Answer
When $x \gt 0$, then we have: $arctan \dfrac{9}{x}=\arcsin \dfrac{9}{\sqrt {x^2+81}} $
When $x \lt 0$, then we have: $arctan \dfrac{9}{x}=\arcsin \dfrac{-9}{\sqrt {x^2+81}} $
Work Step by Step
In order to compute the length of the hypotenuse we will use Pythagorean Theorem.
Consider $\theta=arctan \dfrac{9}{x}$
$h=\sqrt {x^2+y^2}=\sqrt {x^2+9^2 }=\sqrt {x^2+81}$
When $x \gt 0$, then we have: $\sin \theta=\dfrac{9}{\sqrt {x^2+81}} $
When $x \lt 0$, then we have: $\sin \theta=\dfrac{-9}{\sqrt {x^2+81}} $
When $x \gt 0$, then we have: $arctan \dfrac{9}{x}=\arcsin \dfrac{9}{\sqrt {x^2+81}} $
When $x \lt 0$, then we have: $arctan \dfrac{9}{x}=\arcsin \dfrac{-9}{\sqrt {x^2+81}} $
