Denoting the coefficient of static friction by μs, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane.
Use defined quantities and g in your expression ΣFx = 25%

Σ Fₓ =  mg (  μ  cos θ - sin θ) =0

Explanation:

To solve the problem of an inclined plane, we must use Newton's second law.

In these problems the selected reference system is where the x-axis is parallel to the plane and the y-axis is perpendicular to the plane.

The force that exists is the weight, the normal and the friction force, in our reference system the only one that we must break down the weight, let's use trigonometry.

           sin θ = Wx / W

           cos θ = / W

Where the angle of the plane is the same angle of the y-axis

              Wₓ = W sin θ

              = W cos θ

Let's write the equations, just before starting the movement

Y Axis

           N - W_{y} = 0

           N = W cos θ

X axis

          fr - Wₓ = 0

The equation for the force of friction is

          fr = μ N

we replace

           μ N = W sin θ

          μ  W cos θ = W sin θ

         μ = tan θ

The answer is

    ∑ Fₓ = μ mg cos θ - mg sin θ= 0

    Σ Fₓ =  mg (  μ  cos θ - sin θ) =0

i don't know that i agree with that but what's the question ?