If $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}$ are three unit vectors such that $\overrightarrow{\mathrm{b}}$ is not parallel to $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\frac{1}{2}\overrightarrow{\mathrm{b}}$, then the angle between $\overline{a}\text{ and }\stackrel{ }{\overline{c}}$ is

It is given that $\overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\frac{1}{2}\overrightarrow{\mathrm{b}}$

Note: $\overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}}$

$\Rightarrow \frac{1}{2}\overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}}$

$\Rightarrow \left(\overrightarrow{a}\cdot \overrightarrow{c}-\frac{1}{2}\right)\overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{c}=0 ....\left(\mathrm{i}\right)$

Since $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are non-parallel, therefore for the existence of relation (i), the coeff. of $\overrightarrow{\mathrm{b}}$and $\overrightarrow{\mathrm{c}}$ should vanish separately. Therefore, we get 

$\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}}-\frac{1}{2}=0\text{ i.e., }\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}}=\frac{1}{2}\text{ and }\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=0$

Since $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}$ are the unit vectors therefore

$\overrightarrow{\mathrm{a}}\cdot \overline{c}\to =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}\cdot \left(\because {\cos }^{-1} \left(\frac{1}{2}\right)=\theta \right)$