Ignore the book and find the missing mg yourself. Immediately draw a picture of your free body diagram (always always always DO THIS!) which I will describe for you as you draw it. There should be a box representing the homeowners junk of mass $m$. From this box, there should be a vector for each force acting on the box. This will include one straight down with the tag $mg$ which was likely one of the first things you were drawn in class. Next there should be a vector indicating the force of the homeowner which we can call $\vec{F_{}}$ having magnitude $F$ and should be pointing to (I chose) the right and downward labeling the angle between it and your x-axis, the angle $\theta$. Are we done? No, We haven't labeled the friction since we know it's present. There will be a frictional force proportional to the normal force, ah! lets write the normal force first. The normal force, typically indicated as $\vec{F}_N$ will always point... you know where... perpendicular to the surface. It should have a magnitude (which you will write to the side as not to clutter your pretty picture) of... 766
well as it seems... Some component of the homeowners force + the $mg$ that points opposite to it. It seems $\vec{F}_N = Fsin(\theta)+mg$ and is pointing straight up. Now we write down friction since we know $\vec{F}_N$, so the friction will always oppose the applied force and be along the axis of the motion (here the $x$ axis) and so I have mine pointing to the left. You should have something to the side that says that $F_f=\mu F_N$ There! all done. Now add up all the x direction forces you have written down and repeat for the y direction separately. Each will be equal to the acceleration in the respective direction, and since at current there is no acceleration in the y, your y forces should sum to zero. Now for the x-direction (which answers your question), you should get something like $Fcos(\theta) - F\mu sin(\theta) -mg \mu = ma_x $ but we want to know how the $a_x$ could be zero so let's set it equal to zero and see what this gives us. Should give something like this
$$cos(\theta) - \mu sin(\theta) = \frac{mg \mu}{F} $$ where I've put all the angle content on one side because that's what we're interested in. Now if the homeowner were to push infinitely hard, the right side would go to zero. So if we now set that side equal to zero, we are looking at the condition for 1-no acceleration in the $x$ and 2-an infinitely strong homeowner. well it seems the only way that could happen is if $cos(\theta) = \mu sin(\theta)$ which looks a whole lot like what you've presented in equation (not labeled). By dividing both sides by $sin(\theta)$, you'll see that gives the condition $cot(\theta) = \mu$ and so to find $\theta$ you take the inverse tangent of $\frac{1}{\mu}$ 766