Answer
$\arccos \dfrac{3}{\sqrt {x^2-2x+10}}=\arcsin \dfrac{|x-1|}{\sqrt {x^2-2x+10}}$
Work Step by Step
In order to compute the length of the opposite we will use the Pythagorean Theorem.
Consider $\theta=\arccos \dfrac{3}{\sqrt {x^2-2x+10}}$
$h=\sqrt {x^2+y^2} \implies (\sqrt {x^2-2x+10})^2=\sqrt {(x-1)^2+9^2 }=|x-1|$
We can see that $\dfrac{3}{\sqrt {x^2-2x+10}}$ is a positive quantity, and $\theta$ must lie in the first quadrant.
So, we have: $\sin \theta=\dfrac{|x-1|}{\sqrt {x^2-2x+10}} $
and $\arccos \dfrac{3}{\sqrt {x^2-2x+10}}=\arcsin \dfrac{|x-1|}{\sqrt {x^2-2x+10}}$
