So I got this problem:
So let $\triangle ABC$ be a right triangle at vertex $A$ such that $BC=2AB$. Find $\angle{ACB}$
So I found that $\cos(\angle ABC)=\frac{1}{2}$
But when I want to find $\angle ABC$, should I do $\cos^{-1}$ in radians, or degrees ?
$$\cos(60^\circ) = \cos(\pi/3) = \frac 12$$
Since we are speaking of the angle of a triangle, we can narrow the solutions to $\cos^{-1}(\angle ABC)$: We can express $\cos^{-1}(\angle ABC)$ in either degrees ($60^\circ$) or in radians $\left(\dfrac \pi 3\right)$.