A $$2$$ kg brick begins to slide over a surface which is inclined at an angle of $$45°$$ with respect to horizontal axis. The co-efficient of static friction between their surfaces is:

Given that the block begins to slide 

so $$f_l=Force\ acting\ down\ the\ incline$$

We know that $$f_l=\mu\ _s\times\ N$$ , Where N is the normal force acting on the block

Here the force acting down the incline is $$Mg\sin\theta\ $$

so $$f_l=Mg\sin\theta\ $$

Now 

$$f_l=Mg\sin\theta\ =\mu\ _s\times\ N$$

N is normal force which is, $$N=Mg\cos\theta\ $$

$$f_l=Mg\sin\theta\ =\mu\ _s\times\ Mg\cos\theta\ $$

$$\mu\ _s=\tan\ \theta\ $$

Its given that $$\theta\ =45^{\circ\ }$$

so 

$$\mu\ _s=1$$