First draw out the free body diagram for the problem, marking on all forces acting on the objects ie. the mass of the block and the reaction force normal to the plane and the forces acting due to friction. After drawing the diagram the normal force can be relabelled as mgcos(𝛳) and the component of the mass acting down the slope can be re written as mgsin(𝛳).
As the frictional force is proportional to the reaction force F = µR therefore frictional force = mgµcos(𝛳). We can now use Newtons second law F = ma to calculate the forces at limiting equilibrium (a=0) so mgsin(𝛳)=mgµcos(𝛳), rearrange to find tan(𝛳)=µ, therefore 𝛳 = arctan(µ).