Answer
$\arccos \dfrac{(x-2)}{2}=\arcsin \dfrac {\sqrt {4x-x^2}}{(x-2)}$
Work Step by Step
In order to compute the length of the opposite we will use the Pythagorean Theorem.
Consider $\theta=\arccos \dfrac{x-2}{2}$
$h=\sqrt {x^2+y^2} \implies (2)^2=((x-2))^2+(opposite )^2= \implies opposite =\sqrt {4x-x^2}$
We can see that $|x-2| \leq 2 \implies 0 \leq x \leq 4$ and $\theta$ must lie in the first quadrant.
So, we have: $\tan \theta=\dfrac{\sqrt {4x-x^2}}{(x-2)} $
and $\arccos \dfrac{(x-2)}{2}=\arcsin \dfrac {\sqrt {4x-x^2}}{(x-2)}$
