Question

how find the degree directly from the given vectors magnitude A and B

Answer 1

If you are asking for the angle between vectors, we can find it by using the scalar product of the vectors.
$$\begin{gathered} \overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=\mathrm{ABcos\theta } \\ \mathrm{where}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{vectors}. \\ \mathrm{\theta }={\cos }^{-1}\left[\frac{\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}}{\mathrm{AB}}\right] \\ ={\cos }^{-1}\left[\frac{{\mathrm{A}}_1{\mathrm{B}}_1+{\mathrm{A}}_2{\mathrm{B}}_2+{\mathrm{A}}_3{\mathrm{B}}_3}{\sqrt{{{\mathrm{A}}_1}^2+{{\mathrm{A}}_2}^2+{{\mathrm{A}}_3}^2}\sqrt{{{\mathrm{B}}_1}^2+{{\mathrm{B}}_2}^2+{{\mathrm{B}}_3}^2}}\right] \\ \mathrm{If}\overrightarrow{\mathrm{A}}={\mathrm{A}}_1\hat{\mathrm{i}}+{\mathrm{A}}_2\hat{\mathrm{j}}+{\mathrm{A}}_3\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{B}}={\mathrm{B}}_1\hat{\mathrm{i}}+{\mathrm{B}}_2\hat{\mathrm{j}}+{\mathrm{B}}_3\hat{\mathrm{k}} \end{gathered}$$