$$\begin{aligned} & \text { Given, }|\overline{\mathrm{a}}|=1,|\overline{\mathrm{~b}}|=1 \text { and }|\overline{\mathrm{c}}|=2 \\ & \text { Also, } \overline{\mathrm{a}} \times(\overline{\mathrm{a}} \times \overline{\mathrm{c}})+\overline{\mathrm{b}}=\overline{0} \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}) \overline{\mathrm{c}}+\overline{\mathrm{b}}=\overline{0} \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}+\overline{\mathrm{b}}=\overline{0} \quad \ldots\left[\because \overline{\mathrm{a}} \cdot \overline{\mathrm{a}}=|\overline{\mathrm{a}}|^2=1\right] \end{aligned}$$

$$\begin{aligned} & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}=-\overline{\mathrm{b}} \\ & \Rightarrow|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}|=|-\overline{\mathrm{b}}| \\ & \Rightarrow|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}|^2=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow\left|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^{-} \overline{\mathrm{a}}\right|^2+|\overline{\mathrm{c}}|^2-2\{(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}\}=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2|\overline{\mathrm{a}}|^2+|\overline{\mathrm{c}}|^2-2(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2\left\{|\overline{\mathrm{a}}|^2-2\right\}+|\overline{\mathrm{c}}|^2=|\overline{\mathrm{b}}|^2 \end{aligned}$$

$\Rightarrow-(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2+4=1 \quad \ldots\left[\because|\overline{\mathrm{~b}}|^2=1,|\overline{\mathrm{c}}|^2=4\right]$

$$\begin{aligned} & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2=3 \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}= \pm \sqrt{3} \\ & \Rightarrow|\overline{\mathrm{a}}||\overline{\mathrm{c}}| \cos \theta=\sqrt{3}, \end{aligned}$$

$$\begin{aligned} &\text { where } \theta \text { is an acute angle between } \bar{a} \text { and } \bar{c}\\ &\Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} \end{aligned}$$

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