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.

USE: $z = re^{i\theta}$
$2+5i$ = ?
Using modulus which is $r = \sqrt{x^2+y^2}$
$r = \sqrt{(2)^2+(5)^2}$ = $\sqrt{29}$
$\theta = \arctan({\frac{5}{2}})$
$2+5i = \sqrt{29}e^{i\arctan({\frac{5}{2}})}$
