Finding acute angle between line and plane (Vectors)

This is a very very simple question about vectors:

To find the acute angle between a line and a plane, you use the formula

cosx = (scalar product between normal of plane and directional vector of line)/(product of modulus of normal and directional vector)

After that, do 90 degrees minus x to get the answer.

(By the way, I realise that using sinx=.... would spare me the effort of minusing - but is that way recommended?)

In this question of plane 2x-y+4z=9 and directional vector (10,5,-5), after doing all the working, I get cosx= -0.089... and x=95.11 degrees.

But the question wants acute angle and 90-x gives me -5.1 degrees. What should I do? The answer is 5.1 deg but I don't see how 90-x gets me that...

NB: using sinx gives me -5.1 degrees straightaway.

Thank you.

Honestly, between the direction vector $v=(10,5,-5)$ and the normal vector $( 2,-1,4)$ the angle is not acute, in fact it is obtuse (you can see that by a simple plot).
In you case, to find the angle $\theta$ you can do the following : when finding $\cos x$, apply $\arccos$ to find an angle $\phi$. Then subtract $\phi$ from $180$ to get $\alpha=180 - \phi$ . Now $\theta = 90 - \alpha$.