Solution

State of forces equilibrium for the wedge for the horizontal axis: $$N_3+\mu_1N_1=\left(\mu_2\cos\alpha +\sin\alpha\right)N_2$$ In the set mode, the force that will act on the left wall will be equal to zero ($N_3=0$) $$\frac{N_1}{N_2}=\frac{\mu_2\cos\alpha +\sin\alpha}{\mu_1}\quad(1)$$ Similarly, for the vertical axis $$N_1+\mu_2N_2\sin\alpha =N_2\cos\alpha$$ $$\frac{N_1}{N_2}=\cos\alpha -\mu_2\sin\alpha\quad(2)$$ Substituing $(2)$ into $(1)$ $$\mu_1\cos\alpha -\mu_1\mu_2\sin\alpha =\mu_2\cos\alpha +\sin\alpha$$ Divide the left and right parts by $\cos\alpha$ $$\left(1+\mu_1\mu_2\right)\tan\alpha =\mu_1-\mu_2$$ Whence the required angle $$\boxed{\tan\alpha =\frac{{\mu}_{1}-\mu_{2}}{1+\mu_{1}\mu_{2}}}$$