If  $\overline{\mathrm{a}}\text{ and }\overline{\mathrm{b}}$ are two unit vectors such that $\overline{\mathrm{a}}+2\overline{\mathrm{b}}$ and $\overline{5\mathrm{a}}-4\overline{\mathrm{b}}$ are perpendicular to each other, then the angle between $\overline{\mathrm{a}}$ and $\overline{b}$  is 

$\overline{\mathrm{a}}\text{ and }\overline{\mathrm{b}}$ are unit vectors 

$\left|\overline{\mathrm{a}}\right|=\mathrm{a}=1$

$\left|\overline{\mathrm{b}}\right|=\mathrm{b}=1$

$(\overline{a}+2\overline{\mathrm{b}})\perp (5\overline{\mathrm{a}}-4\overline{\mathrm{b}})$

$\Rightarrow (\overline{a}+2\overline{\mathrm{b}})·(5\overline{\mathrm{a}}-4\overline{\mathrm{b}})=0$

$(\overline{a}·5\overline{a})·(\overline{\mathrm{a}}-4\overline{\mathrm{b}})+(2\overline{\mathrm{b}}·5\overline{\mathrm{a}})+2\overline{\mathrm{b}}·(-5\overline{\mathrm{b}})=0$

$5(\overline{a}·\overline{a})-4(\overline{a}·\overline{b})+10(\overline{b}·\overline{a})-8(\overline{b}·\overline{b})=0$

$5\left(a^2\right)-4ab\cos \theta +10ab\cos \theta -8\left(b^2\right)=0$

$5\left(1^2\right)+6ab\cos \theta -8\left(1^2\right)=0$

$5+6\left(1\right)\left(1\right)\cos \theta =3\cos \theta =3/6=1/2$

$\theta =60°=\text{ Angle between }\overline{\mathrm{A}}\text{ and }\overline{\mathrm{B}}$