# How do you find a vector in the form <a,b> 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.

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$\bigl(\,3\cos(80^\circ), 3\sin(80^\circ)\,\bigr)$ –  David Mitra May 24 '12 at 23:16

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 $\left(3 \cos(80^{\circ}), 3 \cos(10^{\circ})\right)$.

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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 $r=\sqrt{x^2+y^2}$.