Need explanation

How can you find the angles of a triangle if you have all the sides?

$\begin{array}{|rcll|} \hline \cos{(B)} &=& \frac{1}{2c}\cdot \left(a+\frac{c^2-b^2}{a} \right) \\ \cos{(C)} &=& \frac{1}{2b}\cdot \left(a-\frac{c^2-b^2}{a} \right) \\ A &=& 180^{\circ}-(B+C) \\ \hline \end{array}$

Using law of sines   and law of cosines     yes

You can use the Cosine rule which is

CosA= c² + b2- a²/ 2bc 

and the same with B and C

If the angles of the triangle are A, B, and C with the opposite sides a, b, and c,

then  C  =  cos^-1(  (a² + b² - c²) / (2·a·b)  ).

Similar formulas for the other angles.