Calculate the inner angles of the triangle $A(2,-3,5),B(0,1,4),C(-2,5,2)$ I want to calculate the inner angles of this triangle.
$$A(2,-3,5),B(0,1,4),C(-2,5,2)$$
I know that for calculate the angle I need to do the following thing:
$$\cos(\alpha)=\frac{A\cdot B}{|A||B|}$$
I need to calculate AB with BC and AC with AB and AC with BC?

Thanks!

EDIT
$$AB(-2,4,-1),AC(-4,8,-3),BC(-2,4,-2)$$
I found the angle between AB and AC = $5.94$
the angle between AC and BC = $5.55$
the angle between AB and BC is = $11.49$
the two other angles are right but the third not, what I did wrong?
 A: At first calculate the vectors: $\overrightarrow{AB}$,  $\overrightarrow{BC}$, and  $\overrightarrow{AC}$, and their norms. After that use the formula you posted.
For example:
$$\overrightarrow{AB} = (0, 1,4)- (2,-3,5) = (-2, 4, -1)$$
$$|\overrightarrow{AB}|=\sqrt{(-2)^2+4^2+(-1)^2}.$$
I think you can conclude now.
Edit
The third angle you got is in fact an exterior angle. Using the dot product formula, you will get the angle between two vectors, when their representations are set sharing a common origin or a common end. See the picture bellow:

A: Remember that the angle between any two vectors say $\overrightarrow a$ & $\overrightarrow b$ originated from the same point (i.e. tail-points coincident & heads directed away from the origin point) is calculated as $$\cos\alpha=\frac{\overrightarrow a \cdot \overrightarrow b  }{|\overrightarrow a||\overrightarrow b|}$$ Thus, keeping in mind the arrow directions of verctors,  all the angles are very easily calculated as follows 
Angle $A$: between vectors $\overrightarrow {AB}=(-2, 4, -1)$ & $\overrightarrow {AC}=(-4, 8, -3)$ (heads directed away from the (origin) vertex A) is given as $$\angle A=\cos^{-1}\left(\frac{(-2)(-4)+(4)(8)+(-1)(-3)}{\left(\sqrt{(-2)^2+(4)^2+(-1)^2}\right)\left(\sqrt{(-4)^2+(8)^2+(-3)^2}\right)}\right)$$$$=\cos^{-1}\left(\frac{43}{\sqrt{1869}}\right)\approx 5.94^o$$
Angle $B$: between vectors $\overrightarrow {BA}=(2, -4, 1)$ & $\overrightarrow {BC}=(-2, 4, -2)$ (heads directed away from the (origin) vertex B) is given as  $$\angle B=\cos^{-1}\left(\frac{(2)(-2)+(-4)(4)+(1)(-2)}{\left(\sqrt{(2)^2+(-4)^2+(1)^2}\right)\left(\sqrt{(-2)^2+(4)^2+(-2)^2}\right)}\right)$$$$=\cos^{-1}\left(\frac{-22}{\sqrt{504}}\right)=180^o-\cos^{-1}\left(\frac{22}{\sqrt{504}}\right)\approx 168.51^o$$
Angle $C$: between vectors $\overrightarrow {CA}=(4, -8, 3)$ & $\overrightarrow {CB}=(2, -4, 2)$ (heads directed away from the (origin) vertex C) is given as  $$\angle C=\cos^{-1}\left(\frac{(4)(2)+(-8)(-4)+(3)(2)}{\left(\sqrt{(4)^2+(-8)^2+(3)^2}\right)\left(\sqrt{(2)^2+(-4)^2+(2)^2}\right)}\right)$$$$=\cos^{-1}\left(\frac{46}{\sqrt{2136}}\right)\approx 5.55^o$$  
