Let's say that we've completed Step #1, and we're trying to decide what to do next. At this point, we know all the sides of the triangle, we know one of the angles, and we're trying to find the other angles. Let's suppose $\theta$ is one of the angles, and we're trying to find $\theta$.
When you use the law of cosines to find some angle $\theta$ of a triangle, first you try to get $\cos\theta$ by itself, and you end up with an equation of the form
$$\cos\theta=\text{BLAH}$$
($\text{BLAH}$ is just some number). Since $\theta$ is the angle of triangle, you know that $0<\theta<180^{\circ}$, so you get that
$$\theta=\cos^{-1}\text{BLAH}.$$
Now the formula for the law of sines is a bit less complicated than the formula for the law of cosines, so perhaps it would be better to use the law of sines. If we use the law of sines, there will be another issue that we have to worry about, as we will see below.
When you use the law of sines to find some angle $\theta$ of a triangle, first you try to get $\sin\theta$ by itself, and you end up with an equation of the form
$$\sin\theta=\text{BLAH}$$
(Again, $\text{BLAH}$ is just some number). And again, we have that $0<\theta<180^{\circ}$, since $\theta$ is an angle of triangle. Note that the sine function is positive in quadrants one and two. Hence there are two possibilities:
$$\theta=\sin^{-1}\text{BLAH}$$
$$\theta=180^{\circ}-\sin^{-1}\text{BLAH}.$$
One of these gives the correct answer for $\theta$. Which one is it?
Your book has a clever way around this issue. Of the two given sides, the angle opposite the shorter given side will be shorter than the angle opposite the other given side. Hence the angle opposite the shorter given side will not be the largest angle of the triangle. So it must be acute. So if $\theta$ is the angle opposite the shorter given side, then $0<\theta<90^{\circ}$. Hence, we'll have just one answer for $\theta$:
$$\theta=\sin^{-1}\text{BLAH}.$$
So we should first try to find the angle that's opposite the shorter given side. That way, we can use law of sines (which is easier than law of cosines), and we won't have to worry about getting more than one answer.
Given a straight line with gradient $m,$ $$\arctan m$$ gives the acute angle measured anticlockwise from the positive $x$-direction (so, clockwise measurements are negative).
($AY$ and $PX$ are just horizontal reference lines.)
$$\measuredangle APB=\measuredangle APX+\measuredangle XPB\\=\left(180^\circ-\measuredangle YAP\right)+\measuredangle XPB\\=
\left(180^\circ-(-\arctan m_1\right))+(-\arctan m_2)\\=180^\circ+\arctan\frac{-1}e-\arctan\frac1{e^2}\\=152^\circ.$$
Best Answer
After that you have $\alpha + \beta =145^{\circ}$ and $\alpha - \beta =$ (something numerical or an expression involving the arctangent of an angle). Just solve as linear simultaneous equations. That will give you $\alpha$ and $\beta$ in degrees. Finally, use another appropriate angle pairing to solve for the remaining side using the same method.
Of course, cosine rule from the start is the much more natural way here.