Mountain goats can easily scale slopes angled at 60∘ from horizontal. What is the static friction force acting on a mountain goat that weighs 750 N and is standing on such a slope?

I know the formula for static friction is Fs = coef of static friction * Normal Force
But I can't find the formula for the coeffecient.

Let's assume that the x-axis is aligned with the slope and the (+)-direction is pointing down the slope. Now let's apply Newton's 2nd law to this problem:

[tex]x:\:\:\:\:W\sin60 - f_s = 0[/tex] (1)

[tex]y:\:\:\:\:N - W\cos60 = 0[/tex] (2)

where W is the weight of the mountain goat, N is the normal force and [tex]f_s = \mu_sN[/tex] is the frictional force. Using Eqn(1), we see that the frictional force is

[tex]f_s = W\sin60 = (750\:\text{N})\sin60 = 649.5\:\text{N}[/tex]

Bonus:

If you want to solve for the coefficient of static friction [tex]\mu_s,[/tex] rewrite Eqn(1) as

[tex]f_s = \mu_sN = W\sin60 \Rightarrow \mu_s = \dfrac{W\sin60}{N}[/tex] (3)

From Eqn(2), we see that

[tex]N = W\cos60[/tex]

Put this result back into Eqn(3) and we get

[tex]\mu_s = \dfrac{W\sin60}{W\cos60} = \tan60 = 1.73[/tex]