The underlying issue you have here is that the inverse cosine function's domain. Consider the graph below:
![enter image source here]()
The domain of the f(x) = cos^-1(x)f(x)=cos−1(x) is [-1, 1][−1,1] (i.e. all real numbers between -1−1 & 11, inclusive), and the range is [1, pi][1,π] (i.e. all real numbers between 11 and piπ inclusive).
Since 2 > 12>1, you cannot plug it into the inverse cosine function.
You can also think of why this is not possible using a triangle:
![enter image source here]()
We know that cos(theta) = y/hcos(θ)=yh, which means that:
theta = cos^-1(y/h)θ=cos−1(yh)
For y/hyh - the ratio of the length of one side to that of the hypotenuse - to be 2, it implicitly implies that y > hy>h, or that that one side is greater in length than that of the hypotenuse. This is not physically possible in a right triangle,
Hope that helped :)
