If  $\left|\overline{a}\right|=2,\left|\overline{b}\right|=3$  and $|2\overline{a}-\overline{b}|=5$   then $|2\overline{a}+\overline{b}|$  equals to 

Given  ${|2\overline{a}-\overline{b}|}^2=5$   ${|2\overline{a}-\overline{b}|}^2=25$   $2{\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2-2(2\overline{a}.\overline{b})=25$  

$$\begin{gathered} {\left|2\overline{a}\right|}^2+{\left|\overline{b}\right|}^2-2\left|2\overline{a}\right|.\left|\overline{b}\right|\cos (2\overline{a},\overline{b})=25 \\ \left(16\right)+9-4\left(2\right)\left(3\right)\cos (2\overline{a},\overline{b})=25 \\ 16+9-24\cos (\overline{a},\overline{b})=25 \\ 25-24\cos (\overline{a},\overline{b})=25 \\ -24\left|\overline{a}\right|\left|\overline{b}\right|\cos \theta =0 \cos \theta =0 \theta =\frac{\pi }{2} \\ {|2\overline{a}+\overline{b}|}^2={\left|2\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+2(2\overline{a}.\overline{b})=4{\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+2\left|2\overline{a}\right|\left|\overline{b}\right|\cos \theta =4\left(4\right)+9+2\left(4\right)\left(3\right)\cos \frac{\pi }{2}=16+9=25 \end{gathered}$$