By rotating the axes at an angle $\alpha$ if (4, 2) in the new system formed as $(2+\sqrt{3} , 1–2 \sqrt{3} )$ then $\alpha$ =

$$\begin{gathered} \mathrm{Given} \mathrm{\theta }=\mathrm{\alpha }, (\mathrm{x}, \mathrm{y})=(4, 2), (X, Y)=(2+\sqrt{3}, 1-2\sqrt{3}) \\ X=\mathrm{x} \mathrm{cos\theta }+\mathrm{y} \mathrm{sin\theta } \\ \Rightarrow 2+\sqrt{3}=4 \mathrm{cos\alpha }+2\sin \alpha - \left(1\right) \\ \mathrm{and} \mathrm{y}=-\mathrm{x} \mathrm{sin\theta }+\mathrm{y} \mathrm{cos\theta } \\ \Rightarrow 1-2\sqrt{3}=-4\mathrm{sin\alpha }+2\mathrm{cos\alpha } - \left(2\right) \\ \left(1\right)\times 2 \Rightarrow 4+2\sqrt{3}=8\cos \mathrm{\alpha }+4\sin \mathrm{\alpha } - \left(3\right) \\ \left(3\right)+\left(2\right)\Rightarrow 10 \cos \mathrm{\alpha }=5 \\ \Rightarrow \cos \mathrm{\alpha }=\frac{1}{2}=\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \boxed{\mathrm{\alpha }=\frac{\mathrm{\pi }}{3}} \end{gathered}$$