
and $r=\sqrt{x^2+y^2}$.
For more information you can watch Polar Coordinates.
