Derivation of gyroscopic precession equation: Why is $d\theta = \frac{dL}{L\sin\theta}$? My textbook gives the following derivation of the angular momentum of gyroscopic precession:

The magnitude of the torque is $$\tau = r M g\sin\theta.$$
Thus $$dL = r M g\sin\theta\, dt.$$
The angle the top precesses through in time $dt$ is $$d\phi=\frac{dL}{L\sin\theta}
=\frac{rMg\sin\theta}{L\sin\theta}dt=\frac{rMg}{L}dt.$$
The precession angular velocity is $\omega_P = \frac{d\phi}{dt}$ and from this equation we see that $$\omega_P = \frac{rMg}{L}= \frac{rMg}{I \omega}.$$

(source)
The derivation on Wikipedia is similar.
I am confused by the use of differentials here. I can see that in the first two lines, the authors have used the fact that $\tau = \frac{dL}{dt}$. But why, in the third line, is $d\phi = \frac{dL}{L\sin\theta}$?
There are many questions on this site seeking a conceptual understanding of precession. I think I understand the concept (how the angular momentum vector changes under the influence of gravity), but I am confused by this derivation.
 A: The length of an arc on a circumference $\delta \ell$ is related to the circle's radius $r$ and angle subtended $\delta\theta$ by:
$$ \delta \theta = \frac{\delta\ell}{r} \qquad \Rightarrow \qquad \mathrm{d}\theta = \frac{\mathrm{d}\ell}{r},$$
where on the right I took the limit of the $\delta$ quantities becoming infinitesimally small.
The circumference here is the precession orbit.  In this case, the small angle subtended is $\mathrm{d}\theta$ (usually called $\mathrm{d}\Omega$), the arc length is the small increase in angular momentum causes by the torque $\mathrm{d} L$ $(= \Gamma \mathrm{d}t)$, and the radius of the orbit is the (projected) angular momentum associated with the gyroscope spinning $L\sin\theta = I\omega\sin\theta$.
Visual aid:

A: I recommend that you first do a derivation for a more symmetrical case.
In the source you linked to the derivation takes on a more general case right away. Instead you can first consider the case where the spin axis of the gyro wheel is at right angles to the vertical. 

Image source: wikimedia
Once you have a derivation for that most symmetrical case you can proceed to generalize. 

Incidentally, note that the description of gyroscopic precession in the source that you linked to is (like most sources) not complete.
"The top precesses around a vertical axis, since the torque is always horizontal and perpendicular to L. If the top is not spinning, it acquires angular momentum in the direction of the torque, and it rotates around a horizontal axis, falling over just as we would expect."
What is not addressed is what happens at intermediate rates of spinning. When the gyroscope is spinning fast enough then what we see with our eyes is that the gyroscope goes in precessing motion instead of falling over; when the gyroscope is spinning too slow it will simply fall over. 
The description seems to suggest a hard either-or outcome. Either the gyroscope will start precessing, or it wiil fall over. But we know that in the real world it is a continuum; in the range from non-spinning to spinning fast there is nowhere a hard cut-off.
