Suppose angle of friction = alpha and angle of repose = theta
Let us suppose a body is placed on an inclined plane as in the above figure.
Various forces involved are:-
1. weight,mg of the body,acting vertically downwards.
2.normal reaction,R,acting perpendicular to inclined plane.
3.Force of friction,F, acting up the plane.
Now, mg can be resolved in two components:-
mgcos theta opposite to R
and mgsin theta opposite to F
In equilibrium,
F = mgsin theta -------- eq.1
R = mgcos theta ---------eq.2
On dividing eq.1 by eq.2 we get,
F/R = mgsin theta/mgcos theta
mu = tan theta
Where mu = coefficient of limiting friction.This is equal to the tangent of the angle of repose between two surfaces in contact.
Now, since mu = tan alpha
i.e coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them.
Thus,
Theta = alpha
i.e angle of friction is equal to angle of repose.