How can I write $2+5i$ in polar form? Original problem:

Write $2+5i$ in polar form.


My attempt:
$|2+5i|=\sqrt{4+25}=\sqrt{29}$
$\arg(2+5i)=\arctan{\frac{5}{2}}=\theta$
$2+5i=\sqrt{29}(\cos(\theta)+i\sin(\theta))$
But how do I calculate $\arctan({\frac{5}{2}})$ in radians?
I know $\tan(\theta)=\frac{5}{2}$ and this implies that if we have a rectangle triangle $ABC$ then $\overline{AB}=5$ and $\overline{BC}=2$.
Here I'm stuck here.
 A: USE: $z = re^{i\theta}$
$2+5i$ = ?
Using modulus which is $r = \sqrt{x^2+y^2}$
$r = \sqrt{(2)^2+(5)^2}$ = $\sqrt{29}$
$\theta = \arctan({\frac{5}{2}})$
$2+5i = \sqrt{29}e^{i\arctan({\frac{5}{2}})}$
A: As mentioned in comments above, you do not need to find the value of $\theta$, namely, you don't need to find the value of $\arctan \frac{5}{2}$. Instead, what you need here are the following two quantities:
$$
\cos\theta,\quad  \sin\theta.
$$
In your right triangle, 
$$
\sin\theta=\frac{5}{\overline{AC}},\quad \cos\theta=\frac{2}{\overline{AC}}.
$$
But you could find the value of $\overline{AC}$ using the Pythagorean theorem.
A: What we need is the values of $\cos\arctan(5/2)$ and $\sin\arctan(5/2)$, and not $\arctan(5/2)$ itself. This can be done easily in general: draw a right angled triangle with hypotenuse $\sqrt{1+x^2}$ and one of the other sides of length $x$, then the other side has length $1$ and it is easy to identify the angle whose tangent is $x$. So the $\sin,\cos$ of that angle can be found as a simple exercise in trigonometric ratios. In general it is easy to see that
$$\sin\arctan(x)=\frac{x}{\sqrt{x^2+1}},\cos\arctan(x)=\frac1{\sqrt{x^2+1}}.$$
